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Algebraic Topology

Lectures by Haynes MillerNotes based on liveTEXed record made by Sanath Devalapurkar

Images created by John Ni

March 4, 2018

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iii

Preface

Over the 2016–2017 academic year, I ran the standard algebraic topology se-quence at MIT. The first semester deals with singular homology and cohomol-ogy, and Poicaré duality; the second builds up basic homotopy theory, spectralsequences, and characteristic classes.

I was lucky enough to have in the audience a student, Sanath Devalapurkar,who spontaneously decided to liveTEX the entire course. This resulted in aremarkably accurate record of what happened in the classroom – right down torandom alarms ringing and embarassing mistakes on the blackboard. Sanath’sTEX forms the basis of these notes.

My goal was to give a standard classical approach to these subjects. In thefirst semester, I tried to give an honest account of the relative cup productsneeded in the proof of Poincaré duality.

Contents

Contents iv

I 18.905: an introduction to algebraic topology 1

1 Homology and CW-complexes 31 Introduction: singular simplices and chains . . . . . . . . . . . . . 32 More about homology . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Categories, functors, natural transformations . . . . . . . . . . . . 94 More about categories . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Homotopy, star-shaped regions . . . . . . . . . . . . . . . . . . . . . 136 Homotopy invariance of homology . . . . . . . . . . . . . . . . . . . 177 Homology cross product . . . . . . . . . . . . . . . . . . . . . . . . . 198 Relative homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 The homology long exact sequence . . . . . . . . . . . . . . . . . . 2310 Excision and applications . . . . . . . . . . . . . . . . . . . . . . . . 2711 The Eilenberg Steenrod axioms and the locality principle . . . . . 3012 Subdivision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3313 Proof of the Locality Principle . . . . . . . . . . . . . . . . . . . . . 3514 CW-complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3715 CW-complexes II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4016 Homology of CW-complexes . . . . . . . . . . . . . . . . . . . . . . 4217 Real projective space . . . . . . . . . . . . . . . . . . . . . . . . . . 4518 Euler characteristic, and homology approximation . . . . . . . . . 4719 Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5020 Tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5221 Tensor and Tor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5722 The fundamental lemma of homological algebra . . . . . . . . . . 5923 Hom and Lim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6224 Universal coefficient theorem . . . . . . . . . . . . . . . . . . . . . . 6725 Künneth and Eilenberg-Zilber . . . . . . . . . . . . . . . . . . . . . 69

2 Cohomology and duality 7526 Coproducts, cohomology . . . . . . . . . . . . . . . . . . . . . . . . 7527 Ext and UCT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

iv

CONTENTS v

28 Products in cohomology . . . . . . . . . . . . . . . . . . . . . . . . . 8329 Cup product, continued . . . . . . . . . . . . . . . . . . . . . . . . . 8530 Surfaces and nondegenerate symmetric bilinear forms . . . . . . . 8831 A plethora of products . . . . . . . . . . . . . . . . . . . . . . . . . 9232 Cap product and “Čech” cohomology . . . . . . . . . . . . . . . . . 9533 H∗ as a cohomology theory, and the fully relative ∩ product . . . 9934 H∗ as a cohomology theory . . . . . . . . . . . . . . . . . . . . . . . 10435 Finish off the proof of Hp excision, topological manifolds, fun-

damental classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10736 Fundamental class . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10937 Covering spaces and Poincaré duality . . . . . . . . . . . . . . . . . 11238 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

II 18.906 – homotopy theory 119

3 Homotopy groups 12339 Limits, colimits, and adjunctions . . . . . . . . . . . . . . . . . . . 12340 Compactly generated spaces . . . . . . . . . . . . . . . . . . . . . . 12741 “Cartesian closed”, Hausdorff, Basepoints . . . . . . . . . . . . . . 13042 Fiber bundles, fibrations, cofibrations . . . . . . . . . . . . . . . . 13143 Fibrations and cofibrations . . . . . . . . . . . . . . . . . . . . . . . 13544 Homotopy fibers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13945 Barratt-Puppe sequence . . . . . . . . . . . . . . . . . . . . . . . . . 14346 Relative homotopy groups . . . . . . . . . . . . . . . . . . . . . . . 14747 Action of π1, simple spaces, and the Hurewicz theorem . . . . . . 15048 Examples of CW-complexes . . . . . . . . . . . . . . . . . . . . . . 15549 Relative Hurewicz and J. H. C. Whitehead . . . . . . . . . . . . . 15650 Cellular approximation, cellular homology, obstruction theory . . 15951 Conclusions from obstruction theory . . . . . . . . . . . . . . . . . 163

4 Vector bundles 16752 Vector bundles, principal bundles . . . . . . . . . . . . . . . . . . . 16753 Principal bundles, associated bundles . . . . . . . . . . . . . . . . . 17154 I-invariance of BunG, and G-CW-complexes . . . . . . . . . . . . 17455 Classifying spaces: the Grassmann model . . . . . . . . . . . . . . 17756 Simplicial sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18057 Properties of the classifying space . . . . . . . . . . . . . . . . . . . 18258 Classifying spaces of groups . . . . . . . . . . . . . . . . . . . . . . 18459 Classifying spaces and bundles . . . . . . . . . . . . . . . . . . . . . 186

5 Spectral sequences 19160 The spectral sequence of a filtered complex . . . . . . . . . . . . . 19161 Exact couples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19562 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19763 Edge homomorphisms, transgression . . . . . . . . . . . . . . . . . 202

vi CONTENTS

64 Serre classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20565 Mod C Hurewicz, Whitehead, cohomology spectral sequence . . 20866 A few examples, double complexes, Dress sseq . . . . . . . . . . . 21067 Dress spectral sequence, Leray-Hirsch . . . . . . . . . . . . . . . . 21368 Integration, Gysin, Euler, Thom . . . . . . . . . . . . . . . . . . . . 216

6 Characteristic classes 22169 Grothendieck’s construction of Chern classes . . . . . . . . . . . . 22170 H∗(BU(n)), splitting principle . . . . . . . . . . . . . . . . . . . . 22471 The Whitney sum formula . . . . . . . . . . . . . . . . . . . . . . . 22772 Stiefel-Whitney classes, immersions, cobordisms . . . . . . . . . . 23073 Oriented bundles, Pontryagin classes, Signature theorem . . . . . 234

Bibliography 239

Part I

18.905: an introduction toalgebraic topology

1

Chapter 1

Homology and CW-complexes

1 Introduction: singular simplices and chains

This is a course on algebraic topology. We’ll discuss the following topics.

1. Singular homology

2. CW-complexes

3. Basics of category theory

4. Homological algebra

5. The Künneth theorem

6. UCT, cohomology

7. Cup and cap products, and

8. Poincaré duality.

The objects of study are of course topological spaces, and the machinery wedevelop in this course is designed to be applicable to a general space. But weare really interested in geometrically natural spaces. Here are some examples.

• The most basic example is n-dimensional Euclidean space, Rn.

• The n-sphere Sn = x ∈ Rn+1 ∶ ∣x∣ = 1, topologized as a subspace ofRn+1.

• Identifying antipodal points in Sn gives real projective space RPn =Sn/(x ∼ −x), i.e. the space of lines through the origin in Rn+1.

• Call an ordered collection of k orthonormal vectors an orthonormal k-frame. The space of orthonormal k-frames in Rn forms the Stiefel man-ifold Vk(Rn), which is topologized as a subspace of (Sn−1)k.

3

4 CHAPTER 1. HOMOLOGY AND CW-COMPLEXES

• The Grassmannian Grk(Rn) is the space of k-dimensional linear sub-spaces ofRn. Forming the span gives us a surjection Vk(Rn)→ Grk(Rn),and the Grassmannian is given the quotient topology. For example,Gr1(Rn) =RPn−1.

All these examples are manifolds; that is, they are Hausdorff spaces locallyhomeomorphic to Euclidean space. Aside from Rn itself, the preceding exam-ples are also compact. Such spaces exhibit a hidden symmetry, which is theculmination of 18.905: Poincaré duality.

As the name suggests, the central aim of algebraic topology is the usage ofalgebraic tools to study topological spaces. A common technique is to probetopological spaces via maps to them from simpler spaces. In different ways,this approach gives rise to singular homology and homotopy groups. We nowdetail the former; the latter takes stage in 18.906.

Definition 1.1. For n ≥ 0, the standard n-simplex ∆n is the convex hull ofthe standard basis e0, . . . , en in Rn+1:

∆n = ∑ tiei ∶∑ ti = 1, ti ≥ 0 ⊆Rn+1.

The ti are called barycentric coordinates.

The standard simplices are related by face inclusions di∶∆n−1 → ∆n for0 ≤ i ≤ n, where di is the affine map that sends verticies to vertices, in order,and omits the vertex ei.

Figure 1.1: The face inclusions for n = 1 (left) and n = 2 (right).

Definition 1.2. Let X be any topological space. A singular n-simplex in Xis a continuous map σ ∶ ∆n →X. Denote by Sinn(X) the set of all n-simplicesin X.

This seems like a rather bold construction to make, as Sinn(X) is huge.But be patient!

1. INTRODUCTION: SINGULAR SIMPLICES AND CHAINS 5

For 0 ≤ i ≤ n, precomposition by the face inclusion di produces a mapdi∶Sinn(X) → Sinn−1(X) sending σ ↦ σ di. This is the “ith face” of σ. Thisallows us to make sense of the “boundary” of a simplex, and we are particularlyinterested in simplices for which that boundary vanishes.

For example, if σ is a 1-simplex that forms a closed loop, then d1σ = d0σ.To express the condition that the boundary vanishes, we would like to writed0σ − d1σ = 0 – but this difference is no longer a simplex. To accommodatesuch formal sums, we will enlarge Sinn(X) further by forming the free abeliangroup it generates.

Definition 1.3. The abelian group Sn(X) of singular n-chains in X is thefree abelian group generated by n-simplices

Sn(X) = ZSinn(X).

So an n-chain is a finite linear combination of simplices,

k

∑i=0

aiσi , ai ∈ Z , σi ∈ Sinn(X) .

If n < 0, Sinn(X) is declared to be empty, so Sn(X) = 0.We can now define the boundary operator

d∶Sinn(X)→ Sn−1(X),

by

dσ =n

∑i=0

(−1)idiσ.

This extends to a homomorphism d∶Sn(X)→ Sn−1(X) by additivity.We use this homomorphism to obtain something more tractable than the

entirety of Sn(X). First we restrict our attention to chains with vanishingboundary.

Definition 1.4. An n-cycle in X is an n-chain c with dc = 0. Denote Zn(X) =ker(d ∶ Sn(X)→ Sn−1(X)).

For example, if σ is a 1-simplex forming a closed loop, then σ ∈ Z1(X) sincedσ = d0σ − d1σ = 0.

It turns out that there’s a cheap way to produce a cycle:

Theorem 1.5. Any boundary is a cycle; that is, d2 = 0.

We’ll leave the verification of this important result as a homework problem.What we have found, then, is that the singular chains form a “chain complex,”as in the following definition.

Definition 1.6. A graded abelian group is a sequence of abelian groups, in-dexed by the integers. A chain complex is a graded abelian group An togetherwith homomorphisms d ∶ An → An−1 with the property that d2 = 0.


The group of n-dimensional boundaries is

Bn(X) = im(d ∶ Sn+1(X)→ Sn(X)) ,

and the theorem tells us that this is a subgroup of the group of cycles: the“cheap” ones. If we quotient by them, what’s left is the “interesting ones,”captured in the following definition.

Definition 1.7. The nth singular homology group of X is:

Hn(X) = Zn(X)Bn(X)

= ker(d ∶ Sn(X)→ Sn−1(X))im(d ∶ Sn+1(X)→ Sn(X))

.

We use the same language for any chain complex: it has cycles, boundaries,and homology groups. The homology forms a graded abelian group.

Both Zn(X) and Bn(X) are free abelian groups because they are subgroupsof the free abelian group Sn(X), but the quotient Hn(X) isn’t necessarilyfree. While Zn(X) and Bn(X) are uncountably generated, Hn(X) is finitelygenerated for the spaces we are interested in. If T is the torus, for example,then we will see that H1(T ) ≅ Z ⊕ Z and σ as described previously is one ofthe two generators. We will learn to compute the homology groups of a widevariety of spaces. The n-sphere has the following homology groups:

Hq(Sn) =

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

Z if q = n > 0

Z if q = 0, n > 0

Z⊕Z if q = n = 0

0 otherwise

2 More about homology

In the last lecture we introduced the standard n-simplex ∆n ⊆Rn+1. Singularsimplices in a spaceX are maps σ∶∆n →X and constitute the set Sinn(X). Forexample, Sin0(X) consists of points of X. We also described the face inclusionsdi ∶ ∆n−1 →∆n, and the induced “face maps”

di ∶ Sinn(X)→ Sinn−1(X) ,0 ≤ i ≤ n ,

given by precomposing with face inclusions: diσ = σ di. For homework youestablished some quadratic relations satisfied by these maps. A collection ofsets Kn, n ≥ 0, together with maps di ∶ Kn → Kn−1 related to each other inthis way, is a semi-simplicial set. So we have assigned to any space X a semi-simplicial set S∗(X).

To the semi-simplicial set Sinn(X), di we then applied the free abeliangroup functor, obtaining a semi-simplicial abelian group. Using the dis, weconstructed a boundary map d which makes S∗(X) a chain complex – that is,

2. MORE ABOUT HOMOLOGY 7

d2 = 0. We capture this process in a diagram:

spaces

Sin∗

H∗ // graded abelian groups

semi-simplicial sets

Z(−)

semi-simplicial abelian groups // chain complexes

take homology

OO

Example 2.1. Suppose we have σ∶∆1 → X. Define φ∶∆1 → ∆1 which sends(t,1 − t) ↦ (1 − t, t). Precomposing σ with φ gives another singular simplex σwhich reverses the orientation of σ. It is not true that σ = −σ in S1(X).

However, we claim that σ ≡ −σ mod B1(X), meaning there is a 2-chain inX whose boundary is σ+σ. If d0σ = d1σ so that σ ∈ Z1(X), then σ and −σ arehomologous: [σ] = −[σ] in H1(X).

To construct an appropriate boundary, consider the projection map π ∶∆2 → ∆1 that is the affine extension of the map sending e0 and e2 to e0 ande1 to e1.

Figure 1.2: If σ is a 1-simplex in X, then precomposing by π gives a 2-simplexσ π in X.

We’ll compute d(σ π). Some of the terms will be constant singular sim-plicies. Let’s write cnx ∶ ∆n →X for the constant map with value x ∈X. Then

d(σ π) = σπd0 − σπd1 + σπd2 = σ − c1σ(0) + σ .


The constant simplex c1σ(0) is an error term, and we wish to eliminate it. Toachieve this we can use the constant 2-simplex c2σ(0) at σ(0); its boundary is

c1σ(0) − c1σ(0) + c

1σ(0) = c

1σ(0) .

Soσ + σ = d(σ π + c2σ(0)) ,

and σ ≡ −σ mod B1(X) as claimed.Some more language: two cycles that differ by a boundary dc are said to

be homologous, and the chain c is a homology between them.

Let’s compute the homology of the very simplest spaces, ∅ and ∗. For thefirst , Sinn(∅) = ∅, so S∗(∅) = 0. Hence ⋯ → S2 → S1 → S0 is the zero chaincomplex. This means that Z∗(∅) = B∗(∅) = 0. The homology in all dimensionsis therefore 0.

For ∗, we have Sinn(∗) = cn∗ for all n ≥ 0. Consequently Sn(∗) = Zfor n ≥ 0 and 0 for n ≤ 0. For each i, dicn∗ = cn−1

∗ , so the boundary mapsd∶Sn(∗)→ Sn−1(∗) in the chain complex depend on the parity of n as follows:

d(cn∗) =n

∑i=0

(−1)icn−1∗ =

⎧⎪⎪⎨⎪⎪⎩

cn−1∗ for n even, and

0 for n odd.

This means that our chain complex is:

0← 0← Z← 1Z← 1Z← 1⋯ .

The boundaries coincide with the cycles except in dimension zero, whereB0(∗) =0 while Z0(∗) = Z. Therefore H0(∗) = Z and Hi(∗) = 0 for i ≠ 0.

We’ve defined homology groups for each space, but haven’t considered whathappens to maps between spaces. A continuous map f ∶X → Y induces a mapf∗∶Sinn(X)→ Sinn(Y ) by composition:

f∗ ∶ σ ↦ f σ .

For f∗ to be a map of semi-simplicial sets, it needs to commute with face maps.Explicitly, we need f∗ di = di f∗. A diagram is said to be commutative if allcomposites with the same source and target are equal, so this is equivalent tocommutativity of the below.

Sinn(X)f∗ //

di

Sinn(Y )

di

Sinn−1(X)

f∗ // Sinn−1(Y )

We see that dif∗σ = (f∗σ)di = f σ di, and f∗(diσ) = f∗(σ di) = f σ di asdesired. The diagram remains commutative when we pass to the free abeliangroups of chains.

3. CATEGORIES, FUNCTORS, NATURAL TRANSFORMATIONS 9

If C∗ and D∗ are chain complexes, a chain map f ∶C∗ → D∗ is a collectionof maps fn∶Cn →Dn such that the following diagram commutes for every n:

Cnfn //

dC

Dn

dD

Cn−1

fn−1 // Dn−1

For example, if f ∶X → Y is a continuous map, then f∗∶S∗(X) → S∗(Y ) is achain map as discussed above.

A chain map induces a map in homology f∗ ∶Hn(C)→Hn(D). The methodof proof is a so-called “diagram chase” and it will be the first of many. We checkthat we get a map Zn(C) → Zn(D). Let c ∈ Zn(C), so that dCc = 0. ThendDfn(c) = fn−1dCc = fn−1(0) = 0, because f is a chain map. This means thatfn(c) is also an n-cycle, i.e., f gives a map Zn(C)→ Zn(D).

Similarly, we also get a map Bn(C)→ Bn(D). Let c ∈ Bn(C), so that thereexists c′ ∈ Cn+1 such that dCc′ = c. Then fn(c) = fndCc′ = dDfn+1(c′). Thusfn(c) is the boundary of fn+1(c′), and f gives a map Bn(C)→ Bn(D).

The two maps Zn(C) → Zn(D) and Bn(C) → Bn(D) give a map on ho-mology f∗ ∶Hn(X)→Hn(Y ), as desired.

3 Categories, functors, natural transformations

From spaces and continuous maps, we constructed graded abelian groups andhomomorphisms. We now cast this construction in the more general languageof category theory.

Our discussion of category theory will be interspersed throughout the text,introducing new concepts as they are needed. Here we begin by introducingthe basic definitions.

Definition 3.1. A category C consists of the following data.

• a class ob(C) of objects;

• for every pair of objects X and Y , a set of morphisms C(X,Y );

• for every object X a identity morphism 1X ∈ C(X,X); and

• for every triple of objects X,Y,Z, a composition map C(X,Y )×C(Y,Z)→C(X,Z), written (f, g)↦ g f .

These data are required to satisfy the following:

• 1Y f = f , and f 1X = f .

• Composition is associative: (h g) f = h (g f).


Note that we allow the collection of objects to be a class. This enables usto talk about a “category of all sets” for example. But we require each C(X,Y )to be set, and not merely a class. Some interesting categories have a set ofobjects; they are called small categories.

We will often write X ∈ C to mean X ∈ ob(C), and f ∶X → Y to meanf ∈ C(X,Y ).

Definition 3.2. If X,Y ∈ C, then f ∶X → Y is an isomorphism if there existsg∶Y →X with f g = 1Y and g f = 1X . We may write

f ∶X ≅Ð→ Y

to indicate that f is an isomorphism.

Example 3.3. Many common mathematical structures can be arranged incategories.

• Sets and functions between them form a category Set.

• Abelian groups and homomorphisms form a category Ab.

• Topological spaces and continuous maps form a category Top.

• Simplicial sets and their maps form a category sSet.

• A monoid is the same as a category with one object, where the elementsof the monoid are the morphisms in the category. It’s a small category.

• The sets [n] = 0, . . . , n for n ≥ 0 together with weakly order-preservingmaps between them form the simplex category ∆, another small category.It contains as a subcategory the semi-simplex category ∆inj with the sameobjects but only injective weakly order-preserving maps.

• A poset forms a category in which there is a morphism from x to y iffx ≤ y. A small category with this property comes from a poset providedthat the only isomorphisms are identities.

Categories may be related to each other by rules describing effect on bothobjects and morphisms.

Definition 3.4. Let C,D be categories. A functor F ∶C → D consists of thedata of

• an assignment F ∶ ob(C)→ ob(D), and

• for all x, y ∈ ob(C), map F ∶ C(x, y)→ C(F (x), F (y)) .

These data are required to satisfy the following two properties:

• For all X ∈ obC, F (1X) = 1F (X) ∈ D(F (X), F (X)), and

• For all composable pairs of morphisms f, g in C, F (g f) = F (g) F (f).

4. MORE ABOUT CATEGORIES 11

We have defined quite a few functors already:

Sinn ∶ Top→ Set , Sn ∶ Top→Ab , Hn ∶ Top→Ab ,

for example. We also have defined, for each X, a morphism d ∶ Sn(X) →Sn−1(X). This is a “morphism between functors.” This property is capturedby another definition.

Definition 3.5. Let F,G∶C → D. A natural transformation or natural mapθ∶F → G consists of maps θX ∶F (X) → G(X) for all X ∈ ob(C) such that forall f ∶X → Y the following diagram commutes.

F (X)

F (f)

θX // G(X)

G(f)

F (Y ) θY // G(Y )

So for example the boundary map d∶Sn → Sn−1 is a natural transformation.Natural transformations are so . . . well, so natural that their occurance is

indicated by a variety of terms: a natural or canonicalmap is precisely a naturaltransformation.

Example 3.6. Suppose that C and D are two categories, and assume that Cis small. We may then form the category of functors Fun(C,D). Its objects arethe functors fron C to D, and given two functors F,G, Fun(C,D)(F,G) is theset of natural transformations from F to G. We let the reader define the restof the structure of this category, and check the axioms. We needed to assumethat C is small in order to guarantee that there is no more than a set of naturaltransformations between functors.

For example, letG be a group (or a monoid) viewed as a one-object category.Any element F ∈ Fun(G,Ab) is simply a group action of G on F (∗) = A,i.e., a representation of G in abelian groups. Given another F ′ ∈ Fun(G,Ab)with F ′(∗) = A′, then a natural transformation from F → F ′ is precisely aG-equivariant homomorphism A→ A′.

4 More about categories

Let Vectk be the category of vector spaces over a field k, and linear transfor-mations between them. Given a vector space V , you can consider the dualV ∗ = Hom(V, k). Does this give us a functor? If you have a linear transforma-tion f ∶ V →W , you get a map f∗ ∶W ∗ → V ∗, so this is like a functor, but theinduced map goes the wrong way. This operation does preserve compositionand identities, in an appropriate sense. This is an example of a contravariantfunctor.

We’ll leave it to you to spell out the definition, but notice that there is auniveral example of a contravariant functor out of a category C: C → Cop, where


Cop has the same objects as C, but Cop(X,Y ) is declared to be the set C(Y,X).The identity morphisms remain the same. To describe the composition in Cop,we’ll write fop for f ∈ C(Y,X) regarded as an element of Cop(X,Y ); thenfop gop = (g f)op.

Then a contravariant functor from C to D is the same thing as a (“covariant”)functor from Cop to D.

Let C be a category, and let Y ∈ ob(C). We get a map Cop → Set that takesX ↦ C(X,Y ), and takes a map X → W to the map defined by compositionC(W,Y ) → C(X,Y ). This is called the functor that is represented by Y . It isvery important to note that C(−, Y ) is contravariant, while, on the other hand,for any fixed X, C(Y,−) is a covariant functor.

Example 4.1. Recall that the simplex category ∆ has objects the totallyordered sets [n] = 0,1, . . . , n, with order preserving maps as morphisms. The“standard simplex” gives us a functor ∆∶∆ → Top. Now fix a space X, andconsider

[n]↦ Top(∆n,X) .This gives us a contravariant functor ∆ → Top, or a covariant functor ∆op →Top. This functor carries in it all the face and degeneracy maps we discussedearlier, and their compositions. Let us make a definition.

Definition 4.2. Let C be any category. A simplicial object in C is a functor K ∶∆op → C. Simplicial objects in C form a category with natural transformationsas morphisms. Similarly, semi-simplicial object in C is a functor ∆op

inj → C,

So the singular functor Sin∗ gives a functor from spaces to simplicial sets(and so, by restriction, to semi-simplicial sets).

In want to interject one more bit of categorical language that will often beuseful to us.

Definition 4.3. A morphism f ∶X → Y in a category C is a split epimorphism(“split epi” for short) if there exists g ∶ Y → X (called a section or a splitting)

such that YgÐ→X

fÐ→ Y is the identity.

Example 4.4. In the category of sets, a map f ∶X → Y is a split epimorphismexactly when, for every element of Y there exists some element of X whoseimage in Y is the original element. So f is surjective. Is every surjective map asplit epimorphism? This reduces to the axiom of choice! because proving thatif y ∈ Y , the map g ∶ Y →X can be constructed by choosing some x ∈ f−1(y).

Every categorical definition is accompanied by a dual definition.

Definition 4.5. A map g ∶ Y → X is a split monomorphism (“split mono” forshort) if there is f ∶X → Y such that f g = 1Y .

Example 4.6. Again let C = Set. Any split monomorphism is an injection:If y, y′ ∈ Y , and g(y) = g(y′), we want to show that y = y′. Apply f , to getf(g(y)) = y = f(g(y′)) = y′. But the injection ∅ → Y is a split monomorphismonly if Y = ∅. So there’s an assymetry in the category of sets.

5. HOMOTOPY, STAR-SHAPED REGIONS 13

Lemma 4.7. A map is an isomorphism if and only if it is both a split epimor-phism and a split monomorphism.

Proof. Standard!

The importance of these definitions is this: Functors will not in generalrespect “monomorphisms” or “epimorphisms,” but:

Lemma 4.8. If f ∶X → Y is a split epi or mono in C, and you have a functorF ∶ C → D, then so is F (f) in D.

Proof. Apply F to the diagram establishing f as a split epi or mono.

Example 4.9. Suppose C = Ab, and you have a split epi f ∶ A → B. Letg ∶ B → A be a section. We also have the inclusion i ∶ ker f → A, and hence amap

[ f i ] ∶ B ⊕ ker f → A.

We leave it to you to check that this map is an isomorphism, and to formulatea dual statement.

5 Homotopy, star-shaped regions

We’ve computed the homology of a point. Let’s now compare the homology ofa general space X to this example. There’s always a unique map X → ∗: ∗ isa “terminal object” in Top. We have an induced map

Hn(X)→Hn(∗) =⎧⎪⎪⎨⎪⎪⎩

Z n = 0

0 else.

Any formal linear combination c = ∑aixi of points of X is a 0-cycle. The mapto ∗ sends c to ∑ai ∈ Z. This defines an “augmentation” ε ∶ H∗(X) → H∗(∗).If X is nonempty, the map X → ∗ is split by any choice of point in X, so theaugmentation is also split. The kernel of ε is the reduced homology H∗(X) ofX, and we get a canonical splitting

H∗(X) ≅ H∗(X)⊕Z

Actually, it’s useful to extend the definition to the empty space by thefollowing device. Extend the singular chain complex for any space to include Zin dimension −1, with d ∶ S0(X)→ S−1(X) given by the augmentation ε. Let’swrite S∗(X) for this chain complex. When X ≠ ∅, ε is surjective and you getthe same answer as above. But

Hq(∅) =⎧⎪⎪⎨⎪⎪⎩

Z for q = −1

0 for q ≠ −1 .

What other spaces have trivial homology? A slightly non-obvious way toreframe the question is this:


When do two maps X → Y induce the same map in homology?

For example, when do 1X ∶ X → X and X → ∗ → X induce the same map inhomology? If they do, then ε ∶H∗(X)→ Z is an isomorphism.

The key idea is that homology is a discrete invariant, so it should be un-changed by deformation. Here’s the definition that makes “deformation” pre-cise.

Definition 5.1. Let f0, f1 ∶ X → Y be two maps. A homotopy from f0 to f1

is a map h ∶ X × I → Y (continuous, of course) such that h(x,0) = f0(x) andf(x,1) = f1(x). We say that f0 and f1 are homotopic, and that h is a homotopybetween them. This relation is denoted by f0 ∼ f1. It is an equivalence relationon maps from X to Y .

Transitivity follows from the gluing lemma of point set topology. We denoteby [X,Y ] the set of homotopy classes of maps from X to Y . In the next lecturewe’ll prove the following result.

Theorem 5.2 (Homotopy invariance of homology). If f0 ∼ f1, then H∗(f0) =H∗(f1): homology cannot distinguish between homotopic maps.

Suppose I have two maps f0, f1 ∶ X → Y with a homotopy h ∶ f0 ∼ f1, anda map g ∶ Y → Z. Composing h with g gives a homotopy between g f0 andg f1. Precomposing also works: If g ∶ W → X is a map and f0, f1 ∶ X → Yare homotopic, then f0g ∼ f1g. This lets us compose homotopy classes: we cancomplete the diagram:

Top(Y,Z) ×Top(X,Y )

// Top(X,Z)

[Y,Z] × [X,Y ] // [X,Z]

Definition 5.3. The homotopy category (of topological spaces) Ho(Top) hasthe same objects as Top, but Ho(Top)(X,Y ) = [X,Y ] = Top(X,Y )/ ∼.

The first key property of homology is this:

Theorem 5.4 (Homotopy invariance of homology). The homology functor H∗ ∶Top→Ab factors as Top→ Ho(Top)→Ab; it is a “homotopy functor.”

We will prove this in the next lecture, but let’s stop now and think aboutsome consequences.

Definition 5.5. A map f ∶X → Y is a homotopy equivalence if [f] ∈ [X,Y ] isan isomorphism in Ho(Top). In other words, there is a map g ∶ Y → X suchthat fg ∼ 1Y and gf ∼ 1X .

5. HOMOTOPY, STAR-SHAPED REGIONS 15

Homotopy equivalences don’t preserve most topological properties. For ex-ample, compactness is not a homotopy-invariant property: Consider the inclu-sion Sn−1 ⊆Rn−0. A homotopy inverse p ∶Rn−0→ Sn−1 can be obtainedby dividing a (always nonzero!) vector by its length. Clearly p i = 1Sn−1 . Wehave to find a homotopy ip ∼ 1Rn−0. This is t a map (Rn−0)×I →Rn−0,and we can use (v, t)↦ tv + (1 − t) v

∣∣v∣∣ .On the other hand:

Corollary 5.6. Homotopy equivalences induce isomorphisms in homology.

Definition 5.7. A space X is contractible if the map X → ∗ is a homotopyequivalence.

Corollary 5.8. Let X be a contractible space. The augmentation ε ∶H∗(X)→Z is an isomorphism.

Homotopy equivalences in general may be somewhat hard to visualize. Aparticularly simple and important class of homotopy equivalences is given bythe following definition.

Definition 5.9. An inclusion A X is a deformation retract provided thatthere is a map h ∶X × I →X such that h(x,0) = x and h(x,1) ∈ A for all x ∈Xand h(a, t) = a for all a ∈ A and t ∈ I.

For example, Sn−1 is a deformation retract of Rn − 0.

We now set about constructing a proof of homotopy invariance of homology.The first step is to understand the analogue of homotopy on the level of chaincomplexes.

Definition 5.10. Let C,D be chain complexes, and f0, f1 ∶ C → D bechain maps. A chain homotopy h ∶ f0 ∼ f1 is a collection of homomorphismsh ∶ Cn →Dn+1 such that dh + hd = f1 − f0.

It’s a really weird relation. Here’s a picture.

Cn+2f1−f0 //

d

Dn+2

d

Cn+1

h

;;

f1−f0 //

d

Dn+2

d

Cn

h

;;

f1−f0

// Dn

Lemma 5.11. If f0, f1 ∶ C → D are chain homotopic, then f0,∗ = f1,∗ ∶H(C)→H(D).


Proof. We want to show that for every c ∈ Zn(C), the difference f1c− f0c is aboundary. Well,

f1c − f0c = (dh + hd)c = dhc + hdc = dhc .

So homology invariance will follow from

Proposition 5.12. Let f0, f1 ∶ X → Y be homotopic. Then f0∗, f1∗ ∶ S(X)→S(Y ) are chain homotopic.

To prove this we will begin with a special case.

Definition 5.13. A subset X ⊆ Rn is star-shaped with respect to b ∈ X if forevery x ∈X the interval

tb + (1 − t)x ∶ t ∈ [0,1]

lies in X.

Any nonempty convex region is star shaped. Any star-shaped region X iscontractible: A homotopy inverse to X → ∗ is given by sending ∗ ↦ b. Onecomposite is perforce the identity. A homotopy from the other composite tothe identity 1X is given by (x, t)↦ tb + (1 − t)x.

So we should expect that ε ∶ H∗(X) → Z is an isomorphism if X is star-shaped. In fact, using a piece of language that the reader can interpret:

Proposition 5.14. S∗(X)→ Z is a chain homotopy equivalence.

Proof. We have maps S∗(X) εÐ→ ZηÐ→ S∗(X) where η(1) = c00. Clearly εη = 1,

and the claim is that ηε ∼ 1 ∶ S∗(X)→ S∗(X). The chain map ηε concentrateseverything at the origin: ηεσ = cn0 for all σ ∈ Sinn(X). Our chain homotopyh ∶ Sq(X)→ Sq+1(X) will actually send simplices to simplices. For σ ∈ Sinq(X),define the chain homotopy evaluated on σ by means of the following “coneconstruction”:

(b ∗ σ)(t0, . . . , tq+1) = t0b + (1 − t0)σ ((t1, . . . , tq+1)

1 − t0) .

Explanation: The denominator 1− t0 makes the entries sum to 1, as they mustif we are to apply σ to this vector. When t0 = 1, this isn’t defined, but itdoesn’t matter since we are multiplying by 1− t0. So (b∗σ)(0, t1, . . . , tq+1) = b;this is the vertex of the cone.

picture neededSetting t0 = 0, we find

d0b ∗ σ = σ .

Setting ti = 0 for i > 0, we find

dib ∗ σ = hdi−1σ .

6. HOMOTOPY INVARIANCE OF HOMOLOGY 17

Using the formula for the boundary operator, we find

db ∗ σ = σ − b ∗ dσ

. . . unless q = 0, whendb ∗ σ = σ − c0b .

This can be assembled into the equation

db ∗ +b ∗ d = 1 − ηε

which is what we wanted.

6 Homotopy invariance of homology

We now know that the homology of a star-shaped region is trivial: in such aspace, every cycle with augmentation 0 is a boundary. We will use that fact,which is a special case of homotopy invariance of homology, to prove the generalresult, which we state in somewhat stronger form:

Theorem 6.1. A homotopy h ∶ f0 ∼ f1 ∶ X → Y determines a natural chainhomotopy f0,∗ ∼ f1,∗ ∶ S∗(X)→ S∗(Y ).

The proof uses naturality (a lot). For a start, notice that if k ∶ g0 ∼ g1 ∶C∗ →D∗ is a chain homotopy, and j ∶D∗ → E∗ is another chain map, then thecomposites j kn ∶ Cn → En+1 give a chain homootpy j g0 ∼ j g1. So if wecan produce a chain homotopy homotopy between the chain maps induced bythe two inclusions i0, i1 ∶ X → X × I, we can get a chain homotopy k betweenf0∗ = h∗ i0∗ and f1∗ = h∗ i1∗ in the form h∗ k.

So now we want to produce a natural chain homotopy, with components kn ∶Sn(X)→ Sn+1(X × I). The unit interval hosts a natural 1-simplex given by anidentification ∆1 → I, and we should imagine k as being given by “multiplying”by that 1-chain. This “multiplication” is a special case of a chain map

× ∶ S∗(X) × S∗(Y )→ S∗(X × Y ) ,

defined for any two spaces X and Y , with lots of good properties. It willultimately be used to compute the homology of a product of two spaces interms of the homology groups of the factors.

Here’s the general result.

Theorem 6.2. There exists a map Sp(X) × Sq(Y )→ Sp+q(X × Y ) that is:

• Natural, in the sense that if f ∶ X → X ′ and g ∶ Y → Y ′, and a ∈ Sp(X)and b ∈ Sp(Y ) so that a×b ∈ Sp+q(X×Y ), then f∗(a)×g∗(b) = (f ×g)∗(a×b).

• Bilinear, in the sense that (a+a′)× b = (a× b)+ (a′ × b), and a× (b+ b′) =a × b + a × b′.


• The Leibniz rule is satisfied, i.e., d(a × b) = (da) × b + (−1)pa × db.

• Normalized, in the following sense. Let x ∈ X and y ∈ Y . Write jx ∶ Y →X × Y sending y ↦ (x, y), and write iy ∶ X → X × Y sending x ↦ (x, y).If b ∈ Sq(Y ), then c0x × b = (jx)∗b ∈ Sq(X × Y ), and if a ∈ Sp(X), thena × c0y = (iy)∗a ∈ Sp(X × Y ).

The Leibniz rule contains the first occurence of the “topologists sign rule”;we’ll see these signs appearing often. Watch for when it appears in our proof.

Proof. We’re going to use induction on p+ q; the normalization axiom gives usthe cases p+ q = 0,1. Let’s assume that we’ve constructed the cross-product intotal dimension p+ q−1. We want to define σ× τ for σ ∈ Sp(X) and τ ∈ Sq(Y ).

Note that there’s a universal example of a p-simplex, namely the identitymap ιp ∶ ∆p →∆p. It’s universal in the sense that given any p-simplex σ ∶ ∆p →X, you get σ = σ∗(ιp) where σ∗ ∶ Sinp(∆p) → Sinp(X) is the map induced byσ. To define σ × τ in general, then, it suffices to define ιp × ιq ∈ Sp+q(∆p ×∆q);we can (and must) then take σ × τ = (σ × τ)∗(ιp × ιq).

Our long list of axioms is useful in the induction. For one thing, if p = 0 orq = 0, normalization provides us with a choice. So now assume that both p andq are positive. We want the cross-product to satisfy the Leibnitz rule:

d(ιp × ιq) = (dιp) × ιq + (−1)pιp × dιq ∈ Sp+q−1(∆p ×∆q)

Since d2 = 0, a necessary condition for ιp × ιq to exist is that d((dιp) × ιq +(−1)pιp × dιq) = 0. Let’s compute what this is, using the Leibnitz rule indimension p + q − 1 where we have it by the inductive assumption:

d((dιp)×ιq+(−1)pιp×(dιq)) = (d2ιp)×ιq+(−1)p−1(dιp)×(dιq)+(−1)p(dιp)×dιq+(−1)qιp×(d2ιq) = 0

because d2 = 0. Note that this calculation would not have worked without thesign!

The subspace ∆p × ∆q ⊆ Rp+1 × Rq+1 is convex, so by translation, it’shomeomorphic to a star-shaped region. Therefore we know that Hp+q−1(∆p ×∆q) = 0 (remember, p + q > 1), which means that every cycle is a boundary. Inother words, our necessary condition is also sufficient! So, choose any elementwith the right boundary and declare it to be ιp × ιq.

The induction is now complete provided we can check that this choice sat-isfies naturality, bilinearity, and the Leibniz rule. We leave this as a relaxingexercise for the reader.

The essential point here is that the space supporting the universal pair ofsimplices – ∆p ×∆q – has trivial homology. Naturality transports the result ofthat fact to the general situation.

The cross-product that this procedure constructs is not unique; it dependson a choice a choice of the chain ιp×ιq for each pair p, q with p+q > 1. The coneconstruction in the proof that star-shaped regions have vanishing homology

7. HOMOLOGY CROSS PRODUCT 19

provids us with a specific choice; but it turns out that any two choices areequivalent up to natural chain homotopy.

We return to homotopy invariance. To define our chain homotopy hX ∶Sn(X) → Sn+1(X × I), pick any 1-simplex ι ∶ ∆1 → I such that d0ι = 1 andd1ι = 0, and define

hXσ = (−1)nσ × ι .

Let’s compute:

dhXσ = (−1)nd(σ × ι) = (−1)n(dσ) × ι + σ × (dι)

But dι = c01 − c00 ∈ S0(I), which means that we can continue (remembering that∣∂σ∣ = n − 1):

= −hXdσ + (σ × c01 − σ × c00) = −hXdσ + (ι1∗σ − ι0∗σ) ,

using the normalization axiom of the cross-product. This is the result.

7 Homology cross product

In the last lecture we proved homotopy invariance of homology using the con-struction of a chain level bilinear cross-product

× ∶ Sp(X) × Sq(Y )→ Sp+q(X × Y )

that satisfied the Leibniz formula

d(a × b) = (da) × b + (−1)pa × (db)

What else does this map give us?Let’s abstract a little bit. Suppose we have three chain complexes A∗, B∗,

and C∗, and suppose we have maps × ∶ Ap ×Bq → Cp+q that satisfy bilinearityand the Leibniz formula. What does this induce in homology?

Lemma 7.1. This determines a bilinear map Hp(A) ×Hq(B) ×Ð→Hp+q(C).

Proof. Let a ∈ Zp(A) and b ∈ Zq(A). We want to define [a] × [b] ∈ Hp+q(C).We hope that [a] × [b] = [a × b]. We need to check that a × b is a cycle. ByLeibniz, d(a× b) = da× b+(−1)pa×db. Because a, b are boundaries, this is zero.Now we need to check that this thing is well-defined. Let’s pick other cyclesa′ and b′ in the same homology classes. We want [a × b] = [a′ × b′]. In otherwords, we need to show that a × b differs from a′ × b′ by a boundary. We canwrite a′ = a + da and b′ = b + db, and compute, using bilinearity:

a′ × b′ = (a + da) + (b + db) = a × b + a × db + (da) × b + (da) × (db)

We need to deal with the last three terms here. But since da = 0,

d(a × b) = (−1)pa × (db) .


Since db = 0,d((a) × b) = (da) × b .

And since d2b = 0,d(a × b) = (da) × (db) .

This means that ad × bd and a × b differ by

d((−1)p(a × b) + a × b + a × db) ,

and so are homologous.The last step is to check bilinearity, which is left to the reader.

This gives the following result.

Theorem 7.2. There is a map

× ∶Hp(X) ×Hq(Y )→Hp+q(X × Y )

that is natural, bilinear, and normalized.

This map is also uniquely defined by these conditions, unlike the chain-levelcross product.

I just want to mention an explicit choice of ιp × ιq. This is called theEilenberg-Zilber chain. You’re highly encouraged to think about this yourself.It comes from a triangulation of the prism.

The simplices in this triangulation are indexed by order preserving injec-tions

ω ∶ [p + q]→ [p] × [q]Injectivity forces ω(0) = (0,0) and ω(p+q) = (p, q). Each such map determinesan affine map ∆p+q → ∆p × ∆q of the same name. These will be the singularsimplices making up ιp×ιq. To specify the coefficients, think of ω as a staircasein the rectangle [0, p] × [0, q]. Let A(ω) denote the area under that staircase.Then the Eilenberg-Zilber chain is given by

ιp × ιq =∑(−1)A(ω)ω

This description is in a paper by Eilenberg and Moore. It’s very pretty, butit’s combinatorially annoying to check that this satisfies the conditions of thetheorem. It provides an explicit chain map

βX,Y ∶ S∗(X) × S∗(Y )→ S∗(X × Y )

that satisfies many good properties on the nose and not just up to chain ho-motopy. For example, it’s associative –

S∗(X) × S∗(Y ) × S∗(Z)βX,Y ×1 //

1×βY,Z

S∗(X × Y ) × S∗(Z)

βX×Y,Z

S∗(X) × S∗(Y ×Z)

βX,Y ×Z // S∗(X × Y ×Z)

8. RELATIVE HOMOLOGY 21

commutes – and commutative –

S∗(X) × S∗(Y )βX,Y //

T

S∗(X × Y )

S∗(T )

S∗(Y ) × S∗(X)βY,X //// S∗(X × Y )

commutes, where on spaces T (x, y) = (y, x), and on chain complexes T (a, b) =(−1)pq(b, a) when a has degree p and b has degree q.

We will see that these properties hold up to chain homotopy for any choiceof chain-level cross product.

8 Relative homology

An ultimate goal of algebraic topology is to find means to compute the setof homotopy classes of maps from one space to another. This is importantbecause many geometrical problems can be rephrased as such a computation.It’s a lot more modest than wanting to characterize, somehow, all continuousmaps from X to Y ; but the very fact that it still contains a great deal ofinteresting information means that it is still very challenging to compute.

Homology is in a certain sense the best “additive” approximation to thisproblem; and its additivity makes it much more computable. To justify this,we want to describe the sense in which homology is “additive.” Here are tworelated aspects of this claim.

1. If A ⊆X is a subspace, then H∗(X) a combination of H∗(A) and H∗(X−A).

2. The homology H∗(A ∪B) is like H∗(A) +H∗(B) −H∗(A ∩B).

The first hope is captured by the long exact sequence of a pair, the secondby the Meyer-Vietoris Theorem. Both facts show that homology behaves likea measure. The precise statement of both facts uses the machinery of exactsequences. We’ll use the following language.

Definition 8.1. A sequence of abelian groups is a diagram of abelian groupsof the form

⋯→ Cn+1fnÐ→ Cn

fn−1ÐÐ→ Cn−1 → ⋯ ,

finite or infinite in either or direction or both directions, in which all compositesare zero; that is, im fn ⊆ fn−1 for all n. It is exact at Cn provided that thisinequality is an equality.

Example 8.2. A sequence infinite in both directions is just a chain complex;it is exact at Cn if and only if Hn(C∗) = 0. So homology measures the failureof exactness.


Example 8.3. 0 → AiÐ→ B is exact iff i is injective, and B

pÐ→ C → 0 is exactiff p is surjective.

Exactness is a key property in the development of algebraic topology, and“exact” is a great word for the concept. A foundational treatment of algebraictopology was published by Sammy Eilenberg and Norman Steenrod publishedin 1952. The story is that in the galleys for the book they left a blank spacewhenever the word representing this concept was used, and filled it in at thelast minute.

Definition 8.4. An exact sequence that’s infinite in both directions is a longexact sequence. A short exact sequence is an exact sequence of the form

0→ AiÐ→ B

pÔ⇒ C → 0 .

Any sequence of the form 0→ A→ B → C → 0 expands to a diagram

ker(p)

""A

OO

i // Bp //

##

C

coker(i)

OO

It is short exact if and only if A→≅ kerp and B →≅ coker(i).As suggested above, we will study the homology of a space X by comparing

it to the homology of a subspace A and a complement or quotient construction.Note that S∗(A) injects into S∗(X). This suggests considering the quotientgroup

Sn(X)Sn(A)

.

This is the group of relative n-chains of the pair (X,A).Let’s formalize this a bit. Along with the category Top of spaces, we have

the category Top2 of pairs of spaces. An object of Top2 is a space X togetherwith a subspace A. A map (X,A) → (Y,B) is a continuous map X → Y thatsends A into B.

There are four obvious functors relating Top and Top2:

X ↦ (X,∅) , X ↦ (X,X) ,

(X,A)↦X , (X,A)↦ A.

Do the relative chains form themselves into a chain complex?

Lemma 8.5. Let A∗ be a subcomplex of the chain complex B∗. There is aunique structure of chain complex on the quotient graded abelian group C∗ withentries Cn = Bn/An such that B∗ → C∗ is a chain map.

9. THE HOMOLOGY LONG EXACT SEQUENCE 23

Proof. To define d ∶ Cn → Cn−1, represent c ∈ Cn by b ∈ Bn and hope that[db] ∈ Bn−1/An−1 is well defined. If we replace b by b + a for a ∈ An, we find

d(b + a) = db + da ≡ da mod An−1 ,

so our hope is justified. Then d2[b] = [d2b] = 0.

Definition 8.6. The relative singular chain complex of the pair (X,A) is

S∗(X,A) = S∗(X)S∗(A)

.

This is a functor from pairs of spaces to chain complexes. Of course

S∗(X,∅) = S∗(X) , S∗(X,X) = 0 .

Definition 8.7. The relative singular homology of the pair (X,A) is the ho-mology of the relative singular chain complex:

Hn(X,A) =Hn(S∗(X,A)) .

One of the nice features of the absolute chain group Sn(X) is that it is freeas an abelian group. Is that also the case for its quotent Sn(X,A)? The mapSinn(A) → Sinn(X) is an injection. As long as A ≠ ∅, this injection admitsa splitting. (If A = ∅, then Sn(X,A) = Sn(X) is indeed free.) So when weapply the free abelian group functor we obtain a split monomorphism. Thenthe induced map S∗(X) → S∗(A) ⊕ S∗(X,A) is an isomorphism, so S∗(X,A)is again free.

Example 8.8. Consider ∆n, relative to its boundary ∂∆n ∶= ⋃ imdi ≅ Sn−1.We have the identity map ιn ∶ ∆n →∆n, the universal n-simplex, in Sinn(∆n) ⊆Sn(∆n). It is not a cycle; its boundary dιn ∈ Sn−1(∆n) is the alternating sumof the faces of the n-simplex. Each of these singular simplices lies in ∂∆n, sodιn ∈ Sn−1(∂∆n), and [ιn] ∈ Sn(∆n, ∂∆n) is a relative cycle. We will see thatthe relative homology Hn(∆n, ∂∆n) is infinite cyclic, with generator [ιn].

9 The homology long exact sequence

A pair of spaces (X,A) gives rise to a short exact sequence of chain complexes:

0→ S∗(A)→ S∗(X)→ S∗(X,A)→ 0 .

In homology, this will relate H∗(A), H∗(X), and H∗(X,A).To investigate what happens, let’s suppse we have a general short exact

sequence of chain complexes,

0→ A∗ → B∗ → C∗ → 0 ,

and investigate what happens in homology.


Here is an expanded part of this short exact sequence:

0 // An+1f //

d

Bn+1g //

d

Cn+1//

d

0

0 // Anf //

d

Bng //

d

Cn //

d

0

0 // An−1f // Bn−1

g // Cn−1// 0

Clearly the composite H(A∗) → H∗(X) → H∗(C) is trivial. Is this sequenceexact? Let [b] ∈ Hn(B) such that g([b]) = 0. It’s determined by some b ∈ Bnsuch that d(b) = 0. If g([b]) = 0, then there is some c ∈ Cn+1 such that dc = gb.Now, g is surjective, so there is some b ∈ Bn+1 such that g(b) = c. Then we canconsider db ∈ Bn, and g(d(b)) = d(c) ∈ Cn. What is b−db? This maps to zero inCn, so by exactness there is some a ∈ An such that f(a) = b − db. Is a a cycle?Well, f(da) = d(fa) = d(b − db) = db − d2b = db, but we assumed that db = 0, sof(da) = 0. This means that da is zero because f is an injection by exactness.Therefore a is a cycle. What is [a] ∈ Hn(A)? Well, f([a]) = [b − db] = [b]because db is a cycle. Is the composite Hn(A) → Hn(B) → Hn(C) zero?Yes, because the composite factors through zero. This proves exactness ofHn(A)→Hn(B)→Hn(C).

On the other hand, H∗(A)→H∗(B) may fail to be injective, and H∗(B)→H∗(C) may fail to be surjective. Instead:

Theorem 9.1 (homology long exact sequence). Let 0→ A∗ → B∗ → C∗ → 0 bea short exact sequence of chain complexes. Then there is a natural homomor-phism ∂ ∶Hn(C)→Hn−1(A) such that the sequence

⋯ // Hn+1(C)

ttHn(A) // Hn(B) // Hn(C)

∂

ttHn−1(A) // ⋯

is exact.

9. THE HOMOLOGY LONG EXACT SEQUENCE 25

Proof. We’ll construct ∂, and leave the rest as an exercise. Again:

0 // An+1f //

d

Bn+1g //

d

Cn+1//

d

0

0 // Anf //

d

Bng //

d

Cn //

d

0

0 // An−1f // Bn−1

g // Cn−1// 0

Let c ∈ Cn such that dc = 0. The map g is surjective, so pick a b ∈ Bn suchthat g(b) = c. Then consider db ∈ Bn−1. But g(d(b)) = 0 = d(g(b)) = dc. So byexactness, there is some a ∈ An−1 such that f(a) = db. How many choices arethere of picking a? One, because a is injective. We need to check that a is acycle. What is d(a)? Well, d2b = 0, so da maps to 0 under f . But because f isinjective, da = 0, i.e., a is a cycle. This means we can define ∂[c] = [a].

To make sure that this is well-defined, let’s make sure that this choice ofhomology class a didn’t depend on the b that we chose. Pick some other b′ suchthat g(b′) = c. Then there is a′ ∈ An−1 such that f(a′) = db′. We want a− a′ tobe a boundary, so that [a] = [a′]. We want a ∈ An such that da = a − a′. Well,g(b − b′) = 0, so by exactness, there is a ∈ An such that f(a) = b − b′. What isda? Well, da = d(b−b′) = db−db′. But f(a−a′) = b−b′, so because f is injective,da = a − a′, i.e., [a] = [a′]. What else do I have to check? It’s an exercise tocheck that ∂ as defined here is a homomorphism. Also, left as an exercise tocheck that this doesn’t depend on c ∈ [c], and that ∂ actually makes the exactsequence above exact.

Example 9.2. A pair of spaces (X,A) gives rise to a natural long exact se-quence in homology:

⋯ // Hn+1(X,A)∂

ttHn(A) // Hn(X) // Hn(X,A)

∂

ttHn−1(A) // ⋯

.

Example 9.3. Let’s think again about the pair (Dn, Sn−1). By homotopyinvariance we know that Hq(Dn) = 0 for q > 0, since Dn is contractible. So

∂ ∶Hq(Dn, Sn−1)→Hq−1(Sn−1)


is an isomorphism for i > 1. The bottom of the long exact sequence looks likethis:

0 // H1(Dn, Sn−1)

ssH0(Sn−1) // H0(Dn) // H0(Dn, Sn−1) // 0

When n > 1, both Sn−1 and Dn are path-connected, so the map H0(Sn−1) →H0(Dn) is an isomorphism, and

H1(Dn, Sn−1) =H0(Dn, Sn−1) = 0 .

When n = 1, we discover that

H1(D1, S0) = Z and H0(D1, S0) = 0 .

The generator of H1(D1, S0) is represented by any 1-simplex ι1 ∶ ∆1 →D1 suchthat d0ι = 1 and d1ι = 0 (or vice versa). To go any further in this analysis, we’llneed another tool, known as “excision.”

We can set this up for reduced homology (as in Lecture 5) as well. Note thatany map induces an isomorphism in S−1, so to a pair (X,A) we can associatea short exact sequence

0→ S∗(A)→ S∗(X)→ S∗(X,A)→ 0

and hence a long exact sequence

⋯ // Hn+1(X,A)∂

ttHn(A) // Hn(X) // Hn(X,A)

∂

ttHn−1(A) // ⋯

.

The homology long exact sequence is often used in conjunction with anelementary fact about a map between exact sequences known as the five lemma.Suppose you have two exact sequences of abelian groups and a map betweenthem – a “ladder”:

A4d //

f4

A3d //

f3

A2d //

f2

A1d //

f1

A0

f0

B4

d // B3d // B2

d // B1d // B0

When can we guarantee that f2 is an isomorphism? We’re going to “diagramchase.” Just follow your nose, making assumptions as necessary.

10. EXCISION AND APPLICATIONS 27

Surjectivity: Let b2 ∈ B2. We want to show that there is something in A2

mapping to b2. We can consider db2 ∈ B1. Let’s assume that f1 is surjective.Then there’s a1 ∈ A1 such that f1(a1) = db2. What is da1? Well, f0(da1) =d(f1(a1)) = d(db) = 0. So we want f0 to be injective. Then da1 is zero, soby exactness of the top sequence, there is some a2 ∈ A2 such that da2 = a1.What is f2(a2)? To answer this, begin by asking: What is d(f2(a2))? Bycommutativity, d(f2(a2)) = f1(d(a2)) = f1(a1) = db2. Let’s consider b2−f2(a2).This maps to zero under d. So by exactness, there is b3 ∈ B3 such that d(b3) =b2 − f2(a2). If we assume that f3 is surjective, then there is a3 ∈ A3 such thatf3(a3) = b3. But now, d(a3) ∈ A2, and f2(d(a3)) = d(f3(a3)) = b2 − f2(a2).This means that b2 = f(a2 + d(a3)), which guarantees surjectivity of f2.

This proves the first half of the following important fact.

Proposition 9.4 (Five lemma). In the map of exact sequences above,

• If f0 is injective and f1 and f3 are surjective, then f2 is surjective.

• If f4 is surjective and f3 and f1 are injective, then f2 is injective.

Very commonly one knows that f0, f1, f3, and f4 are all isomorphisms, andconcludes that f2 is also an isomorphism. For example:

Corollary 9.5. Let

0 // A′∗ //

f

B′∗ //

g

C ′∗ //

h

0

0 // A∗ // B∗ // C∗ // 0

be a map of short exact sequences of chain complexes. If two of the three mapsinduced in homology by f, g, and h are isomorphisms, then so is the third.

Here’s an application.

Proposition 9.6. Let (A,X)→ (B,Y ) be a map of pairs, and assume that antwo of A→ B, X → Y , and (X,A)→ (Y,B) induce isomorphims in homology.Then the third one does as well.

Proof. Just apply the five lemma to the map between the two homology longexact sequences.

10 Excision and applications

We have found two general properties of singular homology: homotopy invari-ance and the long exact sequence of a pair. We also claimed that H∗(X,A)“depends only on X − A.” You have to be careful about this. The followingdefinition gives conditions that will capture the sense in which the relativehomology of a pair (X,A) depends only on the complement of A in X.


Definition 10.1. A triple (X,A,U) where U ⊆ A ⊆X, is excisive if U ⊆ Int(A).The inclusion (X −U,A −U) ⊆ (X,A) is then called an excision.

Theorem 10.2. An excision induces an isomorphism in homology,

H∗(X −U,A −U) ≅Ð→H∗(X,A) .

So you can cut out the bits of the interior of A without changing the relativehomology. The proof will take us a couple of days. Before we give applications,let me pose a different way to interpret the motto “H∗(X,A) depends only onX −A.” Collapsing the subspace A to a point gives us a map of pairs

(X,A)→ (X/A,∗) .

When does this map induce an isomorphism in homology? Excision has thefollowing consequence.

Corollary 10.3. Assume that there is a subspace B of X such that (1) A ⊆IntB and (2) A→ B is a deformation retract. Then

H∗(X,A)→H∗(X/A,∗)

is an isomorphism.

Proof. The diagram of pairs

(X,A)

i // (X,B)

(X −A,B −A)

k

joo

(X/A,∗) ı // (X/A,B/A) (X/A − ∗,B/A − ∗)oo

commutes. We want the left vertical to be a homology isomorphism, and willshow that the rest of the perimeter consists of homology isomorphisms. Themap k is a homeomorphism of pairs while j is an excision by assumption (1).The map i induces an isomorphism in homology by assumption (2), the longexact sequences, and the five-lemma. Since I is a compact Hausdorff space, themap B×I → B/A×I is again a quotient map, so the deformation B×I → B, thatrestricts to the constant deformation on A, descends to show that ∗ → B/A isa deformation retract. So the map ı is also a homology isomorphism. Finally,∗ ⊆ Int(B/A) in X/A, by definition of the quotient topology, so induces anisomorphism by excision.

Now what are some consequences? For a start, we’ll finally get aroundto computing the homology of the sphere. It happens simultaneously with acomputation of H∗(Dn, Sn−1). (Note that S−1 = ∅.) To describe generators,for each n ≥ 0 pick a homeomorphism

(∆n, ∂∆n)→ (Dn, Sn−1) ,

10. EXCISION AND APPLICATIONS 29

and writeιn ∈ Sn(Dn, Sn−1)

for the corresponding relative n-chain.

Proposition 10.4. Let ∗ ∈ Sn−1 be any point (for n > 0).

Hq(Sn) =

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

Z = ⟨[∂ιn+1]⟩ if q = n > 0

Z = ⟨[c0∗]⟩ if q = 0, n > 0

Z⊕Z = ⟨[c0∗], [∂ι1]⟩ if q = n = 0

0 otherwise

Hq(Dn, Sn−1) =⎧⎪⎪⎨⎪⎪⎩

Z = ⟨[ιn]⟩ if q = n0 otherwise

Proof. The division into cases for Hq(Sn) can be eased by employing reducedhomology. Then the claim is merely that for n ≥ 0

Hq(Sn−1) =⎧⎪⎪⎨⎪⎪⎩

Z if q = n − 1

0 if q ≠ n − 1

and the map∂ ∶Hq(Dn, Sn−1)→ Hq−1(Sn−1)

is an isomorphism. The second statement follows from the long exact sequencein reduced homology together with the fact that H∗(Dn) = 0 since Dn iscontractible. The first uses induction and pair of isomorphisms

Hq−1(Sn−1) ≅←ÐHq(Dn, Sn−1) ≅Ð→Hq(Dn/Sn−1,∗)

since Dn/Sn−1 ≅ Sn. The right hand arrow is an isomorphism since Sn−1 is adeformation retract of a neighborhood in Dn.

Why should you care about this complicated homology calculation?

Corollary 10.5. If m ≠ n, then Sm and Sn are not homotopy equivalent.

Proof. Their homology groups are not isomorphic.

Corollary 10.6. If m ≠ n, then Rm and Rn are not homeomorphic.

Proof. If m or n is zero, this is clear, so let m,n > 0. Assume we have ahomeomorphism f ∶ Rm → Rn. This restricts to a homeomorphism Rm −0 → Rn − 0. But these are homotopy equivalent to spheres, of differentdimension.

Theorem 10.7 (Brouwer fixed-point theorem). If f ∶Dn →Dn is continuous,then there is some point x ∈Dn such that f(x) = x.


Proof. Suppose not. Then you can draw a ray from x through f(x). It meetsthe boundary of Dn at a point g(x) ∈ Sn−1. Check that g is continuous. If xis on the boundary, then x = g(x), so g is a null-homotopy of the identity mapon Sn−1. This is inconsistent with our computation because the identity mapinduces the identity map on Hn−1(Sn−1) ≅ Z.

picture neededOur computation of the homology of a sphere also implies that there are

many non-homotopic self-maps of Sn, for any n ≥ 1. We will distinguish themby means of the “degree”: A map f ∶ Sn → Sn induces an endomorphism of theinfinite cyclic group Hn(Sn). Any endomorphism of an infinite cyclic group isgiven by multiplication by an integer. This integer is well defined (independentof a choice of basis), and any integer occurs. Thus End(Z) = Z×, the monoidof integers under multiplication. The homotopy classes of self-maps of Sn alsoform a monoid, under composition, and:

Theorem 10.8. Let n ≥ 1. The degree map provides us with a surjectivemonoid homomorphism

deg ∶ [Sn, Sn]→ Z× .

Proof. Degree is multiplicative by functoriality of homology.Construction of a map of degree k: If n = 1, this is just the winding number;

an example is given by regardeing S1 as unit complex numbers and sending zto zk. The proof that this has degree k is an exercise.

Suppose we’ve constructed a map fk ∶ Sn−1 → Sn−1 of degree k. Extend itto a map fk ∶Dn →Dn by defining fk(tx) = tfk(x) for t ∈ [0,1]. We may thencollapse the sphere to a point and identify the quotient with Sn. This gives usa new map gk ∶ Sn → Sn making the diagram below commute.

Hn−1(Sn−1)

fk∗

Hn(Dn, Sn−1)oo //

Hn(Sn)

gk∗

Hn−1(Sn−1) Hn(Dn, Sn−1)oo // Hn(Sn)

The horizontal maps are isomorphisms, so deg gk = k as well.

We will see (in 18.906) that this map is in fact an isomorphism.

11 The Eilenberg Steenrod axioms and the localityprinciple

Before we proceed to prove the excision theorem, let’s review the properties ofhomology theory as we have developed them. They are captured by a set ofaxioms, due to Sammy Eilenberg and Norman Steenrod.

Definition 11.1. A homology theory (on Top) is:

11. THE EILENBERG STEENROD AXIOMS AND THE LOCALITYPRINCIPLE 31

• a sequence of functors hn ∶ Top2 →Ab for all n ∈ Z and

• a sequence of natural transformations ∂ ∶ hn(X,A)→ hn−1(A,∅)

such that:

• if f0, f1 ∶ (X,A) → (Y,B) are homotopic, then f0,∗ = f1,∗ ∶ hn(X,A) →hn(Y,B).

• excisions induce isomorphisms.

• for any pair (X,A), the sequence

⋯→ hq+1(X,A) ∂Ð→ hq(A)→ hq(X)→ hq(X,A) ∂Ð→ ⋯

is exact, where we have written hq(X) for hq(X,∅).

• (the dimension axiom): hn(∗) is nonzero only in dimension zero.

We add the following “Milnor axiom” to our definition. To state it, let I bea set and suppose that for each i ∈ I we have a space Xi. We can form theirdisjoint union or coproduct ∐Xi. The inclusion maps Xi →∐Xi induce mapshn(Xi) → hn(∐Xi), and these in turn induce a map from the direct sum, orcoproduct.

α ∶⊕i

hn(Xi)→ hn (∐i∈IXi) .

Then:

• The map α is an isomorphism for all n.

Ordinary singular homology satisfies these, with h0(∗) = Z. We will soonadd “coefficents” to homology, producing a homology theory whose value on apoint is any prescribed abelian group. In later developments, it emerges thatthe dimension axiom is rather like the parallel postulate in Euclidean geometry:it’s “obvious,” but, as it turns out, the remaining axioms accomodate extremelyinteresting alternatives, in which hn(∗) is nonzero for infinitely many values ofn (both positive and negative).

Excision is a statement that homology is “localizable.” To make this precise,we need some definitions.

Definition 11.2. Let X be a topological space. A family A of subsets of X isa cover if X is the union of the interiors of elements of A.

Definition 11.3. Let A be a cover of X. An n-simplex σ is A-small if thereis A ∈ A such that the image of σ is entirely in A.


Notice that if σ ∶ ∆n →X isA-small, then so is diσ; in fact, for any simplicialoperator φ, φ∗σ is again A-small. Let’s denote by SinA

∗ (X) the graded set ofA-small simplices. This us a sub-simplicial set of Sin∗(X). Applying the freeabelian group functor, we get the subcomplex

SA∗ (X)

of A-small simplices. Write HA∗ (X) for its homology.

Theorem 11.4. The inclusion SA∗ (X) ⊆ S∗(X) is a chain homotopy equiva-

lence, so HA∗ (X)→H∗(X) is an isomorphism.

This will take a little time to prove. Let’s see right now how it impliesexcision.

Suppose X ⊃ A ⊃ U is excisive, so that U ⊆ IntA, or Int(X −U)∪ IntA =X.This if we let B =X −U , then A = A,B is a cover of X. Rewriting in termsof B,

(X −U,A −U) = (B,A ∩B) ,so we aim to show that

S∗(B,A ∩B)→ S∗(X,A)

induces an isomorphism in homology. We have the following diagram of chaincomplexes with exact rows:

0 // S∗(A)

=

// SA∗ (X)

// SA∗ (X)/S∗(A)

// 0

0 // S∗(A) // S∗(X) // S∗(X,A) // 0

The middle vertical induces an isomorphism in homology by the locality prin-ciple, so the homology long exact sequences combine with the five-lemma toshow that the right hand vertical is also a homology isomorphism. But

SAn (X) = Sn(A) + Sn(B) ⊆ Sn(X)

and a simple result about abelian groups provides an isomorphism

Sn(B)Sn(A ∩B)

= Sn(B)Sn(A) ∩ Sn(B)

≅Ð→ Sn(A) + Sn(B)Sn(A)

= SAn (X)Sn(A)

,

so excision follows.This case of a cover with two elements leads to another expression of exci-

sion, known as the “Mayer-Vietoris sequence.” In describing it we will use thefollowing notation for the various inclusion.

A ∩Bj1 //

j2

A

i1

Bi2// X

12. SUBDIVISION 33

Theorem 11.5 (Mayer-Vietoris). Assume that A = A,B is a cover of X.There are natural maps ∂ ∶Hn(X)→Hn−1(A ∩B) such that the sequence

⋯β // Hn+1(X)

∂

ssHn(A ∩B) α // Hn(A)⊕Hn(B)

β // Hn(X)∂

ssHn−1(A ∩B) α // ⋯

is exact, where

α = [ j1∗−j2∗

] , β = [ i1∗ i2∗ ] .

Proof. This is the homology long exact sequence associated to the short exactsequence of chain complexes

0→ S∗(A ∩B) αÐ→ S∗(A)⊕ S∗(B) βÐ→ SA∗ (X)→ 0 ,

combined with the locality principle.

12 Subdivision

We will begin the proof of the locality principle today, and finish it in the nextlecture. The key is a process of subdivision of singular simplices. It will usethe “cone construction” b∗ from Lecture 5. The cone construction dealt witha region X in Euclidean space, star-shaped with respect to b ∈ X, and gave achain-homotopy between the identity and the “constant map” on S∗(X):

db ∗ +b ∗ d = 1 − ηε

where ε ∶ S∗(X) → Z is the augmentation and η ∶ Z → S∗(X) sends 1 to theconstant 0-chain c0b .

Let’s see how the cone construction can be used to “subdivide” an “affinesimplex.” An affine simplex is the convex hull of a finite set of points in Eu-clidean space. To make this non-degenerate, assume that the points a0, a1, . . . , an,have the property that a1 − a0, . . . , an − a0 is linearly independent. Thebarycenter of this simplex is the center of mass of the vertices,

b = 1

n + 1∑ai .

Start with n = 1. To subdivide a 1-simplex, just cut it in half. For the2-simplex, look at the subdivision of each face, and form the cone of them withthe barycenter of the 2-simplex. This gives us a decomposition of the 2-simplexinto six sub-simplices.


We want to formalize this process, and extend it to singular simplices (usingnaturality, of course). Define a natural transformation

$ ∶ Sn(X)→ Sn(X)

by defining it on standard n-simplex, namely by specifying what $(ιn) is whereιn ∶ ∆n →∆n is the universal n-simplex, and then extending by naturality:

$(σ) = σ∗$(ιn) .

Here’s the definition. When n = 0, define $ to be the identity; i.e., $ι0 = ι0. Forn > 0, define

$ιn ∶= bn ∗ $dιn

where bn is the barycenter of ∆n. This makes a lot of sense if you drawout a picture, and it’s a very clever definition that captures the geometry wedescribed. Here’s what we’ll prove.

Proposition 12.1. $ is a natural chain map S∗(X)→ S∗(X) that is naturallychain-homotopic to the identity.

Proof. Let’s try to prove that it’s a chain map. We’ll use induction on n. It’senough to show that d$ιn = $dιn, because then, for any n-simplex σ,

d$σ = d$σ∗ιn = σ∗d$ιn = σ∗$dιn = $dσ∗ιn = $dσ .

Dimension zero is easy: since S−1 = 0, d$ι0 and $dι0 are both zero and henceequal.

For n ≥ 1, we want to compute d$ιn. This is:

d$ιn = d(bn ∗ $dιn)= (1 − ηbε − bn ∗ d)($dιn)

We’ll use induction on n. What happens when n = 1? Well,

ηbε$dι1 = ηbε$(c01 − c00) = ηbε(c01 − c00) = 0 ,

since ε takes sums of coefficients. So the ηbε term drops out for any n ≥ 1. Let’scontinue, using the inductive hypothesis:

d$ιn = (1 − bn ∗ d)($dιn)= $dιn − bn ∗ d$dιn= $dιn − bn$d2ιn

= $dιn

because d2 = 0.

13. PROOF OF THE LOCALITY PRINCIPLE 35

To define the chain homotopy T , we’ll just write down a formula and notjustify it. We just need to define Tιn by naturality. So define:

Tιn = bn ∗ ($ιn − ιn − Tdιn) ∈ Sn+1(∆n) .

Once again, we’re going to check that T is a chain homotopy by induction,and, again, we need to check only on the universal case.

When n = 0, the formula gives Tι0 = 0, so it’s true that dT ι0−Tdι0 = $ι0−ι0.Now let’s assume that dTc−Tdc = $c− c for every (n− 1)-chain. Let’s start bycomputing dT ιn:

dT ιn = dn(bn ∗ ($ιn − ιn − Tdιn))= (1 − bn ∗ d)($ιn − ιn − Tdιn)= $ιn − ιn − Tdιn − bn ∗ (d$ιn − dιn − dTdιn)

All we want now is that bn ∗ (d$ιn −dιn −dTdιn) = 0. We can do this using theinductive hypothesis, because dιn is in dimension n − 1.

dTdιn = −Td(dιn) + $dιn − dιn= $dιn − dιn= d$ιn − dιn .

This means that d$ιn − dιn − dTdιn = 0, so T is indeed a chain homotopy.

13 Proof of the Locality Principle

We have constructed the subdivision operator $ ∶ S∗(X) → S∗(X), with theidea that it will shink chains and by iteration eventually render any chain A-small. Does $ succeed in making simplices smaller? Let’s look first at the affinecase. Recall that the “diameter” of a subset X of a metric space is given by

diam(X) = supd(x, y) ∶ x, y ∈X .

Lemma 13.1. Let σ be an affine n-simplex, and τ a simplex in $σ. Thendiam(τ) ≤ n

n+1diam(σ).

Proof. Suppose that the vertices of σ are v0, v1, . . . , vn. Let b be the barycenterof σ, and write the vertices of τ as w0 = b,w1, . . . ,wn. We compute

∣b − vi∣ = ∣v0 +⋯ + vn − (n + 1)vin + 1

∣ = ∣ (v0 − vi) + (v1 − vi) +⋯ + (vn − vi)n + 1

∣ .

One of the terms in the numerator is zero, so we can continue:

∣b − vi∣ ≤n

n + 1maxi,j

∣vi − vj ∣ =n

n + 1diam(σ)

Since wi ∈ σ,∣b −wi∣ ≤ max

i∣b − vi∣ ≤

n

n + 1diam(σ) .


For the other cases, well, we use induction:

∣wi −wj ∣ ≤ diam(simplex in $dσ) ≤ n − 1

ndiam(dσ) ≤ n

n + 1diam(σ) .

Now let’s transfer this calculation to singular simplices in a spaceX equippedwith a cover A.

Lemma 13.2. For any singular chain c, some iterate of the subdivision oper-ator sends c to an A-small chain.

Proof. We may assume that c is a single simplex σ ∶ ∆n → X, because ingeneraly you just take the largest iterate of $ needed to send each simplex in cto an A-small chain. We now encounter another of the great virtues of singularhomology: we pull A back to a cover of the standard simplex. Define an opencover σ−1A of ∆n defined by

U ∶= σ−1(Int(A)) ∶ A ∈ A .

The space ∆n is a compact Hausdorff space, and so is subject to the Lebseguecovering lemma, which we apply to the open cover IntB ∶ B ∈ σ−1A.

Lemma 13.3 (Lebsegue covering lemma). Let M be a compact metric space,and let U be an open cover. Then there is ε > 0 such that for all x ∈ M ,Bε(x) ⊆ U for some U ∈ U.

To apply this, we will have to understand iterates of the subdivision oper-ator.

Lemma 13.4. For any k ≥ 1, $k ∼ 1 ∶ S∗(X)→ S∗(X).

Proof. We construct Tk such that dTk +Tkd = $k −1. To begin, we take T1 = T ,since dT + Td = $ − 1. Let’s apply $ to this equation. We get $dT + $Td =$2 − $.Sum up these two equations to get

dT + Td + $dT + $Td = $2 − 1 ,

which simplifies tod($ + 1)T + ($ + 1)Td = $2 − 1

since $d = d$.So define T2 = ($ + 1)T , and continuing, you see that we can define

Tk = ($k−1 + $k−2 +⋯ + 1)T .

14. CW-COMPLEXES 37

We are now in position to prove the Locality Principle, which we recall:

Theorem 13.5. Let A be a cover of a space X. The inclusion SA∗ (X) ⊆ S∗(X)

is a chain homotopy equivalence, so HA∗ (X)→H∗(X) is an isomorphism.

Proof. To prove surjectivity let c be an n-cycle in X. We want to find anA-small n-cycle that is homologous to c. There’s only one thing to do. Pick ksuch that $kc is A-small. This is a cycle because because $k is a chain map. Iwant to compare this new cycle with c. That’s what the chain homotopy Tk isdesigned for:

$kc − c = dTkc + Tkdc = dTkc

since c is a cycle. So $kc and c are homologous.Now for injectivity. Suppose c is a cycle in SAn (X) such that c = db for some

b ∈ Sn+1(X). We want c to be a boundary of an A-small chain. Use the chainhomotopy Tk again: Suppose that k is such that $kc is A-small. Compute:

d$kb − c = d($k − 1)b = d(dTk + Tkd)b = dTkc

soc = d$kb − dTkc = d($kb − Tkc) .

Now, $kb is A-small, by choice of k. Is Tkc also A-small? I claim that it is.Why? It is enough to show that Tkσ is A-small if σ is. We know that σ = σ∗ιn.Because σ is A-small, we know that σ ∶ ∆n → X is the composition i∗σ whereσ ∶ ∆n → A and i ∶ A → X is the inclusion of some A ∈ A. By naturality, then,Tkσ = Tki∗σ = i∗Tkσ, which certainly is A-small.

This completes the proof of the Eilenberg Mac Lane axioms for singularhomology. In the next chapter, we will develop a variety of practical tools,using these axioms to compute the singular homology of many spaces.

14 CW-complexes

There are various ways to model geometrically interesting spaces. Manifoldsprovide one important model, well suited to analysis. Another model, one wehave not talked about, is given by simplicial complexes. It’s very combinatorial,and constructing a simplicial complex model for a given space involves makinga lot of choices that are combinatorial rather than topological in character.A more flexible model, one more closely reflecting topological information, isgiven by the theory of CW-complexes.

In building up a space as a CW-complex, we will successively “glue” cellsonto what has been already built. It’s a general construction.


Suppose we have a pair (B,A), and a map f ∶ A → X. Define a spaceX ∪f B (or X ∪A B) in the diagram

Af // _

X

B // X ∪f B

byX ∪f B =X ⊔B/ ∼

where the equivalence relation is generated by requiring that a ∼ f(a) for alla ∈ A. We say that we have “attached B to X along f (or along A).”

There are two kinds of equivalence classes in X ∪f B: (1) singletons con-taining elements of B − A, and (2) x ⊔ f−1(x) for x ∈ X. The topology onX ∪f B is the quotient topology, and is characterized by a universal property:any solid-arrow commutative diagram

Af // _

X

j

j

B //

g))

X ∪f B

##Y

can be filled in. It’s a “push-out.”

Example 14.1. If X = ∗, then ∗ ∪f B = B/A.

Example 14.2. If A = ∅, then X ∪f B is the coproduct X ⊔B.

Example 14.3. If both,

B/∅ = ∗ ∪∅ B = ∗ ⊔B .

For example, ∅/∅ = ∗. This is creation from nothing. We won’t get into thereligious ramifications.

Example 14.4 (Attaching a cell). A basic collection of pairs of spaces is givenby the disks relative to their boundaries: (Dn, Sn−1). (Recall that S−1 = ∅.)In this context, Dn is called an “n-cell,” and a map f ∶ Sn−1 → X allows us toattach an n-cell to X, to form

Sn−1 f // _

X

Dn // X ∪f Dn

14. CW-COMPLEXES 39

You might want to generalize this a little bit, and attach a bunch of n-cells allat once:

∐α∈A Sn−1α

f // _

X

∐α∈AD

nα

// X ∪f ∐α∈ADnα

What are some examples? When n = 0, (D0, S−1) = (∗,∅), so you are justadding a discrete set to X:

X ∪f ∐α∈A

D0 =X ⊔A

More interesting:

S0 ⊔ S0 f // _

∗

D1 ⊔D1 // ∗ ∪f (D1 ⊔D1)

Again there’s just one choice for f , and ∗ ∪f (D1 ⊔D1) is a figure 8, becauseyou start with two 1-disks and identify the four boundary points together. Letme write S1 ∨ S1 for this space. We can go on and attach a single 2-cell tomanufacture a torus. Think of the figure 8 as the perimeter of a square withopposite sides identified. Then the inside of the square is a 2-cell, attached tothe perimeter by a map I’ll denote by aba−1b−1:

S1 aba−1b−1// _

S1 ∨ S1

D2 // (S1 ∨ S1) ∪f D2 = T 2 .

This example illuminates the following definition.

Definition 14.5. A CW-complex is a space X equipped with a sequence ofsubspaces

∅ = Sk−1X ⊆ Sk0X ⊆ Sk1X ⊆ ⋯ ⊆Xsuch that

• X is the union of the SknX’s, and

• for all n, there is a pushout diagram like this:

∐α∈An Sn−1α

fn // _

Skn−1X

∐α∈AnD

nα

gn // SknX

.


The subspace SknX is the n-skeleton of X. Sometimes it’s convenent to usethe alternate notation Xn for the n-skeleton. The first condition is intendedtopologically, so that a subset of X is open if and only if its intersection witheach SknX is open; or, equivalently, a map f ∶X → Y is continuous if and onlyif its restriction to each SknX is continuous. The maps fn are the attachingmaps and the maps gn arecharacteristic maps.

Example 14.6. We just constructed the torus as a CW complex with Sk0T2 =

∗,Sk1T2 = S1 ∨ S1, and Sk2T

2 = T 2.

Definition 14.7. A CW-complex is finite-dimensional if SknX = X for somen; of finite type if each An is finite, i.e., finitely many cell in each dimension;and finite if it’s finite-dimensional and of finite type.

The dimension of a CW complex is the largest n for which there are n cells.This is not obviously a topological invariant, but, have no fear, it turns outthat it is.

In “CW,” the “C” is for cell, and the “W” is for weak, because of the topologyon a CW-complex. This definition is due to J. H. C. Whitehead. Here are acouple of important facts about them.

Theorem 14.8. Any CW-complex is Hausdorff, and it’s compact if and onlyif it’s finite.Any compact smooth manifold admits a CW structure.

15 CW-complexes II

We have a few more general things to say about CW complexes.Suppose X is a CW complex, with skeleton filtration ∅ = X−1 ⊆ X0 ⊆ X1 ⊆

⋯ ⊆X and cell structure

∐α∈An Sn−1α

fn // _

Xn−1

∐α∈AnD

nα

gn // Xn

.

In each case, the boundary of a cell gets identified with part of the previousskeleton, but the “interior”

IntDn = x ∈Dn ∶ ∣x∣ < 1

does not. (Note that IntD0 =D0.) Thus as sets – ignoring the topology –

X = ∐n≥0

∐α∈An

Int(Dnα) .

The subsets IntDnα are called “open n-cells,” despite the fact that they not

generally open in the topology on X, and (except when n = 0) they are nothomeomorphic to compact disks.

15. CW-COMPLEXES II 41

Definition 15.1. Let X be a CW-complex with a cell structure gα ∶ Dnα →

Xn∣α ∈ An. A subcomplex is a subspace Y ⊆ X such that for all n, there is asubset Bn of An such that Yn = Y ∩Xn provides Y with a CW-structure withcharacteristic maps gβ ∣β ∈ Bn.

Example 15.2. SknX ⊆X is a subcomplex.

Proposition 15.3 (Bredon, p. 196). Let X be a CW-complex with a chosencell structure. Any compact subspace of X lies in some finite subcomplex.

Remark 15.4. For fixed cell structures, unions and intersections of subcom-plexes are subcomplexes.

The n-sphere Sn (for n > 0) admits a very simple CW structure: Let ∗ =Sk0(Sn) = Sk1(Sn) = ⋯ = Skn−1(Sn), and attach an n-cell using the uniquemap Sn−1 → ∗. This is a minimal CW structure – you need at least two cellsto build Sn.

This is great – much simpler than the simplest construction of Sn as asimplicial complex – but it is not ideal for all applications. Here’s anotherCW-structure on Sn. Regard Sn ⊆Rn+1, filter the Euclidean space by leadingsubspaces

Rk = ⟨e1, . . . , ek⟩ .

and defineSkkS

n = Sn ∩Rk+1 = Sk .

picture neededNow there are two k-cells for each k with 0 ≤ k ≤ n, given by the two

hemispheres of Sk. For each k there are two characteristic maps,

u, ` ∶Dk → Sk

defining the upper and lower hemispheres:

u(x) = (x,√

1 − ∣x∣2) , `(x) = (x,−√

1 − ∣x∣2) .

Note that if ∣x∣ = 1 then ∣u(x)∣ = ∣`(x)∣ = 1, so each characteristic map restrictson the boundary to a map to Sk−1, and serves as an attaching map. This cellstructure has the advantage that Sn−1 is a subcomplex of Sn.

The case n = ∞ is allowed here. Then R∞ denotes the countably infinitedimensional inner product space that is the topological union of the leadingsubspacesRn. The CW-complex S∞ is of finite type but not finite dimensional.It has the following interesting property. We know that Sn is not contractible(because the identity map and a constant map have different behavior in ho-mology), but:

Proposition 15.5. S∞ is contractible.


Proof. This is an example of a “swindle,” making use of infinite dimensionality.Let T ∶ R∞ → R∞ send (x1, x2, . . .) to (0, x1, x2, . . .). This sends S∞ to itself.The location of the leading nonzero entry is different for x and Tx, so the linesegment joining x to Tx doesn’t pass through the origin. Therefore

x↦ tx + (1 − t)Tx∣tx + (1 − t)Tx∣

provides a homotopy 1 ∼ T . On the other hand, T is homotopic to the constantmap with value (1,0,0, . . .), again by an affine homotopy.

This “inefficient” CW structure on Sn has a second advantage: it’s “equiv-ariant” with respect to the antipodal involution. This provides us with a CWstructure on the orbit space for this action.

Recall that RPk = Sk/ ∼ where x ∼ −x. The quotient map Sk → RPk is adouble cover, identifying upper and lower hemispheres. The inclusion of onesphere in the next is compatible with this equivalence relation, and gives us“linear” embeddings RPk−1 ⊆RPk. This suggests that

∅ ⊆RP0 ⊆RP1 ⊆ ⋯ ⊆RPn

might serve as a CW filtration. Indeed, for each k,

Sk−1 //

Dk

u

RPk−1 // RPk

is a pushout: A line in Rk+1 either lies in Rk or is determined by a uniquepoint in the upper hemisphere of Sk.

16 Homology of CW-complexes

The skeleton filtration of a CW complex leads to a long exact sequence inhomology, showing that the relative homology H∗(Xk,Xk−1) controls how thehomology changes when you pass from Xk−1 to Xk. What is this relativehomology? If we pick a set of attaching maps, we get the following diagram.

∐α Sk−1 //

f

∐αDkα

//

⋁α Skα

Xk−1

// Xk ∪f B // Xk/Xk−1

where ⋁ is the wedge sum (disjoint union with all basepoints identified): ⋁α Skαis a bouquet of spheres. The dotted map exists and is easily seen to be ahomeomorphism.

16. HOMOLOGY OF CW-COMPLEXES 43

Luckily, the inclusion Xk−1 ⊆Xk satisfies what’s needed to conclude that

Hq(Xk,Xk−1)→Hq(Xk/Xk−1,∗)

is an isomorphism. After all, Xk−1 is a deformation retract of the space youget from Xk by deleting the center of each k-cell.

We know Hq(Xk/Xk−1,∗) very well:

Hq( ⋁α∈Ak

Skα) ≅⎧⎪⎪⎨⎪⎪⎩

Z[Ak] q = k0 q ≠ k

.

Lesson: The relative homology Hq(Xk,Xk−1) keeps track of the k-cells of X.

Definition 16.1. The group of cellular n-chains in a CW complex X is

Ck(X) ∶=Hk(Xk,Xk−1) = Z[Ak] .

If we put the fact that Hq(Xk,Xk−1) = 0 for q ≠ k, k + 1 into the homologylong exact sequence of the pair, we find first that

Hq(Xk−1)≅Ð→Hq(Xk) for1 ≠ k, k − 1 ,

and then that there is a short exact sequence

0→Hk(Xk)→ Ck(X)→Hk−1(Xk−1)→ 0 .

So if we fix a dimension q, and watch how Hq varies as we move throughthe skelata of X, we find the following picture. Say q > 0. Since X0 is discrete,Hq(X0) = 0. Then Hq(Xk) continues to be 0 till you get up to Xq. Hq(Xq is asubgroup of the free abelian group Ck(X) and hence is free abelian. Relationsmay get introduced into it when we pass to Xq+1; but thereafter all the maps

Hq(Xq+1)→Hq(Xq+2)→ ⋯

are isomorphisms. All the q-dimensional homology of X is created on Xq, andall the relations in Hq(X) occur by Xq+1.

This stable value of Hq(Xk) maps isomorphically to Hq(X), even if X isinfinite dimensional. This is because the union of the images of any finiteset of singular simplices in X is compact and so lies in a finite subcomplexand in particular lies in a finite skeleton. So any chain in X is the imageof a chain in some skeleton. Since Hq(Xk)

≅Ð→ Hq(Xk+1) for k > q, we findthat Hq(Xq) → Hq(X) is surjective. Similarly, if c ∈ Sq(Xk) is a boundaryin X, then it’s a boundary in X` for some ` ≥ k. This shows that the mapHq(Xq+1)→Hq(X) is injective. We summarize:

Lemma 16.2. Let q ≥ 0. Then

Hq(Xk) = 0 for q < k

andHq(Xk)

≅Ð→Hq(X) for q > k .In particular, Hq(X) = 0 if q exceeds the dimension of X.


We have defined the cellular n-chains of a CW complex X,

Cn(X) =Hn(Xn,Xn−1) ,

and found that it is the free abelian group on the set of n cells. We claim thatthese abelian groups are related to each other; they form the groups in a chaincomplex.

What should the boundary of an n-cell be? It is represented by a charac-teristic map Dn →Xn whose boundary is the attaching map

α ∶ Sn−1 →Xn−1

This is a lot of information, and hard to interpret because Xn−1 is itself poten-tially a complicated space. But things get much simpler if I pinch out Xn−2.This suggests defining

d ∶ Cn(X) =Hn(Xn,Xn−1)∂Ð→Hn−1(Xn)→Hn−1(Xn−1,Xn−2) = Cn−1(X) .

The fact that d2 = 0 is embedded in the following large diagram, in whichthe two columns and the central row are exact.

Cn+1(X) =Hn+1(Xn+1,Xn)

∂n

d

++

0 =Hn−1(Xn−2)

Hn(Xn) //

jn //

Cn(X) =Hn(Xn,Xn−1)∂n−1 //

d

++

Hn−1(Xn−1)

jn−1

Hn(Xn+1)

Cn−1(X) =Hn−1(Xn−1,Xn−2)

0 =Hn(Xn+1,Xn)

Now, ∂n−1 jn = 0. So the composite of the diagonals is zero, i.e., d2 = 0, andwe have a chain complex! This is the “cellular chain complex” of X.

We should compute the homology of this chain complex, Hn(C∗(X)) =kerd/ imd. Now, kerd = ker(jn−1 ∂n−1). But jn−1 is injective, so

kerd = ker∂n−1 = im jn =Hn(Xn) .

On the other hand

imd = jn(im∂n) = im∂n ⊆Hn(Xn) .

SoHn(C∗(X)) =Hn(Xn)/ im∂n =Hn(Xn+1)

by exactness of the left column; but as we know this is exactly Hn(X)!

17. REAL PROJECTIVE SPACE 45

Theorem 16.3. For a CW complex X, there is an isomorpphism

H∗(C∗(X)) ≅H∗(X)

natural with respect to filtration-preserving maps between CW complexes.

This has an immediate and surprisingly useful corollary.

Corollary 16.4. Suppose that the CW complex X has only even cells – thatis, X2k X2k+1 is an isomorphism. Then d = 0 in the cellular chain complexof X, and

H∗(X) ≅ C∗(X) .

That is, Hn(X) = 0 for n odd, is free abelian for all n, and the rank of Hn(X)for n even is the number of n-cells.

Example 16.5. Complex projective space CPn has a CW structure in which

Sk2kCPn = Sk2k+1CPn =CPk .

The attaching S2k−1 →CPk sends v ∈ S2k−1 ⊆Cn to the complex line throughv. So

Hk(CPn) =⎧⎪⎪⎨⎪⎪⎩

Z for 0 ≤ k ≤ 2n, k even0 otherwise .

Finally, notice that in our proof of Theorem 16.3 we used only propertiescontained in the Eilenberg-Steenrod axioms. As a result, any construction of ahomology theory satisfying the Eilenberg-Steenrod axioms gives you the samevalues on CW complexes as singular homology.

17 Real projective space

Let’s try to compute H∗(RPn). This computation will invoke a second wayto think of the cellular chain group Cn(X). Each cell has a characteristic mapDn →Xn, and we have the diagram

∐(Dn, Sn−1) //

''

(Xn,Xn−1)

(⋁Sn−1,∗).

We’ve shown that the vertical map induces an isomorphism in homology, andthe diagonal does as well. (For example, ∐Dn has a CW structure in whichthe (n − 1)-skeleton is ∐Sn−1.) So

Hn(∐(Dn, Sn−1)) ≅Ð→ Cn(X).


We have a CW structure on RPn with Skk(RPn) = RPk; there is onek-cell – which we’ll denote by ek – for each k between 0 and n. So the cellularchain complex looks like this:

0 C0(RPn)oo C1(RPn)oo ⋯oo Cn(RPn)oo 0oo

0 Z⟨e0⟩oo Z⟨e1⟩d=0oo ⋯oo Z⟨en⟩oo 0oo

The first differential is zero because we know what H0(RPn) is (it’s Z!). Thedifferential in the cellular chain complex is given by the top row in the followingcommutative diagram.

Cn =Hn(RPn,RPn−1) ∂ // Hn−1(RPn−1) //

**

Hn−1(RPn−1,RPn−2) = Cn−1

≅

Hn(Dn, Sn−1)

≅

OO

∂

≅// Hn−1(Sn−1)

π∗

OO

// Hn−1(Dn−1/Sn−2,∗) .

The map π ∶ Sn−1 → RPn−1 is the attaching map of the top cell of RPn; thatis, the double cover. The diagonal composite pinches the subspace RPn−1 toa point. The composite map Sn−1 →Dn−1/Sn−2 factors as follows:

Sn−1 double cover //

((

RPn−1 pinch // Dn−1/Sn−2

Sn−1/Sn−2 = Sn−1 ∨ Sn−1

55

One of the maps Sn−1 → Sn−1 from the wedge is the identity, and the other mapis the antipodal map α ∶ Sn−1 → Sn−1. Write σ for a generator of Hn−1(Sn−1).Then in Hn−1 we have σ ↦ (σ,σ)↦ σ +α∗σ. So we need to know the degree ofthe antipodal map on Sn−1. The antipodal map reverses all n coordinates inRn. Each reversal is a reflection, and acts on Sn−1 by a map of degree −1. So

degα = (−1)n .

Therefore the cellular complex of RPn is as follows:

dim −1 0 1 ⋯ n n + 1 ⋯

0 Z0oo Z

2oo ⋯0oo Z2 or 0oo 0oo ⋯oo

The homology is then easy to read off.

Proposition 17.1. The homology of real projective space is as follows.

Hk(RPn) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

Z k = 0 and k = n oddZ/2Z k odd, 0 < k < n0 otherwise .

18. EULER CHARACTERISTIC, AND HOMOLOGY APPROXIMATION47

Here’s a table. Missing entries are 0.

dim 0 1 2 3 4 5 ⋯

RP0 Z

RP1 Z Z

RP2 Z Z/2

RP3 Z Z/2 0 Z

RP4 Z Z/2 0 Z/2

RP5 Z Z/2 0 Z/2 0 Z

The moral: In real projective space, odd cells create new generators; even cells(except for the zero-cell) create torsion in the previous dimension.

This example illustrates the significance of cellular homology, and, therefore,of singular homology. A CW structure involves attaching maps

∐Sn−1 → SknX .

Knowing these, up to homotopy, determines the full homotopy type of the CWcomplex. Homology does not record all this information. Instead, it recordsonly information about the composite obtaind by pinching out Skn−1X.

∐a∈An Sn−1a

//

''

SknX

⋁b∈An−1

Sn−1b .

In Hn−1, this can be identified with a map

∂ ∶ Z[An]→ Z[An−1]

that is none other than the differential in the cellular chain complex.The moral: homology picks off only the “first order” structure of a CW

complex.On the other hand, we’ll see in the next lecture that it does a very good

job of that.

18 Euler characteristic, and homology approximation

Theorem 18.1. Let X be a finite CW-complex with an n-cells. Then

χ(X) =∞∑k=0

(−1)kak


depends only on the homotopy type of X; it is independent of the choice of CWstructure.

This integer χ(X) is called the Euler characteristic of X. We will prove thistheorem by showing that χ(X) equals a number computed from the homologygroups of X, which are themselves homotopy invariants.

We’ll need a little bit of information about the structure of finitely generatedabelian groups.

Let A be an abelian group. The set of torsion elements of A,

Tors(A)a ∈ A ∶ na = 0 for some n ≠ 0 ,

is a subgroup of A (since A is commutative). A group is torsion free if Tors(A) =0. For any A the quotient group A/Tors(A) is torsion free.

For a general abelian group, that’s about all you can say. But now assumeA is finitely generated. Then Tors(A) is a finite abelian group and A/Tors(A)is a finitely generated free abelian group, isomorphic to Zr for some integer rcalled the rank of A. Pick elements of A that map to a set of generators ofA/Tors(A), and use them to define a map A/TorsA→ A splitting the projectionmap. This shows that if A is finitely generated then

A ≅ Tors(A)⊕Zr .

A finite abelian group A is necessarily of the form

Z/n1 ⊕Z/n2 ⊕⋯⊕Z/nt where n1∣n2∣⋯∣nt .

These are the “torsion coefficients” of A. They are well defined natural numbers.

Lemma 18.2. Let 0 → A → B → C → 0 be a short exact sequence of finitelygenerated abelian groups. Then

rankA − rankB + rankC = 0 .

Theorem 18.3. Let X be a finite CW complex. Then

χ(X) =∑k

(−1)krankHk(X) .

Proof. Pick a CW-structure with, say ak k-cells for each k. We have the cellularchain complex C∗. Write H∗, Z∗, and B∗ for the homology, the cycles, and theboundaries, in this chain complex. From the definitions, we have two familiesof short exact sequences:

0→ Zk → Ck → Bk−1 → 0

and0→ Bk → Zk →Hk → 0 .

18. EULER CHARACTERISTIC, AND HOMOLOGY APPROXIMATION49

Let’s use them and facts about rank rewrite the alternating sum is:

∑k

(−1)kak =∑k

(−1)krank(Ck)

=∑k

(−1)k(rank (Zk) + rank (Bk−1))

=∑k

(−1)k(rank (Bk) + rank (Hk) + rank (Bk−1))

The terms rankBk + rankBk−1 cancel because it’s an alternating sum. Thisleaves ∑k(−1)krankHk. But Hk =Hsing

k (X).

In the early part of the 20th century, “homology groups” were not discussed.It was Emmy Noether who first described things that way. Instead, peopleworked mainly with the sequence of ranks,

βk = rankHk(X) ,

which are known (following Poincaré) as the Betti numbers of X.Given a CW-complex X of finite type, can we give a lower bound on the

number of k-cells in terms of the homology of X? Let’s see. Hk(X) is finitelygenerated because Ck(X) Zk(X)↠Hk(X). Thus

Hk(X) =t(k)⊕i=1

Z/ni(k)Z⊕Zr(k)

where the n1(k)∣⋯∣nt(k)(k) are the torsion coefficients of Hk(X) and r(k) isthe rank.

The minimal chain complex with Hk = Zr and Hq = 0 for q ≠ k is just thechain complex with 0 everywhere else except for Zr in the kth degree. Theminimal chain complex of free abelian groups with Hk = Z/nZ and Hq = 0 forq ≠ k is the chain complex with 0 everywhere else except in dimensions k + 1

and k, where we have ZnÐ→ Z These small complexes are called “elementary

chain complexes.”This implies that a lower bound on the minimal number of k-cells is

r(k) + t(k) + t(k − 1) .

The first two terms give generators for Hk, and the last gives relations for Hk−1.These elementary chain complexes can be realized as the reduced cellular

chains of CW complexes (at least if k > 0). A wedge of r copies of Sk hasreduced cellular chains Zr in dimension k and 0 in other dimensions. To con-struct a CW complex with reduced chains Z

nÐ→ Z in dimensions k + 1 and k,start with Sk as k-skeleton and attach a k + 1-cell by a map of degree n. Forexample, when k = 1 and n = 2, you have RP2. These CW complexes are called“Moore spaces.”

This maximally efficient construction of a CW complex in a homotopy typecan in fact be achieved:


Theorem 18.4 (Wall). Let X be a simply connected CW-complex of finitetype. Then there exists a CW complex Y with r(k)+ t(k)+ t(k − 1) k-cells, forall k, and a homotopy equivalence Y →X.

We will prove this theorem in 18.906.The construction of Moore spaces can be generalized:

Proposition 18.5. For any graded abelian group A∗ with Ak = 0 for k ≤ 0,there exists a CW complex X with H∗(X) = A∗.

Proof. Let A be any abelian group. Pick generators for A. They determinea surjection from a free abelian group F0. The kernel of that surjection isfree, being a subgroup of a free abelian group. Write G0 for minimal set ofgenerators of F0, and G1 for a minimal set of generators for F1.

Let k ≥ 1. DefineXk to be the wedge of ∣G0∣ copies of Sk, soHk(Xk) = Z∣G0∣.Now define an attaching map

α ∶ ∐b∈G1

Skb →Xk

by specifying it on each summand Ska . The generator a ∈ G1 ⊆ F1 is given by alinear combination of the generators of F0, say

b =s

∑i=1

niai .

Now map Sk → ⋁s Sk by pinching (s − 1) longitudes to points. Map the ithsphere in the wedge to Skai ⊆ Xk by a map of degree ni. The map on thesummand Ska is then the composite of these two maps,

Ska →⋁i

Ski →⋁a

Ska

Altogether, we get a map α that realizes F1 → F0 in Hk. So using it as anattaching map produces a CW complex X with Hq(X) = A for q = k and 0otherwise. Write M(A,k) for a CW complex pruduced in this way.

Finally, given a graded abelian group A∗, for the wedge over k of the spacesM(A,k).

Such a space M(A,k), with Hq(M(A,k)) = A for q = k and 0 otherwise,is called a Moore space of type (A,k). The notation is a bit deceptive, sinceM(A,k) cannot be made into a functor Ab→ HoTop.

19 Coefficients

As we have seen, abelian groups can be quite complicated, even finitely gen-erated ones. Vector spaces over a field are so much simpler! A vector space isdetermined up to isomorphism by a single cardinality, its dimension. Wouldn’t

19. COEFFICIENTS 51

it be great to have a version of homology that took values in the category ofvector spaces over a field?

We can do this, and more. Let R be any commutative ring at all. Insteadof forming the free abelian group on Sin∗(X), we could just as form the freeR-module:

S∗(X;R) = RSin∗(X)This gives, first, a simplicial object in the category of R-modules. Formingthe alternating sum of the face maps produces a chain complex of R-modules:Sn(X;R) is an R-module for each n, and d ∶ Sn(X;R) → Sn−1(X;R) is anR-module homomorphism. The homology groups are then again R-modules:

Hn(X;R) = ker(d ∶ Sn(X;R)→ Sn−1(X;R))im(d ∶ Sn+1(X;R)→ Sn(X;R))

.

This is the singular homology of X with coefficients in the commutative ringR. It satisfies all the Eilenberg-Steenrod axioms, but

Hn(∗;R) =⎧⎪⎪⎨⎪⎪⎩

R forn = 0

0 otherwise .

We could actually have replaced the ring R by any abelian group here, but thiswill become much clearer after we have the tensor product as a tool.

The rings that are most important in algebraic topology are simple ones:the integers and the prime fields Fp and Q; typically, a PID.

As an experiment, let’s compute H∗(RPn;R) for various rings R. Let’sstart with R = F2, the field with 2 elements. This is a favorite among algebraictopologists, because using it for coefficients eliminates all sign issues. Thecellular chain complex has Sk = F2 for 0 ≤ k ≤ n, and the differential alternatesbetween multiplication by 2 and by 0. But in F2, 2 = 0: so d = 0, and thecellular chains coincide with the homology:

Hk(RPn;F2) =⎧⎪⎪⎨⎪⎪⎩

F2 for0 ≤ k ≤ n0 otherwise .

On the other hand, suppose that R is a ring in which 2 is invertible. Theuniversal case is Z[1/2], but any subring of the rationals containing 1/2 woulddo just as well, as would Fp for p odd. Now the cellular chain complex (indimensions 0 through n) looks like

R0←Ð R

≅←Ð R0←Ð R

≅←Ð ⋯ ≅←Ð R

R0←Ð R

≅←Ð R0←Ð R

≅←Ð ⋯ 0←Ð R

for n odd. Therefore

Hk(RPn;R) =⎧⎪⎪⎨⎪⎪⎩

R fork = 0

0 otherwise


for n even, and

Hk(RPn;R) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

R fork = 0

R fork = n0 otherwise .

You get a much simpler result! Away from 2, even projective spaces look likea point, and odd projective spaces look like a sphere!

I’d like to generalize this process a little bit, and allow coefficients not justin a commutative ring, but more generally in a module M over a commutativering; in particular, any abelian group. This is most cleanly done using themechinism of the tensor product. That mechanism will also let us address thefollowing natural question:

Question 19.1. Given H∗(X;R), can we deduce H∗(X;M) for an R-moduleM?

The answer is called the “universal coefficient theorem”. I’ll spend a fewdays developing what we need to talk about this.

20 Tensor product

The category of R-modules is what might be called a “categorical ring,” in whichaddition corresponds to the direct sum, the zero element is the zero module, 1is R itself, and multiplication is . . . well the subject for today. We care aboutthe tensor product for two reasons: First, it allows us to deal smoothly withbilinear maps such at the cross-product. Second, and perhaps more important,it will allow us relate homology with coefficients in an any abelian group tohomology with coefficients in Z.

Let’s begin by recalling the definition of a bilinear map over a commutativering R.

Definition 20.1. Given three R-modules, M,N,P , a bilinear map (or, to beexplicit, R-bilinear map) is a function β ∶M ×N → P such that

β(x + x′, y) = β(x, y) + β(x′, y) , β(x, y + y′) = β(x, y) + β(x, y′) ,

andβ(rx, y) = rβ(x, y) , β(x, ry) = rβ(x, y) ,

for x,x′ ∈M , y, y′ ∈ N , and r ∈ R.

Example 20.2. Rn ×Rn →R given by the dot product is an R-bilinear map.The cross product R3×R3 →R3 is R-bilinear. If R is a ring, the multiplicationR ×R → R is R-bilinear, and the multiplication on an R-module M given byR×M →M is R-bilinear. This enters into topology because the cross-productHm(X;R) ×Hn(Y ;R) ×Ð→Hm+n(X × Y ;R) is R-bilinear.

Wouldn’t it be great to reduce stuff about bilinear maps to linear maps?We’re going to do this by means of a universal property.

20. TENSOR PRODUCT 53

Definition 20.3. Let M,N be R-modules. A tensor product of M,N is aR-module P and a bilinear map β0 ∶ M ×N → P such that for every bilinearmap β ∶M ×N → Q there is a unique factorization

M ×Nβ0 //

β

##

P

f

Q

through an R-module homomorphism f .

We should have pointed out that the composition f β0 is indeed againR-bilinear; but this is easy to check.

So β0 is a universal bilinear map out of M ×N . Instead of β0 we’re goingto write ⊗ ∶ M × N → P . This means that β(x, y) = f(x ⊗ y) in the abovediagram. There are lots of things to say about this. When you have somethingthat is defined via a universal property, you know that it’s unique . . . but youstill have to check that it exists!

Construction 20.4. I want to construct a univeral R-bilinear map out ofM ×N . Let β ∶M ×N → Q be any R-bilinear map. This β isn’t linear. Maybewe should first extend it to a linear map. Consider R⟨M × N⟩, the free R-module generated by the setM ×N . Well, at least β is a map of sets, so there’sa unique R-linear homomorphism β ∶ R⟨M ×N⟩→ Q extending it:

M ×Nβ //

[−]

&&

Q

R⟨M ×N⟩

β

::

The map [−] isn’t bilinear. So we should quotient R⟨M ×N⟩ by a submoduleS of relations to make it bilinear. So S is the sub R-module generated by thefour familes of elements (corresponding to the four relations in the definitionof R-bilinearity):

1. [(x + x′, y)] − [(x, y)] − [(x′ − y)]

2. [(x, y + y′)] − [(x, y)] − [(x, y′)]

3. [(rx, y)] − r[(x, y)]

4. [(x, ry)] − r[(x, y)]

for x,x′ ∈M , y, y′ ∈ N , and r ∈ R. Now the composite M ×N → R⟨M ×N⟩/S isbilinear - we’ve quotiented out by all things that made it false! Now the mapR⟨M ×N⟩→ Q factors through via R⟨M ×N⟩→ R⟨M ×N⟩/S fÐ→ Q because β isbilinear, uniquely because the map to the quotient is surjective. This completesthe construction.


You’re never going to use this construction to compute anything. If youfind yourself using this construction, stop and think about what you’re doing.Here’s an example: for any abelian group A,

A ×Z/nZ→ A/nA , (a, b)↦ ba mod nA

is clearly bilinear, and is universal as such. Just look: If β ∶ A × Z/nZ → Q isbilinear then β(na, b) = nβ(a, b) = β(a,nb) = β(a,0) = 0, so β factors throughA/nA. And A ×Z/nZ→ A/nA is surjective.

Remark 20.5. Note that the image of M ×N in R⟨M ×N⟩/S generates it asan R-module. These x⊗ y are called “decomposable tensors.”

What are the properties of such a universal bilinear map?

Property 20.6 (Uniqueness). Suppose β0 ∶M ×N → P and β′0 ∶M ×N → P ′

are both universal. Then there’s a linear map f ∶ P → P ′ such that β′0 = fβ0

and a linear map f ′ ∶ P ′ → P such that β0 = f ′β′0. The composite f ′f ∶ P → Pis a linear map such that f ′fβ0 = f ′β′0 = β0. The identity map is another. Butby universality, there’s only one such linear map, so f ′f = 1P . An identicalargument shows that ff ′ = 1P ′ as well, so they are inverse linear isomorphism.In brief:

The target of a univeral R-bilinear map β0 ∶M ×N → P is uniqueup to a unique linear isomorphism compatible with the map β0.

This entitles us to speak of “the” universal bilinear map out of M × N , andgive the target a symbol: M ⊗R N . If R is the ring of integers, or otherwiseunderstood, we will drop it from the notation.

Property 20.7 (Functoriality). Suppose f ∶ M → M ′ and ∶ N → N ′. Studythe diagram

M ×N

f×g

⊗ //

&&

M ⊗N

f⊗g

M ′ ×N ′ ⊗ // M ′ ⊗N ′

There is a unique R-linear map f ⊗ g because the diagonal map is R-bilinearand the map M ×N →M ⊗N is the universal R-bilinear map out of M ×N .You are invited to show that this construction is functorial.

Property 20.8 (Unitality, associativity, commutativity). I said that this wasgoing to be a “categorical ring,” so we should check various properties of thetensor product. For example, R⊗RM should be isomorphic to M . Let’s thinkabout this for a minute. I have an R-bilinear map R ⊗M → M , given bymultiplication. I just need to check the universal property. Suppose I have anR-bilinear map β ∶ R×M → P . I have to construct a map f ∶M → P such thatβ(r, x) = f(rx) and show it’s unique. Our only choice is f(x) = β(1, x), andthat works.

20. TENSOR PRODUCT 55

Similarly, we should check that there’s a unique isomorphism L⊗(M⊗N) ≅Ð→(L⊗M)⊗N that’s compatible with L×(M ×N) ≅ (L×M)×N , and that there’sa unique isomorphism M ⊗N → N ⊗M that’s compatible with the switch mapM ×N → N ×M . There are a few other things to check, too: have fun.

Property 20.9 (Sums). What happens with M ⊗ (⊕α∈ANα)? This might bea finite direct sum, or maybe an uncountable collection. How does this relateto ⊕α∈A(M ⊗Nα)? Let’s construct a map

f ∶ ⊕α∈A

(M ⊗Nα)→M ⊗ (⊕α∈A

Nα) .

We just need to define mapsM ⊗Nα →M ⊗(⊕α∈ANα) because the direct sumis the coproduct. We can use 1 ⊗ inα where inα ∶ Nα → ⊕α∈ANα. These giveyou a map f .

What about a map the other way? We’ll define a map out of the tensorproduct using the universal property. So we need to define a bilinear mapout of M × (⊕α∈ANα). Send (x, y) (where y ∈ Nβ , say) to x ⊗ inβy, whereinβ ∶ Nβ → ⊕Nα is the incusion of a summand. It’s up to you to check thatthese are inverses.

Property 20.10 (Distributivity). Suppose f ∶ M ′ → M , r ∈ R, and g0, g1 ∶N ′ → N . Then

f ⊗ (g0 + g1) = f ⊗ g0 + f ⊗ g1 ∶M ′ ⊗N ′ →M ⊗N

andf ⊗ rg0 = r(f ⊗ g0) ∶M ′ ⊗N ′ →M ⊗N .

We’ll leave this to you to check.

Our immediate use of this construction is to give a clean definition of “ho-mology with coefficients in M ,” where M is any abelian group. First, endowsingular chains with coefficients in M like this:

S∗(X;M) = S∗(X)⊗M

Then we defineHn(X;M) =Hn(S∗(X;M)) .

Since Sn(X) = ZSinn(X), Sn(X;M) is a direct sum of copies of M indexed byn-simplices in X. If M happens to be a ring, this coincides with the notationused in the last lecture. The boundary maps are just d ⊗ 1 ∶ Sn(X) ⊗M →Sn−1(X)⊗M .

As we have noted, the sequence

0→ Sn(A)→ Sn(X)→ Sn(X,A)→ 0


is split short exact, and therefore applying − ⊗M to it produces another splitshort exact sequence. So

Sn(X,A)⊗M = Sn(A;M)/Sn(X;M) ,

and it makes sense to use the notation Sn(X,A;M) for this, and to define

Hn(X,A;M) =Hn(Sn(X,A;M)) .

Notice that

Hn(∗;M) =⎧⎪⎪⎨⎪⎪⎩

M for n = 0

0 otherwise .

The following result is immediate:

Proposition 20.11. For any abelian group M , (X,A) ↦ H∗(X,A;M) pro-vides a homology theory satisfying the Eilenberg-Steenrod axioms with H0(∗;M) =M .

Suppose R is a commutative ring and A is an abelian group. Then A ⊗Ris naturally an R-module. So S∗(X;R) is a chain complex of R-modules – freeR-modules. We can go a little further: suppose that M is an R-module. ThenA ⊗M is an R-module; and S∗(X;M) is a chain complex of R-modules. Wecan also write

S∗(X;M) = S∗(X;R)⊗RM .

This construction is natural in the R-module M ; and, again using the factthat sums of exact sequences are exact, a short exact sequence of R-modules

0→M ′ →M →M ′′ → 0

leads to a short exact sequence of chain complexes

0→ S∗(X;M ′)→ S∗(X;M)→ S∗(X;M ′′)→ 0

and hence to a long exact sequence in homology, a “coefficient long exact se-quence”:

⋯ // Hn+1(X;M ′′)∂

ssHn(X;M ′) // Hn(X;M) // Hn(X;M ′′)

∂

ssHn−1(X;M ′) // ⋯ .

A particularly important case is when R is a field; then S∗(X;R) is a chaincomplex of vector spaces over R, and H∗(X;R) is a graded vector space overR.

Question 20.12. A reasonable question is this: Suppose we know H∗(X).Can we compute H∗(X;M) for an abelian group M? More generally, supposewe know H∗(X;R) and M is an R-module. Can we compute H∗(X;M)?

21. TENSOR AND TOR 57

21 Tensor and Tor

We continue to study properties of the tensor product. Recall that

A⊗Z/nZ = A/nA .

Consider the exact sequence

0→ Z2Ð→ Z→ Z/2Z→ 0 .

Let’s tensor it with Z/2Z. We get

0→ Z/2Z→ Z/2Z→ Z/2Z→ 0 .

This cannot be a short exact sequence! This is a major tragedy: tensoringdoesn’t preserve exact sequences; one says that Z/nZ⊗ − is not “exact.” Thisis why we can’t form homology with coefficients in A by simply tensoringhomology with A.

Tensoring does respect certain exact sequences:

Proposition 21.1. The functor N ↦M ⊗R N preserves cokernels; it is rightexact.

Proof. Suppose that N → N ′′ is a surjection of R-modules, and M is anyR-module. Then

M ⊗R N // M ⊗R N ′′

M ×N

OO

// // M ×N ′′ .

OO

At least we know that M × − preserves surjections. But the image of M ×N ′′

generates M ⊗R N ′′ as an R-module, so the image of M ×N generates it aswell. This implies that M ⊗R N →M ⊗R N ′′ is surjective.

How about this failure of exactness? What can we do about that? Failureof exactness is bad, so let’s try to repair it. A key observation is that if Mis free, then M ⊗R − is exact. If M = R⟨S⟩, the free R-module on a set S,then M ⊗R N = ⊕SN , since tensoring distributes over direct sums. Then weremember:

Lemma 21.2. If M ′i →Mi →M ′′

i is exact for all i ∈ I, then so is

⊕M ′i →⊕Mi →⊕M ′′

i .

Proof. Clearly the composite is zero. Let (xi ∈ Mi, i ∈ I) ∈ ⊕Mi and supposeit maps to zero. That means that each xi maps to zero in M ′′

i and hence is inthe image of some x′i ∈M ′

i . Just make sure to take x′i = 0 if xi = 0.


To exploit this observation, let’s “resolve” M by free modules. This means:find a surjection from a free R-module, F0 → M . This amounts to specifyingR-module generators. The kernel of F0 → M won’t generally be free. Let’smake sure that it is by assuming that R is a PID, and write F1 for the kernel.The failure of M ⊗− to be exact is measured, at least partially, by the leftmostterm (defined as a kernel) in the exact sequence

0→ TorR1 (M,N)→ F1 ⊗R N → F0 ⊗R N →M ⊗R N → 0 .

The notation suggests that this Tor term is independent of the resolution.This is indeed the case, as we shall show presently. But before we do, let’scompute some Tor groups.

Example 21.3. For any PID R, if M = F is free over R we can take F0 = Fand F1 = 0, and discover that then TorR1 (F,N) = 0 for any N .

Example 21.4. Let R = Z and M = Z/nZ, and N any abelian group. WhenR = Z it is often omitted from the notation for Tor. There is a nice freeresolution staring at us: F0 = F1 = Z, and F1 → F0 given by multiplication byn. The sequence looks like

0→ Tor1(Z/nZ,N)→ Z⊗N n⊗1ÐÐ→ Z⊗N → Z/nZ⊗N → 0 ,

soZ/nZ⊗N = N/nN , Tor1(Z/nZ,N) = ker(n∣N) .

The torsion in this case is the “n-torsion” in N . This accounts for the name.

Functors like Tor1 can be usefully defined for any ring, and moving to thatgeneral case makes their significance clearer and illuminates the reason whyTor1 is independent of choice of generators.

So let R be any ring and M a module over it. By picking R-module gener-ators I can produce a surjection from a free R-module, F0 →M . Write K0 forthe kernel of this map. It is the module of relations among the generators. Wecan no longer guarantee that it’s free, but we can at least find a set of modulegenerators for it, and construct a surjection from a free R-module, F1 → K0.Continuing in this way, we get a diagram like this –

⋯ // F2

d // F1

d // F0

K2

>>

!!

K1

>>

!!

K0

>>

!!

N

0

>>

0

==

0

==

0

>>

0

– in which the upside-down V subdiagrams are short exact sequences and Fs isfree for all s. Splicing these exact sequences gives you an exact sequence in the

22. THE FUNDAMENTAL LEMMA OF HOMOLOGICAL ALGEBRA 59

top row. This is a free resolution of N . The top row, F∗, is a chain complex. Itmaps to the very short chain complex with N in degree 0 and 0 elsewhere, andthis chain map is a homology isomorphism (or “quasi-isomorphism”). We havein effect replaced N with this chain complex of free modules. The module Nmay be very complicated, with generators, relations, relations between relations. . . . All this is laid out in front of us by the free resolution. Generators of F0

map to generators for N , and generators for F1 map to relations among thosegenerators.

Now we can try to define higher Tor functors by tensoring F∗ with N andtaking homology. If R is a PID and the resolution is just F1 → F0, forminghomology is precisely taking cokernel and kernel, as we did above. In general,we define

TorRn (M,N) =Hn(M ⊗R F∗) .

In the next lecture we will check that this is well-defined – independent offree resolution, and functorial in the arguments. For the moment, notice that

TorRn (M,F ) = 0 forn > 0 if F is free ,

since I can take F≅←Ð F ← 0← ⋯ as a free resolution; and that

TorR0 (M,N) =M ⊗R N

since we know that M ⊗R − is right-exact.

22 The fundamental lemma of homological algebra

We will now show that the R-modules TorRn (M,N) are well-defined and func-torial. This will be an application of a very general principle.

Theorem 22.1 (Fundamental theorem of homological algebra). Let M andN be R-modules; let

0←M ← E0 ← E1 ← ⋯

be a sequence in which each En is free, and

0← N ← F0 ← F1 ← ⋯

be an exact sequence; and let f ∶ M → N be a homomorphism. Then we canlift f to a chain map f∗ ∶ E∗ → F∗, uniquely up to chain homotopy.

Proof. Let’s try to construct f0. Consider:

0 // K0 = ker(εM) //

g0

E0εM //

f0

M

f

0 // L0 = ker(εN) // F0

εN // // N // 0


We know that E0 = R⟨S⟩. What we do is push the generators of E into M viaεM and then into F via f , and then lift them to F0 via εN (which is possiblebecause it’s surjective). Then extend to a homomorphism, to get f0. You canrestrict it to kernels to get g0.

Now the map d ∶ E1 → E0 satisifes εM d = 0, and so factors through a mapto K0 = ker εM . Similarly, d ∶ F1 → F0 factors through a map F1 → L1, and thismap must be surjective because the sequence F1 → F0 → N is exact. We findourselves in exactly the same situation:

0 // K1//

g1

E1//

f1

K0

g0

0 // L1

// F1// L0

// 0

So by we construct f∗ by induction.Now we need to prove the chain homotopy claim. So suppose I have f∗, f ′∗ ∶

E∗ → F, both lifting f ∶ M → N . Then f ′n − fn (which we’ll rename `n) isa chain map lifting 0 ∶ M → N . We want to consruct a chain null-homotopyof `∗; that is, we want h ∶ En → Fn+1 such that dh + hd = `. At the bottom,E−1 = 0 so we want h ∶ E0 → F1 such that dh = `0. This factorization happensin two steps.

E0

`0

~~

h

vv

// M

0

F1

// // L0// F0

// N .

First, d`0 = 0 implies that `0 factors through L1 = ker εN . Next, F1 → L0 issurjective, by exactness, and E0 is free, so we can lift generators and extendR-linearly as indicated.

The next step is organized by the diagram

E1d //

`1

~~

h

vv

E0

`0

h

~~F2

// //

d

33L1// F1

d // F0

This diagram doesn’t commute; while dh = `0, we want to construct h ∶ E1 → F2

such that dh = `1 − hd. Since

d(`1 − hd) = `0d − dhd = (`0 − dh)d = 0 .

the map `1 − hd lifts to L1 = kerd. But then it lifts through F2, since F2 → L1

is surjective and E1 is free.Exactly the same process continues.

This proof uses a property of freeness that is shared by a broader class ofmodules.

22. THE FUNDAMENTAL LEMMA OF HOMOLOGICAL ALGEBRA 61

Definition 22.2. An R-module P is projective if any map out of P factorsthrough any surjection:

M

P

>>

// N

Every free module is projective; this is what we have been using; our proofof the fundamental lemma of homogolical algebra uses only that En is pro-jective. Anything that’s a direct summand in a projective is also projective.Any projective module is a direct summand of a free module. Over a PID,every projective is free, because any submodule of a free is free. But there areexamples of nonfree projectives:

Example 22.3. Let k be a field and let R be the product ring k × k. It actson k in two ways, via (a, b)c = ac and via (a, b)c = bc. This are both projectiveR-modules that are not free.

Now we will apply the Fundamental Lemma to verify that our proposedconstruction of Tor is independent of free (or projective!) resolution, and isfunctorial.

Suppose I have f ∶ N ′ → N . Pick arbitrary free resolutions N ′ ← N ′∗ and

N ← N∗, and pick any chain map f∗F ′∗ → F∗ lifting f . We claim that the map

induced in homology by 1 ⊗ f∗ ∶ M ⊗R F ′∗ → M ⊗R F∗ is independent of the

choice of lift. Suppose f ′∗ is another lift, and pick a chain homotopy h ∶ f ∼ f ′.Since M ⊗R − is additive, the relation

1⊗ h ∶ 1⊗ f ∼ 1⊗ f ′

still holds. So 1⊗ f and 1⊗ f ′ induce the same map in homology.For example, suppose that F∗ and F ′

∗ are two projective resolutions of N .Any two lifts of the identity map are chain-homotopic, and so induce the samemap H∗(M ⊗R F∗) → H∗(M ⊗R F ′

∗). So if f ∶ F∗ → F ′∗ and g ∶ F ′

∗ → F∗are chain maps lifing the identity, then f∗ g∗ induces the same self-map ofH∗(M ⊗R F ′

∗) as the identity self-map does, and so (by functoriality) is theidentity. Similarly, g∗ f∗ induces the identity map on H∗(M ⊗R F∗). So theyinduce inverse isomorphisms.

Putting all this together shows that any two projective resolutions of N in-duce canonically isomorphic modules TorRn (M,N), and that a homomorphismf ∶ N ′ → N induces a well defined map TorRn (M,N ′) → TorRn (M,N) thatrenders TorRn (M,−) a functor.

Last comment about Tor is that there’s a symmetry there. Of course,M ⊗RN ≅ N ⊗RM . This uses the fact that R is commutative. This leads righton to saying that TorRn (M,N) ≅ TorRn (N,M). We’ve been computing Tor bytaking a resolution of the second variable. But I could equally have taken aresolution of the first variable. This follows from the fundamental theorem ofhomological algebra.


Example 22.4. I want to give an example when you do have higher Tormodules. Let k be a field, and let R = k[d]/(d2). This is sometimes calledthe “dual numbers”, or the exterior algebra over k. We’re going to considerR-modules. Let’s construct a projective resolution of k. What is an R-moduleM? It’s just a k-vector space M with an operator d (given by multiplicationby e) that satisfies d2 = 0. Even though there’s no grading around, I can stilldefine the “homology” of M :

H(M ;d) = kerd

imd.

This k-algebra is augmented by an algebra map ε ∶ R → k splitting the unit;ε(d) = 0. Let’s construct a free R-module resolution of this module. Here’s apicture.

oo

oo

oo

oo

The vertical lines indicate multiplication by e. We could write this as

0← kε←Ð R

e←Ð Re←Ð R ← ⋯ .

Now tensor this over R with an R-module M ; so M is a vector spaceequipped with an operator d with d2 = 0. Each copy of R gets replaced bya copy of M , and the differential gives multiplication by d on M . So takinghomology gives

TorRn (k,M) =⎧⎪⎪⎨⎪⎪⎩

k ⊗RM =M/dM for n = 0

H(M ;d) forn > 0 .

So for exampleTorRn (k, k) = k for n ≥ 0 .

23 Hom and Lim

We will now deveop more properties of the tensor product: its relationship tohomomorphisms and to direct limits.

The tensor product arose in our study of bilinear maps. Even more naturalare linear maps. Given a commutative ring R and two R-modulesM and N , wecan think about the collection of all R-linear maps fromM to N . Not only doesthis set form an abelian group (under pointwise addition of homomorphisms);it forms an R-module, with

(rf)(y) = f(ry) = rf(y) , r ∈ R, y ∈M .

23. HOM AND LIM 63

The check that this is again an R-module homomorphism uses commutativityof R. We will write HomR(M,N), or just Hom(M,N), for this R-module.

Since Hom(M,N) is an R-module, we are entitled to think about what anR-module homomorphism into it is. Given

f ∶ L→ Hom(M,N)

we can define a new function

f ∶ L ×M → N , f(x, y) = (f(x))(y) ∈ N .

You should check that this new function f is R-bilinear! So we get a naturalmap

Hom(L,Hom(M,N))→ Hom(L⊗M,N) .

Conversely, given a map f ∶ L ⊗M → N and x ∈ L, we can define f(x) ∶M → N by the same formula. These are inverse operations, so:

Lemma 23.1. The natural map Hom(L,Hom(M,N)) → Hom(L ⊗M,N) isan isomorphism.

One says that ⊗ and Hom are adjoint, a word suggested by Sammy Eilenbergto Dan Kan, who first formulated this relationship between functors.

This relationship makes a number of things obvious. For example, it’s clearthat

Hom(⊕α

Mα,N) =∏α

Hom(Mα,N)

and this implies that the tensor product distributes over arbitrary direct sums.We’ll see another example in a minute.

The second thing we will discuss is a generalization of one perspective onhow the rational numbers are constructed from the integers – by a limit process:there are compatible maps in the diagram

Z2 //

1

Z3 //

1/2

Z4 //

1/3!ww

Z5 //

1/4!tt

⋯

Q

and Q is the “universal,” or “initial,” abelian group you can put in that position.We will formalize this process, using partially ordered sets as indexing sets.

Recall that a partially ordered set, or poset, is a small category I such that#I(i, j) ≤ 1 and the only isomorphisms are the identity maps. We will beinterested in a particular class of posets.

Definition 23.2. A poset (I,≤) is directed if for every i, j, there exists a ksuch that i ≤ k and j ≤ k.


Example 23.3. This is a very common condition. For example, the naturalnumbers N with inequality. Another example: if X is a space and I is the setof open subsets of X. It’s directed by saying that U ≤ V if U ⊆ V . This isbecause U,U ′ need not be comparable, but U,U ′ ⊆ U ∪U ′. Another example isthe positive natural numbers, with i ≤ j if i∣j. This is because i, j∣(ij).

Definition 23.4. Let I be a directed set. An I-directed diagram in a categoryC is a functor I → C. This means that for every i ∈ I we are given an object

Xi ∈ C, and for every i ≤ j we are given a map Xi

fi,jÐÐ→ Xj , in such a way thatfi,i = 1Xi and if i ≤ j ≤ k then fi,k = fj,k fi,j ∶Xi →Xk.

Example 23.5. If I = (N,≤), then you get a “linear system” X0f01ÐÐ→ X1

f12ÐÐ→X2 → ⋯.

Example 23.6. Suppose I = (N>0, ∣), i.e., the third example above. You can

consider I → Ab, say assigning to each i the integers Z, and fij ∶ Zj/iÐ→ Z.

This seems like a more “choice-free” way to get at a construction of Q out ofZ.

These directed systems can be a little complicated. But there’s a simpleone, namely the constant one.

Example 23.7. Let I be any directed system. Any object A ∈ C determinesan I-directed set, namely the constant functor cA ∶ I → C.

Not every directed system is constant, but we can try to find a best ap-proximating constant system. To compare systems, we need morphisms. Ofcourse, I-directed systems in C are functors I → C. They have natural trans-formations, and those are the morphisms in the category of I-directed systems.That is to say, a morphism is a choice of map gi ∶ Xi → Yi, for each i ∈ I, suchthat

Xi//

gi

Xj

gj

Yi // Yj

commutes for all i ≤ j.

Definition 23.8. Let X ∶ I → C be a directed system. A direct limit is anobject L and a map X → cL that is initial among maps to constant systems.This means that given any other map to a constant system, say X → cA, thereis a unique map f ∶ L→ A such that

cL

cf

X

77

''cA

commutes.

23. HOM AND LIM 65

This is a universal property. So two different direct limits are canonicallyisomorphic; but a directed system may fail to have a direct limit. For example,the linear directed systems we used to creat the rational numbers exist in thecategory of finitely generated abelian groups; but Q is not finitely generated,and there’s no finitely generated group that will serve as a direct limit in thecategory of finitely generated abelian groups.

Example 23.9. Suppose we have an increasing sequence of subspaces, X0 ⊆X1 ⊆ ⋯ ⊆ X. This gives us a directed system of spaces, directed by the poset(N,≤). It’s pretty clear that as a set the direct limit of this system is theunion of the subspaces. Suppose that union is X. Then saying that X is thedirect limit of this directed system of spaces is saying that the topology onX is determined by the topology on the subspaces; it’s the “weak topology,”characterized by the property that a map f ∶ X → Y is continuous if and onlyif the restriction of f to each Xn is continuous. This is saying that a subset ofX is open if and only if its intersection with each Xn is open in X.

Direct limits may be constructed from the material of coproducts and quo-tients. So suppose X ∶ I → C is a directed system. To construct the directlimit, begin by forming the coproduct over the elements of I,

∐i∈IXi .

There are maps ini ∶Xi →∐Xi, but they are not yet compatible with the orderrelation in I. Form a quotient of the coproduct to enforce that compatibility:

limÐ→i∈I

Xi = (∐i∈IXi) / ∼

where ∼ is the equivalence relation generated by requiring that for any i ∈ Iand any x ∈Xi,

inix ∼ injfij(x) .The process of forming the coproduct and the quotient will depend upon thecategory you are working in, and may not be possible. In sets, coproductis disjoint union and the quotient just forms equivalence classes. In abeliangroups, the coproduct is the direct sum and to form the quotient you divideby the subgroup generated by differences.

Proposition 23.10. Let I be a direct set, and let M ∶ I → ModR be a I-directed system of R-modules. There is a natural isomorphism

(limÐ→I

Mi)⊗R N ≅ limÐ→I

(Mi ⊗R N) .

Proof. Let’s verify that both sides satisfy the same universal property. A mapfrom limÐ→IMi) ⊗R N to an R-module L is the same thing as a linear maplimÐ→IMi → HomR(N,L). This is the same as a compatible family of mapsMi → HomR(N,L), which in turn is the same as a compatible family of mapsMi ⊗R N → L, which is the same as a linear map limÐ→I(Mi ⊗R N)→ L.


Here’s a lemma that lets us identify when a map to a constant functor is adirect limit.

Lemma 23.11. Let X ∶ I → Ab (or ModR). A map f ∶ X → cL (given byfi ∶Xi → L for i ∈ I) is the direct limit if and only if:

1. For every x ∈ L, there exists an i and an xi ∈Xi such that fi(xi) = x.

2. Let xi ∈ Xi be such that fi(xi) = 0 in L. Then there exists some j ≥ isuch that fij(xi) = 0 in Xj.

Proof. Straightforward.

This lemma generalizes the observation that Q is the colimit of the diagramwe drew above for I = (N>0, ∣).

Proposition 23.12. The direct limit functor limÐ→I ∶ Fun(I,Ab)→Ab is exact.

In other words, if XpÐ→ Y

qÐ→ Z is an exact sequence of I-directed systems(meaning that at every degree we get an exact sequence of abelian groups), thenlimÐ→I X → limÐ→I Y → limÐ→I Z is exact.

Proof. First of all, pi ∶ X → Z is zero, which is to say that it factors throughthe constant zero object, so limÐ→I X → limÐ→I Z is certainly the zero map. Lety ∈ limÐ→I Y , and suppose y maps to 0 in limÐ→I Z. By condition (1), there existsi such that y = fi(yi) for some yi ∈ Yi. Then 0 = q(y) = fiq(yi) because q is amap of direct systems. This means that there is j ≥ i such that fijq(yi) = 0in Zj . So qfijyi = 0, again because q is a map of direct systems. We have anelement in Yj that maps to zero under q, so there is some xj ∈ Xj such thatp(xj) = yj . Then fj(xj) ∈ limÐ→I X maps to y.

The exactness of the direct limit has many useful consequences. For exam-ple:

Corollary 23.13. Let i↦ C(i) be a directed system of chain complexes. Thenthere is a natural isomorphism

limÐ→i∈I

H∗(C(i))→H∗(limÐ→i∈I

C(i)) .

Putting together things we have just said:

Corollary 23.14. H∗(X;Q) =H∗(X)⊗Q.

So we can redefine the Betti numbers of a space X as

βn = dimQHn(X;Q)

and discuss the Euler characteristic entirely in terms of the rational vectorspaces making up the rational homology of X.

24. UNIVERSAL COEFFICIENT THEOREM 67

24 Universal coefficient theorem

Suppose that we are given H∗(X;Z). Can we compute H∗(X;Z/2Z)? Thisis non-obvious. Consider the map RP2 → S2 that pinches RP1 to a point.In H2(RP2;Z) = 0, so in H2 this map is zero. But in Z/2Z-coefficients, indimension 2, this map gives an isomorphism. This shows that there’s not afunctorial relationship between H∗(X;Z) and H∗(X;Z/2Z); the effect of amap in integral homology does not determine its effect in mod 2 homology. Sohow do we go between different coefficients? That’s the mystery.

Let R be a commutative ring and M an R-module, and suppose we have achain complex C∗ of R-modules. It could be the singular complex of a space,but it doesn’t have to be. Let’s compare Hn(C∗)⊗M with Hn(C∗⊗M). (Hereand below we’ll just write ⊗ for ⊗R.) The latter thing gives homology withcoefficients in M . How can we compare these two? Let’s investigate, and buildup conditions on R and C∗ as we go along.

First, there’s a natural map

α ∶Hn(C∗)⊗M →Hn(C∗ ⊗M) ,

sending [z]⊗m to [z ⊗m]. We propose to show that it is injective. The mapα fits into a commutative diagram with exact columns like this:

0 0

Hn(C∗)⊗Mα //

OO

Hn(C∗ ⊗M)

OO

Zn(C∗)⊗M //

OO

Zn(C∗ ⊗M)

OO

Cn+1 ⊗M= //

OO

Cn+1 ⊗M .

Now, Zn(C∗⊗M) is a submodule of Cn⊗M , but the map Zn(C)⊗M → Cn⊗Mneed not be . . . unless we impose more restrictions. If we can guarantee that itis, then a diagram chase shows that α is a monomorphism.

So let’s assume that R is a PID and that Cn is a free R-module for all n.Then the submodule Bn−1(C∗) ⊆ Cn−1 is again free, so the short exact sequence

0 // Zn(C∗) // Cn //

$$

Bn−1(C∗) //

0

Cn−1

splits. So Zn(C∗) → Cn is a split monomorphism, and hence Zn(C∗) ⊗M →Cn ⊗M is too.


In fact, a little thought shows that this argument produces a splitting ofthe map α.

Now, α is not always an isomorphism. But it certainly is if M = R, and it’scompatible with direct sums, so it certainly is if M is free. The idea is now toresolve M by frees, and see where that idea takes us.

So let0→ F1 → F0 →M → 0

be a free resolution of M . Again, we’re using the assumption that R is a PID,to guarantee that ker(F0 →M) is free. Again using the assumption that eachCn is free, we get a short exact sequence of chain complexes

0→ C∗ ⊗ F1 → C∗ ⊗ F0 → C∗ ⊗M → 0 .

In homology, this gives a long exact sequence. Unsplicing it gives the left-hand column in the following diagram.

0

0

coker(Hn(C∗ ⊗ F1)→Hn(C∗ ⊗ F0)

≅ // coker(Hn(C∗)⊗ F1 →Hn(C∗)⊗ F0))

Hn(C∗ ⊗M)

∂

= // Hn(C∗ ⊗M)

ker(Hn−1(C∗ ⊗ F1)→Hn−1(C∗ ⊗ F0))

≅ //

ker(Hn−1(C∗)⊗ F1 →Hn−1(C∗)⊗ F0)

0 0

The right hand column occurs because α is an isomorphism when the moduleinvolved is free. But

coker(Hn(C∗)⊗ F1 →Hn(C∗)⊗ F0)) =Hn(C∗)⊗M

and

ker(Hn−1(C∗)⊗ F1 →Hn−1(C∗)⊗ F0) = TorR1 (Hn−1(C∗),M) .

Theorem 24.1 (Universal Coefficient Theorem). Let R be a PID and C∗ achain complex of R-modules such that Cn is free for all n. Then there is anatural short exact sequence of R-modules

0→Hn(C∗)⊗MαÐ→Hn(C∗ ⊗M) ∂Ð→ TorR1 (Hn−1(C∗),M)→ 0

that splits (but not naturally).

25. KÜNNETH AND EILENBERG-ZILBER 69

Example 24.2. The pinch map RP2 → S2 induces the following map ofuniversal coefficient short exact sequences:

0 // H2(RP2)⊗Z/2Z

0

// H2(RP2;Z/2Z)

≅

≅ // Tor1(H1(RP2),Z/2Z)

0

// 0

0 // H2(S2)⊗Z/2Z ≅ // H2(S2;Z/2Z) // Tor1(H1(S2),Z/2Z) // 0

This shows that the splitting of the universal coefficient short exact sequencecannot be made natural, and it explains the mystery that we began with.

Remark 24.3. The hypotheses are essential. Exercise: construct two coun-terexamples: one with R = Z but in which the groups in the chain complex arenot free, and one in which R = k[d]/d2 and the modules in C∗ are free over R.

25 Künneth and Eilenberg-Zilber

We want to compute the homology of a product. Long ago, in Lecture 7, weconstructed a bilinear map Sp(X) × Sq(Y ) → Sp+q(X × Y ), called the crossproduct. So we get a linear map Sp(X)⊗Sq(Y )→ Sp+q(X ×Y ), and it satisfiesthe Leibniz formula, i.e., d(x × y) = dx × y + (−1)px × dy. The method we usedwas really an example of the fundamental theorem of homological algebra. Itworks with any coefficient ring, not just the integers.

Definition 25.1. Let C∗,D∗ be two chain complexes. Their tensor product isthe chain complex with

(C∗ ⊗D∗)n = ⊕p+q=n

Cp ⊗Dq .

The differential (C∗⊗D∗)n → (C∗⊗D∗)n−1 sends Cp⊗Dq into the submoduleCp−1 ⊗Dq⊕Cp ⊗Dq−1 by

x⊗ y ↦ dx⊗ y + (−1)px⊗ dy .

So the cross product is a map of chain complexes S∗(X)⊗S∗(Y )→ S∗(X ×Y ). There are two questions:(1) Is this map an isomorphism in homology?(2) How is the homology of a tensor product of chain complexes related to thetensor product of their homologies?

It’s easy to see what happens in dimension zero, because π0(X) × π0(Y ) =π0(X × Y ) implies that H0(X)⊗H0(Y ) ≅Ð→H0(X × Y ).

Let’s dispose of the purely algebraic question (2) first.

Theorem 25.2. Let R be a PID and C∗,D∗ be chain complexes of R-modules.Assume that Cn is a free R-module for all n. There is a short exact sequence

0→ ⊕p+q=n

Hp(C)⊗Hq(D)→Hn(C∗⊗D∗)→ ⊕p+q=n−1

TorR1 (Hp(C),Hq(D))→ 0

natural in these data.


Proof. This is exactly the same as the proof for the UCT. It’s a good idea towork through this on your own.

Corollary 25.3. Under these conditions, if C∗ → R and D∗ → R are homologyisomorphisms then so is C∗ ⊗D∗ → R.

Our attack on question (1) is via the method of “acyclic models.” This isreally a special case of the fundamental lemma of homological algebra!

Definition 25.4. Let C be a category, and fix a set M of object in C, to becalled the “models.” A functor F ∶ C → Ab is M-free if it is the free abeliangroup of a coproduct of corepresentable functors. That is, F is a direct sum offunctors of the form ZHomC(M,−) where M ∈M.

Example 25.5. Since we are interested in the singular homology of a productof two spaces, it may be sensible to take as C the category of ordered pairs ofspaces, C = Top2, and for M the set of pairs of simplicies, M = (∆p,∆q) ∶p, q ≥ 0. Then

Sn(X×Y ) = Z[HomTop(∆n×X)×HomTop(∆n, Y ) = ZHomTop2((∆n,∆n), (X,Y )) .

isM-free.

Example 25.6. With the same category and models,

(S∗(X)⊗ S∗(Y ))n = ⊕p+q=n

Sp(X)⊗ Sq(Y ) ,

isM-free, since the tensor product has as free basis the set

∐p+q=n

Sinp(X) × Sinq(Y ) = ∐p+q=n

HomTop2((∆p,∆q), (X,Y )) .

Definition 25.7. A natural transformation of functors θ ∶ F → G is an M-epimorphism if θM ∶ F (M)→ G(M) is a surjection of abelian groups for everyM ∈ M. A sequence of natural transformations is a composable pair G′ →G → G′′ with trivial composition. Let K be the objectwise kernel of G → G′′.There is a factorization G′ → K. The sequence is M-exact if G′ → K is aM-epimorphism. Equivalently, G′(M) → G(M) → G′′(M) is exact for allM ∈M.

Example 25.8. We claim that

⋯→ Sn(X × Y )→ Sn−1(X × Y )→ ⋯→ S0(X × Y )→H0(X × Y )→ 0

isM-exact. Just plug in (∆p,∆q): you get an exact sequence, since ∆p ×∆q

is contractible.

Example 25.9. The sequence

⋯→ (S∗(X)⊗S∗(Y ))n → (S∗(X)⊗S∗(Y ))n−1 → ⋯→ S0(X)⊗S0(Y )→H0(X)⊗H0(Y )→ 0 .

is alsoM-exact, by Corollary 25.3.


The terms “M-free” and “M-exact” relate to each other in the expectedway:

Lemma 25.10. Let C be a category with a set of models M and let F,G,G′ ∶C →Ab be functors. Suppose that F isM-free, let G′ → G be aM-epimorphism,and let f ∶ F → G be any natural transformation. Then there is a lifting:

G′

F

f>>

f // G

Proof. Clearly we may assume that F (X) = ZHomC(M,X). Suppose thatX =M ∈M. We get:

G′(M)

ZHomC(M,M)

fM

77

fM // G(M)

Consider 1M ∈ ZHomC(M,M). Its image fM(1M) ∈ G(M) is hit by someelement in cM ∈ G′(M), since G′ → G is anM-epimorphism. Define fM(1M)−cM .

Now we exploit naturality! Any ϕ ∶M →X should produce a commutativediagram

C(M,M)fM //

ϕ∗

G′(M)

ϕ∗

C(M,X)

fX // G′(X)

Chase 1M around the diagram, to see what the value of fX(ϕ) must be:

fX(ϕ) = fX(ϕ∗(1M)) = ϕ∗(fM(1M)) = ϕ∗(cM) .

Now extend linearly. You should check that this does define a natural trans-formation.

This is precisely the condition required to prove the Fundamental Lemmaof Homological Algebra. So we have the

Theorem 25.11 (Acyclic Models). Let M be a set of models in a categoryC. Let θ ∶ F → G be a natural transformation of functors fron C to Ab. LetF∗ and G∗ be functors from C to chain complexes, with augmentations F0 → Fand G0 → G. Assume that Fn is M-free for all n, and that G∗ → G → 0 isanM-exact sequence. Then there is a unique natural chain homotopy of chainmaps F∗ → G∗ covering θ.


Corollary 25.12. Suppose furthermore that θ is a natural isomorphism. Ifeach Gn isM-free and F∗ → F → 0 is anM-exact sequence, then any naturalchain map F∗ → G∗ covering θ is a natural chain homotopy equivalence.

Applying this to our category Top2 with models as before, we get thefollowing theorem that completes work we did in Lecture 7.

Theorem 25.13 (Eilenberg-Zilber theorem). There are unique chain homo-topy classes of natural chain maps:

S∗(X)⊗ S∗(Y ) S∗(X × Y )

covering the usual isomorphism

H0(X)⊗H0(Y ) ≅H0(X × Y ) ,

and they are natural chain homotopy inverses.

Corollary 25.14. There is an isomorphism H(S∗(X)⊗S∗(Y )) ≅H∗(X ×Y ).

Combining this theorem with the algebraic Künneth theorem, we get:

Theorem 25.15 (Künneth theorem). Take coefficients in a PID R. There isa short exact sequence

0→ ⊕p+q=n

Hp(X)⊗Hq(Y )→Hn(X × Y )→ ⊕p+q=n−1

TorR1 (Hp(X),Hq(Y ))→ 0

natural in X, Y . It splits as R-modules, but not naturally.

Example 25.16. If R = k is a field, every module is already free, so the Torterm vanishes, and you get a Künneth isomorphism:

× ∶H∗(X;k)⊗k H∗(Y ;k) ≅Ð→H∗(X × Y ;k)

This is rather spectacular. For example, what is H∗(RP3 ×RP3;k), wherek is a field? Well, if k has characteristic different from 2, RP3 has the samehomology as S3, so the product has the same homology as S3 × S3: the di-mensions are 1,0,0,2,0,0,1. If chk = 2, on the other hand, the cohomologymodules are either 0 or k, and we need to form the graded tensor product:

k k k kk k k kk k k kk k k k

so the dimensions of the homology of the product are 1,2,3,4,3,2,1.


The palindromic character of this sequence will be explained by Poincaréduality. Let’s look also at what happens over the integers. Then we have thetable of tensor products

Z Z/2Z 0 ZZ Z Z/2Z 0 Z

Z/2Z Z/2Z Z/2Z 0 Z/2Z0 0 0 0 0Z Z Z/2Z 0 Z

There is only one nonzero Tor group, namely

TorZ1 (H1(RP3),H1(RP3)) = Z/2Z.

Putting this together, we get the groups

H0 ZH1 Z/2Z⊕Z/2ZH2 Z/2ZH3 Z⊕Z⊕Z/2ZH4 Z/2Z⊕Z/2ZH5 0H6 Z

The failure of perfect symmetry here is interesting, and will also be explainedby Poinaré duality.

Chapter 2

Cohomology and duality

26 Coproducts, cohomology

The next topic is cohomology. This is like homology, but it’s a contravariantrather than covariant functor of spaces. There are three reasons why you mightlike a contravariant functor.(1) Many geometric contructions pull back; that is, they behave contravariantly.For example, if I have some covering space X →X and a map f ∶ Y →X, I geta pullback covering space f∗X. A better example is vector bundles (that we’lltalk about in 18.906) – they don’t push out, they pullback. So if we want tostudy them by means of “natural” invariants, these invariants will have to liein a (hopefully computable) group that also behaves contravariantly. This willlead to the theory of characteristic classes.(2) The structure induced by the diagonal map from a space to its squareinduces stucture in contravariant functors that is more general and easier tostudy.(3) Cohomology turns out to be the target of the Poincaré duality map.

Let’s elaborate on point (2). Every space has a diagonal map

X∆Ð→X ×X .

This induces a map H∗(X;R) → H∗(X ×X;R), for any coefficient group R.Now, if R is a ring, we get a cross product map

× ∶H∗(X;R)⊗R H∗(X;R)→H∗(X ×X;R) .

If R is a PID, the Künneth Theorem tells us that this map is a monomorphism.If the remaining term in the Künneth Theorem is zero, the cross product is anisomorphism. So if H∗(X;R) is free over R (or even just flat over R), we geta “diagonal” or “coproduct”

∆ ∶H∗(X;R)→H∗(X;R)⊗R H∗(X;R) .

This map is universally defined, and natural in X, if R is a field.

75

76 CHAPTER 2. COHOMOLOGY AND DUALITY

This kind of structure is unfamiliar, and at first seems a bit strange. Afterall, the tensor product is defined by a universal property for maps out of it;maps into it just are what they are.

Still, it’s often useful, and we pause to fill in some of its properties.

Definition 26.1. Let R be a ring. A (graded) coalgebra over R is a (graded)R-module M equipped with a comultiplication ∆ ∶M →M ⊗RM and a counitmap ε ∶M → R such that the following diagrams commute.

M

∆

=

&&

=

xxR⊗RM M ⊗RM

ε⊗1oo 1⊗ε // M ⊗R R

M∆ //

∆

M ⊗RM

∆⊗1

M ⊗RM

1⊗∆// M ⊗RM ⊗RM

It is commutative if in addition

M

∆

zz

∆

$$M ⊗RM

τ // M ⊗RM

commutes, where τ(x⊗ y) = (−1)∣x∣⋅∣y∣y ⊗ x is the twist map.

Using acyclic models, we saw that the the Künneth map is coassociativeand cocommutative: The diagrams

S∗(X)⊗ S∗(Y )⊗ S∗(Z) ×⊗1 //

1⊗×

S∗(X × Y )⊗ S∗(Z)

×

S∗(X)⊗ S∗(Y ×Z) × // S∗(X × Y ×Z)

andS∗(X)⊗ S∗(Y ) τ //

×

S∗(Y )⊗ S∗(X)

×

S∗(X × Y ) T∗ // S∗(Y ×X)

commute up to natural chain homotopy, where τ is as defined above on thetensor product and TX × Y → Y ×X is the swap map.

Corollary 26.2. Suppose R is a PID and H∗(X;R) is free over R. ThenH∗(X;R) has the natural structure of a commutative graded coalgebra over R.

26. COPRODUCTS, COHOMOLOGY 77

We could now just go on and talk about coalgebras. But they are lessfamiliar, and available only if H∗(X;R) is free over R. So instead we’re goingto dualize, talk about cohomology, and get an algebra structure. Some saythat cohomology is better because you have algebras, but that’s more of asociological statement than a mathematical one.

Let’s get on with it.

Definition 26.3. Let N be an abelian group. A singular n-cochain on X withvalues in N is a function Sinn(X)→ N .

If N is an R-module, then I can extend linearly to get an R-module homo-morphism Sn(X;R)→ N .

Notation 26.4. Write

Sn(X;N) = Map(Sinn(X),N) = HomR(Sn(X;R),N) .

This is going to give us something contravariant, that’s for sure. But wehaven’t quite finished dualizing. The differential d ∶ Sn+1(X;N) → Sn(X;R)induces a “coboundary map”

d ∶ Sn(X;N)→ Sn+1(X;N)

defined by(df)(σ) = (−1)n+1f(dσ) .

The sign is a little strange, and we’ll see an explanation in a minute. Anyway,we get a “cochain complex,” with a differential that increases degree by 1. Westill have d2 = 0, since

(d2f)(σ) = ±d(f(dσ)) = ±f(d2σ) = ±f(0) = 0 ,

so we can still take homology of this cochain complex.

Definition 26.5. The nth singular cohomology group of X with coefficients inan abelian group N is

Hn(X;N) = ker(Sn(X;N)→ Sn+1(X;N))im(Sn−1(X;N)→ Sn(X;N))

.

Let’s first compute H0(X;N). A 0-cochain is a function Sin0(X) → N ;that is, a function (not required to be continuous!) f ∶ X → N . To computedf , take a 1-simplex σ ∶ ∆1 →X and evaluate f on its boundary:

(df)(σ) = −f(dσ) = −f(σ(e0) − σ(e1)) = f(σ(e1)) − f(σ(e0)) .

So f is a cocycle if it’s constant on path components. That is to say:

Lemma 26.6. H0(X;N) = Map(π0(X),N).


Warning 26.7. Sn(X;Z) = Map(Sinn(X);Z) =∏Sinn(X)Z, which is probablyan uncountable product. An awkward fact is that this is never free abelian.

The first thing a cohomology class does is to give a linear functional onhomology, by “evaluation.” Let’s spin this out a bit.

We want to tensor together cochains and chains. But to do that we shouldmake the differential in S∗(X) go down, not up. Just as a notational matter,let’s write

S∨−n(X;N) = Sn(X;N)

and define a differential d ∶ S∨−n(X) → S∨−n−1(X) to be the differential d ∶Sn(X)→ Sn+1(X). Now S∨∗ (X) is a chain complex, albeit a negatively gradedone. Form the graded tensor product, with

(S∨∗ (X;N)⊗ S∗(X))n = ⊕p+q=n

S∨p (X;N)⊗ Sq(X) .

Now evaluation is a degree zero chain map

⟨−,−⟩ ∶ S∨∗ (X;N)⊗ S∗(X)→ N ,

where N is regarded as a chain complex concentrated in degree 0. We wouldlike this map to be a chain map. So let f ∈ Sn(X;N) and σ ∈ Sn(X), andcompute

0 = d⟨f, σ⟩ = ⟨df, σ⟩ + (−1)n⟨f, dσ⟩ .

This forces(df)(σ) = ⟨df, σ⟩ = −(−1)nf(dσ) .

Here’s the payoff: There’s a natural map

H−n(S∨∗ (X;N))⊗Hn(S∗(X))→H0 (S∨∗ (X;N)⊗ S∗(X))→ N

This gives us the Kronecker pairing

⟨−,−⟩ ∶Hn(X;N)⊗Hn(X)→ N .

We can develop the properties of cohomology in analogy with propertiesof homology. For example: If A ⊆ X, there is a restriction map Sn(X;N) →Sn(A;N), induced by the injection Sinn(A) Sinn(X). And as long as A isnonempty, we can split this injection, so any function Sinn(A)→ N extends toSinn(X) → N . This means that Sn(X;N) → Sn(A;N) is surjective. (This isthe case if A = ∅, as well!)

Definition 26.8. The relative n-cochain group with coefficients in N is

ker (Sn(X;N)→ Sn(A;N)) .

This defines a sub cochain complex of S∗(X;N), and we define

Hn(X,A;N) =Hn(S∗(X,A;N)) .

27. EXT AND UCT 79

The short exact sequence of cochain complexes

0→ S∗(X,A;N)→ S∗(X;N)→ S∗(A;N)→ 0

induces the long exact cohomology sequence

⋯

H1(X,A;N) // H1(X;N) // H1(A;N)

δ

kk

H0(X,A;N) // H0(X;N) // H0(A;N) .

δ

kk

27 Ext and UCT

Let R be a ring (probably a PID) and N an R-module. The singular cochainson X with values in N ,

S∗(X;N) = Map(Sin∗(X),N) ,

then forms a cochain complex of R-modules. It is contravariantly functorial inX and covariantly functorial in N . The Kronecker pairing defines a map

Hn(X;N)⊗R Hn(X;R)→ N

whose adjointβ ∶Hn(X;N)→ HomR(Hn(X;R),N)

gives us an estimate of the cohomology in terms of the homology of X. Here’show well it does:

Theorem 27.1 (Mixed variance Universal Coefficient Theorem). Let R be aPID and N an R-module, and let C∗ be a chain-complex of free R-modules.Then there is a short exact sequence of R-modules,

0→ Ext1R(Hn−1(C∗),N)→Hn(HomR(C∗,N))→ HomR(Hn(C∗),N)→ 0 ,

natural in C∗ and N , that splits (but not naturally).

Taking C∗ = S∗(X;R), we the short exact sequence

0→ Ext1R(Hn−1(X;R),N)→Hn(X;N) βÐ→ HomR(Hn(X;R),N)→ 0

that splits, but not naturally. This also holds for relative cohomology.What is this Ext?The problem that arises is that HomR(−,N) ∶ ModR → ModR is not

exact. Suppose I have an injection M ′ → M . Is Hom(M,N) → Hom(M ′,N)


surjective? Does a map M ′ → N necessarily extend to a map M → N? No!For example, Z/2Z Z/4Z is an injection, but the identity map Z/2Z→ Z/2Zdoes not extend over Z/4Z.

On the other hand, if M ′ iÐ→ MpÐ→ M ′′ → 0 is an exact sequence of R-

modules then

0→ HomR(M ′′,N)→ HomR(M,N)→ HomR(M ′,N)

is again exact. Check this statement!Now homological algebra comes to the rescue to repair the failure of exact-

ness! Pick a free resolution of M ,

⋯→ F2 → F1 → F0 →M → 0 .

Apply Hom to get a chain complex

0→ Hom(F0,N)→ Hom(F1,N)→ Hom(F2,N)→ ⋯ .

Definition 27.2. ExtnR(M,N) =Hn(HomR(F∗,N)).

Remark 27.3. Ext is well-defined and functorial, by the fundamental lemmaof homological algebra. If M is free (or projective) then Extn(M,−) = 0 forn > 0, since we can take M as its own projective resolution. If R is a PID, thenwe can assume F1 = ker(F0 →M) and Fn = 0 for n > 1, so Extn = 0 if n > 1. IfR is a field, then Extn = 0 for n > 0.

Example 27.4. Let R = Z and take M = Z/kZ. This admits a simple freeresolution: 0 → Z

kÐ→ Z → Z/kZ → 0. Apply Hom(−,N) to it, and rememberthat Hom(Z,N) = N , to get the very short cochain complex, with entries indimensions 0 and 1:

0→ NkÐ→ N → 0 .

Taking homology gives us

Hom(Z/kZ,N) = ker(k∣N) Ext1(Z/kZ,N) = N/kN .

Proof. of Theorem 27.1 First of all, notice that

Cn/Zn ≅ Bn−1

is a submodule Cn−1 and hence is free. Thus both of the following short exactsequences split:

0→ Zn → Cn → Cn/Zn → 0 (2.1)

0→ Zn/Bn → Cn/Bn → Cn/Zn → 0 . (2.2)

Note that the second one can be rewritten as

0→Hn → Cn/Bn → Bn−1 → 0 .

27. EXT AND UCT 81

Start with the diagram

0 // BnHom(C∗,N) //

ZnHom(C∗,N)

≅

// Hn(Hom(C∗,N))

// 0

0 // Hom(Bn−1,N) // Hom(Cn/Bn,N) // Hom(Hn,N) // 0

The bottom row arises from (2.2) and is exact because (2.2) splits. The middle

arrow starts with f ∶ Cn → N such that Cn+1dÐ→ Cn

fÐ→ N is zero. This conditionis equivalent to requiring that f kill boundaries, and so it factors through aunique map Cn/Bn → N .

We claim that the composite BnHom(C∗,N)→ Hom(Hn,N) is trivial. Sostart with f ∶ Cn → N such that Cn+1 → Cn → N is trivial. Then f kills Bn andso factors through Cn/Bn, giving an element of Hom(Cn/Bn,N); but it alsokills the larger submodule Zn, and hence factors through Cn/Zn. This impliesthat the composite Hn → Cn/Bn → N is trivial since Hn → Cn/Bn → Cn/Znis.

So we can fill in the maps to get a map of short exact sequences. By thesnake lemma, the right arrow is surjective and its kernel K fits into the shortexact sequence at the top of the following diagram.

0 // BnHom(C∗,N) //

Hom(Bn−1,N)

=

// K //

0

0 // I // Hom(Bn−1,N) // Ext1(Hn−1,N) // 0

The bottom row of this diagram comes from the long exact sequence associatedto the short exact sequence

0→ Bn−1 → Zn−1 →Hn−1 → 0 ,

soI = im (Hom(Zn−1,N)→ Hom(Bn−1,N)) .

We claim that there is a surjection on the left as shown (making the squarecommutative). This completes the proof, since the snake lemma then impliesthat the right arrow is an isomorphism.

The left arrow occurs at the right of the diagram

0 // Zn−1 Hom(C∗,N)

// Hom(Cn−1,N)

// Bn

// 0

0 // Hom(Hn−1,N) // Hom(Zn−1,N) // I // 0

Here the middle surjection is induced from the split injection Zn−1 → Cn−1. Weneed to construct the left arrow; this will finish the proof, since then the rightarrow exists and is surjective again by the snake lemma.


So let f ∶ Cn−1 → N be such that Cn → Cn−1 → N is trivial. Its image underthe vertical surjection is the composite Zn−1 → Cn−1 → N . Now

Cn

// Cn−1// N

Bn−1// Zn−1

OO ;;

// Hn−1

OO

// 0

The composite Bn−1 → N is trivial since Cn → Bn−1 is surjective, and thedesired factorization through Hn−1 follows.

Remark 27.5. Question: Why is Ext called Ext?Answer: It classifies extensions. Let R be a commutative ring, and let M,Nbe two R-modules. I can think about “extensions of M by N , that is, shortexact sequences of the form

0→ N → L→M → 0 .

For example, I have two extensions of Z/2Z by Z/2Z:

0→ Z/2Z→ Z/2Z⊕Z/2Z→ Z/2Z→ 0

and0→ Z/2Z→ Z/4Z→ Z/2Z→ 0 .

We’ll say that two extensions are equivalent if there’s a map of short exactsequences between them is the identity on N and M . The two extensionsabove aren’t equivalent, for example.

Another definition of Ext1R(M,N) is the set of extensions like this modulo

this notion of equivalence. The zero in the group is the split extension.

The universal coefficient theorem is useful in transferring properties of ho-mology to cohomology. For example, if f ∶ X → Y is a map that inducesan isomorphism in H∗(−;R), then it induces an isomorphism in H∗(−;N) forany R-module N , at least provided that R is a PID. (This is true in general,however.)

Cohomology satisfies the appropriate analogues of the Eilenber Steenrodaxioms.Homotopy invariance: If f0 ∼ f1 ∶ (X,A)→ (Y,B), then

f∗0 = f∗1 ∶H∗(Y,B;N)→H∗(X,A;N) .

I can’t use the UCT to address this because the UCT only tells you that thingsare isomorphic. But we did establish a chain homotopy f0,∗ ∼ f1,∗ ∶ S∗(X,A)→S∗(Y,B), and applying Hom converts chain homotopies to cochain homotopies.Excision: If U ⊆ A ⊆ X such that U ⊆ Int(A), then H∗(X,A;N) → H∗(X −U,A − U ;N) is an isomorphism. This follows from excision in homology andby the mixed variance UCT.

28. PRODUCTS IN COHOMOLOGY 83

Mayer-Vietoris sequence: If A,B ⊆X are such that their interiors cover X,then there is a long exact sequence

⋯

rrHn(X;N) // Hn(A;N)⊕Hn(B;N) // Hn(A ∩B;N)

rrHn+1(X;N) // ⋯

Milnor axiom: The inclusions induce an isomorphism

H∗(∐α

Xα;N)→∏α

H∗(Xα;N) .

28 Products in cohomology

We’ll talk about the cohomology cross product first. Actually, the first step isto produce a map on chains that goes in the reverse direction from the crossproduct we constructed in Lecture 7.

Construction 28.1. For each pair of natural numbers p, q, we will define anatural homomorphism

α ∶ Sp+q(X × Y )→ Sp(X)⊗ Sq(Y ) .

It suffices to define this on simplices, so let σ ∶ ∆p+q → X × Y be a singular

(p + q)-simplex in the product. I can write σ = (σ1

σ2) where σ1 ∶ ∆p+q → X and

σ2 ∶ ∆p+q → Y . I have to produce a p-simplex in X and a q-simplex in Y . Firstdefine two maps in the simplex category: the “front face” αp ∶ [p] → [p + q],sending i to i for 0 ≤ i ≤ p, and the “back face” ωq ∶ [q] → [p + q], sending jto j + p for 0 ≤ j ≤ q. Use the same symbols for the affine extensions to maps∆p →∆p+q and ∆q →∆p+q. Now let

α(σ) = σ1 αp ⊗ σ2 ωq .

This seems like a very random construction; but it works! It’s named aftertwo great early algebraic topologists, Alexander and Whitney. For homework,you will show that these maps assemble into a chain map

α ∶ S∗(X × Y )→ S∗(X)⊗ S∗(Y ) .

This works over any ring R. To get a map in cohomology, we should form

Sp(X;R)⊗RSq(Y ;R)→ HomR(Sp(X;R)⊗RSq(Y ;R),R) α∗Ð→ HomR(Sp+q(X×Y ;R),R) = Sp+q(X×Y ) .


The first map goes like this: Given chain complexes C∗ andD∗, we can considerthe dual cochain complexes HomR(C∗,R) and HomR(D∗,R), and construct achain map

HomR(C∗,R)⊗R HomR(D∗,R)→ HomR(C∗ ⊗R D∗,R)

by

f ⊗ g ↦⎧⎪⎪⎨⎪⎪⎩

(x⊗ y ↦ (−1)pqf(x)g(y)) ∣x∣ = ∣f ∣ = p, ∣y∣ = ∣g∣ = q0 otherwise.

Again, I leave it to you to check that this is a chain map.Altogether, we have constructed a natural chain map

× ∶ Sp(X)⊗ Sq(Y )→ Sp+q(X × Y )

From this, we get a homomorphism

H∗(S∗(X)⊗ S∗(Y ))→H∗(X × Y ) .

I’m not quite done! As in the Künneth theorem, there is an evident naturalmap

H∗(X)⊗H∗(Y )→H∗(S∗(X)⊗ S∗(Y )) .

The composite

× ∶H∗(X)⊗H∗(Y )→H∗(S∗(X)⊗ S∗(Y ))→H∗(X × Y )

is the cohomology cross product.It’s not very easy to do computations with this, directly. We’ll find indirect

means. Let me make some points about this construction, though.

Definition 28.2. The cup product is the map obtained by taking X = Y andcomposing with the map induced by the diagonal ∆ ∶X →X ×X:

∪ ∶Hp(X)⊗Hq(X) ×Ð→Hp+q(X ×X) ∆∗

Ð→Hp+q(X), .

These definitions make good sense with any ring for coefficients.Let’s explore this definition in dimension zero. I claim that H0(X;R) ≅

Map(π0(X),R) as rings. When p = q = 0, both α0 and ω0 are the identitymaps, so we are just forming the pointwise product of functions.

There’s a distinguished element inH0(X), namely the the function π0(X)→R that takes on the value 1 on every path component. This is the identity forthe cup product. This comes out because when p = 0 in our above story, thenα0 is just including the 0-simplex, and ωq is the identity.

The cross product is also associative, even on the chain level.

Proposition 28.3. Let f ∈ Sp(X), g ∈ Sq(Y ), and h ∈ Sr(Z), and let σ ∶∆p+q+r →X × Y ×Z be any simplex. Then

((f × g) × h)(σ) = (f × (g × h))(σ) .

29. CUP PRODUCT, CONTINUED 85

Proof. Write σ12 for the composite of σ with the projection map X × Y ×Z →X × Y , and so on. Then

((f × g) × h)(σ) = (−1)(p+q)r(f × g)(σ12 αp+q)h(σ3 ωr) .

But(f × g)(σ12 αp+q) = (−1)pqf(σ1 αp)g(σ2 µq) ,

where µq is the “middle face,” sending ` to ` + p for 0 ≤ ` ≤ q. In other words,

((f × g) × h)(σ) = (−1)pq+qr+rpf(σ1 αp)g(σ2 µq)h(σ3 ωr) .

I’ve used associativity of the ring. But you get exactly the same thing whenyou expand (f × (g × h))(σ), so the cross product is associative.

Of course the diagonal map is “associative,” too, and we find that the cupproduct is associative:

(α ∪ β) ∪ γ = α ∪ (β ∪ γ) .

But this product is obviously not commutative on the level of cochains. Ittreats the two maps completely differently. But we have ways of dealing withthis. You will show for homework that the method of acyclic models showsthat

α ∪ β = (−1)∣α∣⋅∣β∣β ∪ α .So H∗(X;R) forms a commutative graded R-algebra.

29 Cup product, continued

We have constructed an explicit map Sp(X)⊗ Sq(Y ) ×Ð→ Sp+q(Y ) via:

(f × g)(σ) = (−1)pqf(σ1 αp)g(σ2 ωq)

where αp ∶ ∆p → ∆p+q takes k ↦ k for 0 ≤ k ≤ p and ωq ∶ ∆q → ∆p+q sendsh ↦ j + p for 0 ≤ j ≤ q. This is a chain map, and it induces a “cross product”H(Sp(X)⊗ Sq(Y ))→Hp+q(X × Y ) and, by composing with the map inducedby the diagonal embedding, a “cup product”

∪ ∶Hp(X)⊗Hq(X)→Hp+q(X) .

We formalize the structure that this product imposes on cohomology.

Definition 29.1. Let R be a commutative ring. A graded R-algebra is a gradedR-module ⋯,A−1,A0,A1,A2,⋯ equipped with maps Ap ⊗R Aq → Ap+q and amap η ∶ R → A0 that make the following diagram commute.

Ap ⊗R R1⊗η //

=

&&

Ap ⊗R A0

A0 ⊗R Aq

R⊗R Aqη⊗1oo

=

xxAp Aq


Ap ⊗R (Aq ⊗R Ar) //

Ap ⊗R Aq+r

Ap+q ⊗R Ar // Ap+q+r

A graded R-algebra A is commutative if the following diagram commutes:

Ap ⊗R Aqτ //

%%

Aq ⊗R Ap

yyAp+q

where τ(x⊗ y) = (−1)pqy ⊗ x.

We claim that H∗(X;R) forms a commutative graded R-algebra under thecup product. This is nontrivial. On the cochain level, this is clearly not gradedcommutative. We’re going to have to work hard – in fact, so hard that you’regoing to do some of it for homework. What needs to be checked is that thefollowing diagram commutes up to natural chain homotopy.

S∗(X × Y ) T∗ //

αX,Y

S∗(Y ×X)

αY,X

S∗(X)⊗R S∗(Y ) τ // S∗(Y )⊗R S∗(X)

Acyclic models helps us prove things like this.You might hope that there is some way to produce a commutative product

on a chain complex modeling H∗(X). With coefficients in Q, this is possible,by a construction due to Dennis Sullivan. With coefficients in a field of nonzerocharacteristic, it is not possible. Steenrod operations provide the obstruction.

My goal now is to compute the cohomology algebras of some spaces. Somespaces are easy! There is no choice for the product structure on H∗(Sn), forexample. (When n = 0, we get a free module of rank 2 in dimension 0. Thisadmits a variety of commutative algebra structures; but we have already seenthat H0(S0) = Z ×Z as an algebra.) Maybe the next thing to try is a productof spheres. More generally, we should ask whether there is an algebra structureon H∗(X)⊗H∗(Y ) making the cross product an algebra map. If A and B aretwo graded algebras, there is a natural algebra structure on A ⊗ B, given by1 = 1⊗ 1 and

(a′ ⊗ b′)(a⊗ b) = (−1)∣b′∣⋅∣a∣a′a⊗ b′b .

If A and B are commutative, then so is A⊗B with this algebra structure.

Proposition 29.2. The cohomology cross product

× ∶H∗(X)⊗H∗(Y )→H∗(X × Y )

is an R-algebra homomorphism.

29. CUP PRODUCT, CONTINUED 87

Proof. I have diagonal maps ∆X ∶ X → X × X and ∆Y ∶ Y → Y × Y . Thediagonal on X × Y factors as

X × Y ∆X×Y //

∆X×∆Y

''

X × Y ×X × Y

X ×X × Y × Y .

1×T×1

55

Let α1, α2 ∈ H∗(X) and β1, β2 ∈ H∗(Y ). Then α1 × β1, α2 × β2 ∈ H∗(X × Y ),and I want to calculate (α1 × β1) ∪ (α2 × β2). Let’s see:

(α1 × β1) ∪ (α2 × β2) = ∆∗X×Y (α1 × β1 × α2 × β2)

= (∆X ×∆Y )∗(1 × T × 1)∗(α1 × β1 × α2 × β2)= (∆X ×∆Y )∗(α1 × T ∗(β1 × α2) × β2)

= (−1)∣α2∣⋅∣β1∣(∆X ×∆Y )∗(α1 × α2 × β1 × β2)

Now, I have a diagram:

H∗(X × Y ) H∗(X)⊗R H∗(Y )×X×Yoo

H∗(X ×X × Y × Y )

(∆X×∆Y )∗OO

H∗(X ×X)⊗H∗(Y × Y )×X×X,Y ×Yoo

∆×X⊗∆∗

Y

OO

This diagram commutes because the cross product is natural. We learn:

(α1 × β1) ∪ (α2 × β2) = (−1)∣α2∣⋅∣β1∣(∆X ×∆Y )∗(α1 × α2 × β1 × β2)

= (−1)∣α2∣⋅∣β1∣(α1 ∪ α2) × (β1 ∪ β2) .

That’s exactly what we wanted.

Example 29.3. How about H∗(Sp × Sq)? I’ll assume that p and q are bothpositive, and leave the other cases to you. The Künneth theorem guaranteesthat × ∶ H∗(Sp) ⊗H∗(Sq) → H∗(Sp × Sq) is an isomorphism. Write α for agenerator of Sp and β for a generator of Sq; and use the same notations forthe pullbacks of these elements to Sp × Sq under the projections. Then

H∗(Sp × Sq) = Z⟨1, α, β,αβ⟩ ,

andα2 = 0 , β2 = 0 , αβ = (−1)pqβα .

This calculation is useful!

Corollary 29.4. Let p, q > 0. Any map Sp+q → Sp × Sq induces the zero mapin Hp+q(−).


Proof. Let f ∶ Sp+q → Sp × Sq be such a map. It induces an algebra mapf∗ ∶H∗(Sp ×Sq)→H∗(Sp+q). This map must kill α and β, for degree reasons.But then it also kills their product, since f∗ is multiplicative.

The space Sp ∨Sq ∨Sp+q has the same cohomology groups as Sp ×Sq. Bothare built as CW complexes with cells in dimensions 0, p, q, and p+ q. But theyare not homotopy equivalent. We can see this now because there is a mapSp ∨ Sq ∨ Sp+q → Sp+q inducing an isomorphism in Hp+q(−), namely, the mapthat pinches the other two factors to the basepoint.

30 Surfaces and nondegenerate symmetric bilinearforms

We are aiming towards a proof of a fundamental cohomological property ofcompact manifolds.

Definition 30.1. A (topological) manifold is a Hausdorff space such thatevery point has an open neighborhood that is homeomorphic to some (finitedimensional) Euclidean space.

If all these Euclidean space can be chosen to be Rn, we have an n-manifold.We will always make two additional assumptions about our manifolds.

Second countable: There is a countable neighborhood basis.This is to avoid manifolds like an uncountable discrete set.

Good cover: There is a cover by Euclidean opens such that all finite inter-sections are either empty or again Euclidean.

This can be acheived in a smooth manifold by picking a metric and usingconvex neighborhoods.

In this lecture we will state a case of the Poincaré duality theorem andstudy some consequences of it, especially for compact 2-manifolds. This wholelecture will be happening with coefficients in F2.

Theorem 30.2. Let M be a compact manifold of dimension n. There existsa unique class [M] ∈Hn(M), called the fundamental class, such that for everyp, q with p + q = n the pairing

Hp(M)⊗Hq(M) ∪Ð→Hn(M)⟨−,[M]⟩ÐÐÐÐ→ F2

is perfect.

This means that the adjoint map

Hp(M)→ Hom(Hq(M),F2)

is an isomorphism. Since cohomology vanishes in negative dimensions, onething this implies is that Hp(M) = 0 for p > n. Since M is compact, π0(M) isfinite, and

Hn(M) = Hom(H0(M),F2) = Hom(Map(π0(M),F2),F2) = F2[π0(M)] .

30. SURFACES AND NONDEGENERATE SYMMETRIC BILINEARFORMS 89

A vector space V admitting a perfect pairing V ⊗W → F2 is necessarily finitedimensional; so Hp(M) is in fact finite-dimensional for all p.

Combining this pairing with the universal coefficient theorem, we get iso-morphisms

Hn−p(M) ≅Ð→ Hom(Hp(M),F2)≅←ÐHp(M) .

The homology and cohomology classes corresponding to each other under thisisomorphism are said to be “Poincaré dual.”

Using these isomorphisms, the cup product pairing can be rewritten as ahomology pairing:

Hp(M)⊗Hq(M) ⋔ //

≅

Hn−p−q(M)

≅

Hn−p(M)⊗Hn−q(M) ∪ // H2n−p−q(M) .

This is the intersection pairing. Here’s how to think of this. Take homologyclasses α ∈ Hp(M) and β ∈ Hq(M) and represent them (if possible!) as theimage of the fundamental classes of submanifolds of M , of dimensions p andq. Move them if necessary to make them intersect “transversely.” Then theirintersection will be a submanifold of dimension n − p − q, and it will representthe homology class α ⋔ β.

This relationship between the cup product and the intersection pairing isthe source of the symbol for the cup product.

Example 30.3. Let M = T 2 = S1 × S1. We know that

H1(M) = F2⟨α,β⟩

and α2 = β2 = 0, while αβ = βα generates H2(M). The Poincaré duals of theseclasses are represented by cycles wrapping around one or the other of the twofactor circles. They can be made to intersect in a single point. This reflectsthe fact that

⟨α ∪ β, [M]⟩ = 1 .

Similarly, the fact that α2 = 0 reflects the fact that its Poincaré dual cycle canbe moved so as not to intersect itself.

This example exhibits a particularly interesting fragment of the statementof Poincaré duality: In an even dimensional manifold – say n = 2k – the cupproduct pairing gives us a nondegenerate symmetric bilinear form on Hk(M).As indicated above, this can equally well be considered a bilinear form onHk(M), and it is then to be thought of as describing the number of points(mod 2) two k-cycles intersect in, when put in general position relative to oneanother. It’s called the intersection form. We’ll denote it

α ⋅ β = ⟨α ∪ β, [M]⟩ .


Example 30.4. In terms of the basis α,β, the intersection form has matrix

[ 0 11 0

] .

This is a “hyperbolic form.”

Let’s discuss finite dimensional nondegenerate symmetric bilinear formsover F2 in general. A form on V restricts to a form on any subspace W ⊆ V ,but the restricted form may be degenerate. Any subspace has an orthogonalcomplement

W ⊥ = v ∈ V ∶ v ⋅w = 0 for all w ∈W .

Lemma 30.5. The restriction of a nondegenerate bilinear form on V to asubspace W is nondegenerate exactly when W ∩W ⊥ = 0. In that case W ⊥ isalso nondegenerate, and the splitting

V ≅W ⊕W ⊥

respects the forms.

Using this easy lemma, we may inductively decompose a general (finitedimensional) symmetric bilinear form. First, if there is a vector v ∈ V suchthat v ⋅ v = 1, then it generates a nondegenerate subspace and

V = ⟨v⟩⊕ ⟨v⟩⊥ .

Continuing to split off one-dimensional subspaces brings us to the situation ofa nondegenerate symmetric bilinear form such that v ⋅ v = 0 for every vector.Unless V = 0 we can pick a nonzero vector. Since the form is nondegenerate,we may find another vector w such that v ⋅w = 1. The two together generate a2-dimensional hyperbolic subspace. Split it off and continue. We conclude:

Proposition 30.6. Any finite dimensional nondegenerate symmetric bilin-ear form splits as an orthogonal direct sum of forms with matrices [1] and

[ 0 11 0

].

Let Bil be the set of isomorphism classes of finite dimensional nondegener-ate symmetric bilinear forms over F2. I’ve just given a classification of thesethings. This is a commutative monoid under orthogonal direct sum. It can beregarded as the set of nonsingular symmetric matrices modulo the equivalencerelation of “similarity”: Two matrices M and N are similar if N = AMAT forsome nonsingular A.

Claim 30.7. ⎡⎢⎢⎢⎢⎢⎣

11

1

⎤⎥⎥⎥⎥⎥⎦∼⎡⎢⎢⎢⎢⎢⎣

11

1

⎤⎥⎥⎥⎥⎥⎦

30. SURFACES AND NONDEGENERATE SYMMETRIC BILINEARFORMS 91

Proof. This is the same thing as saying that⎛⎜⎝

11

1

⎞⎟⎠= AAT for some non-

singular A. Let A =⎡⎢⎢⎢⎢⎢⎣

1 1 11 0 10 1 1

⎤⎥⎥⎥⎥⎥⎦.

It’s easy to see that there are no further relations; Bil is the commutativemonoid with two generators I and H, subject to the relation I +H = 3I.

Let’s go back to topology. Let n = 2, k = 1 (so that 2k = n). Then you get anintersection pairing on H1(M). Consider RP2. We know that H1(RP2) = F2.This must be the form we labelled I. This says that anytime you have anontrivial cycle on a projective plane, there’s nothing I can do to remove itsself interesections. You can see this. The projective plane is a Möbius bandwith a disk sown on along the boundary. The waist of the Möbius band servesas a generating cycle. The observation is that if this cycle is moved to intersectitself tranversely, it must intersect itself an odd number of times.

We can produce new surfaces from old by a process of “addition.” Giventwo connected surfaces Σ1 and Σ2, cut a disk out of each one and sew themtogether along the resulting circles. This is the connected sum Σ1#Σ2.

Proposition 30.8. There is an isomorphism

H1(Σ1#Σ2) ≅H1(Σ1)⊕H1(Σ2)

compatible with the intersection forms.

Proof. Let’s compute the cohomology of Σ1#Σ2 using Mayer-Vietoris. The twodimensional cohomology of Σi −D2) vanishes because the punctured surfaceretracts onto its 1-skeleton. The relevant fragment is

0→H1(Σ1#Σ2)→H1(Σ1−D2)⊕H1(Σ2−D2)→H1(S1) δÐ→H2(Σ1#Σ2)→ 0 .

The boundary map must be an isomorphism, because the connected sum is acompact connected surface so has nontrivial H2.

The classification of surfaces may now be summarized as folows:

Theorem 30.9. Formation of the intersection bilinear form gives isomorphismof commutative monoids Surf → Bil.

This is a kind of model result of algebraic topology! – a complete algebraicclassification of a class of geometric objects. The oriented surfaces correspondto the bilinear forms of type gH; g is the genus. But it’s a little strange. Wemust have a relation corresponding to H ⊕ I = 3I, namely

T 2#RP2 ≅ (RP2)#3 .

The two-fold connected sum RP2#RP2 is the Klein bottle K. In fact, moregenerally


Claim 30.10. If Σ is a nonoriented surface then Σ#T 2 ≅ Σ#K.

There’s more to be said about this. Away from characteristic 2, symmetricbilinear forms and quadratic forms are interchangeable. But over F2 you canask for a quadratic form q such that

q(x + y) = q(x) + q(y) + x ⋅ y .

This is a “quadratic refinement” of the symmetric bilinear form. Of course itimplies that x ⋅x = 0 for all x, so this will correspond to some further structureon an oriented surface. This structure is a “framing,” a trivialization of thenormal bundle of an embedding into a high dimensional Euclidean space. Thereare then further invariants of this framing; this is the story of the Kervaireinvariant.

31 A plethora of products

Recall that we have the Kronecker pairing

⟨−,−⟩ ∶Hp(X ∶ R)⊗Hp(X ∶ R)→ R .

It’s obviously not “natural,” because Hp is contravariant while homology iscovariant. But given f ∶ X → Y , b ∈ Hp(Y ), and x ∈ Hp(X), we can ask: Howdoes ⟨f∗b, x⟩ relate to ⟨b, f∗x⟩?

Claim 31.1. ⟨f∗b, x⟩ = ⟨b, f∗x⟩.

Proof. This is easy! I find it useful to write out diagrams of where things are.We’re going to work on the chain level.

Hom(Sp(Y ),R)⊗ Sp(X)1⊗f∗ //

f∗⊗1

Hom(Sp(Y ),R)⊗ Sp(Y )

⟨−,−⟩

Hom(Sp(X),R)⊗ Sp(X)⟨−,−⟩ // R

We want this diagram to commute. Suppose [β] = b and [ξ] = x. Then fromthe to left, going to the right and then down gives

β ⊗ ξ ↦ β ⊗ f∗(ξ)↦ β(f∗ξ) .

The other way gives

β ⊗ ξ ↦ f∗(β)⊗ ξ = (β f∗)⊗ ξ ↦ (β f∗)(ξ) .

This is exactly β(f∗ξ).

31. A PLETHORA OF PRODUCTS 93

There’s actually another product around:

µ ∶H(C∗)⊗H(D∗)→H(C∗ ⊗D∗)

given by [c]⊗ [d]↦ [c⊗ d]. I used it to pass from the chain level computationwe did to the homology statement.

We also have the two cross products:

× ∶Hp(X)⊗Hq(Y )→Hp+q(X × Y )

and× ∶Hp(X)⊗Hq(Y )→Hp+q(X × Y ) .

You should think of this as fishy because both maps are in the same direc-tion. This is OK because we used different things to make these constructions:the chain-level cross product (or Eilenberg-Zilber map) for homology and theAlexander-Whitney map for cohomology. Still, they’re related:

Lemma 31.2. Let a ∈Hp(X), b ∈Hq(Y ), x ∈Hp(X), y ∈Hq(Y ). Then:

⟨a × b, x × y⟩ = (−1)∣x∣⋅∣b∣⟨a, x⟩⟨b, y⟩

Proof. Look at the chain-level cross product and the Alexander-Whitney map

× ∶ S∗(X)⊗ S∗(Y ) S∗(X × Y ) ∶ α

Both of them are the identity in dimension 0, and both sides are projectiveresolutions with respect to the models (∆p,∆q); so by acyclic models they arenatural chain homotopy inverses.

Say [f] = a, [g] = b, [ξ] = x, [η] = y. Write fg for the composite

Sp(X)⊗ Sq(Y ) ×Ð→ Sp+q(X × Y ) f⊗gÐÐ→ R⊗R → R .

Then:

(f × g)(ξ × η) = (fg)α(ξ × η) ∼ (fg)(ξ ⊗ η) = (−q)pqf(ξ)g(η) .

We can use this to prove a restricted form of the Künneth theorem incohomology.

Theorem 31.3. Let R be a PID. Assume that Hp(X) is a finitely generatedfree R-module for all p. Then

× ∶H∗(X;R)⊗R H∗(Y ;R)→H∗(X × Y ;R)

is an isomorphism.


Proof. Write M∨ for the linear dual of an R-module M . By our assumptionabout Hp(X), the map

Hp(X)∨ ⊗Hq(Y )∨ → (Hp(X)⊗Hq(Y ))∨ ,

sending f ⊗ g to (x⊗ y ↦ (−1)pqf(x)g(y)), is an isomorphism. The homologyKünneth theorem guarantees that the bottom map in the following diagram isan isomorphism.

⊕p+q=nHp(X)⊗Hq(Y ) × //

≅

Hn(X × Y )

≅

⊕p+q=nHp(X)∨ ⊗Hq(Y )∨ ≅ // (⊕p+q=nHp(X)⊗Hq(Y ))∨ Hn(X × Y )∨≅ oo

Commutativity of this diagram is exactly the content of Lemma 31.2.

We saw before that × is an algebra map, so under the conditions of the theo-rem it is an isomorphism of algebras. You do need some finiteness assumption,even if you are working over a field. For example let T be an infinite set, re-garded as a space with the discrete topology. Then H0(T ;R) = Map(T,R).But

Map(T,R)⊗Map(T,R)→Map(T × T,R)

sending f ⊗ g to (s, t) → f(s)g(t) is not surjective; the characteristic functionof the diagonal is not in the image, for example.

There are more products around. For example, there is a map

Hp(Y )⊗Hq(X,A)→Hp+q(Y ×X,Y ×A) .

Constructing this is on your homework. Suppose Y =X. Then I get

∪ ∶H∗(X)⊗H∗(X,A)→H∗(X ×X,X ×A) ∆∗

Ð→H∗(X,A)

where ∆ ∶ (X,A) → (X ×X,X × A) is the “relative diagonal.” This “relativecup product” makes H∗(X,A) into a module over the graded algebra H∗(X).The relative cohomology is not a ring – it doesn’t have a unit, for example –but it is a module. And the long exact sequence of the pair is a sequence ofH∗(X)-modules.

I want to introduce you to one more product, which will enter into ourexpression of Poincaré duality. This is the cap product. What can I do withSp(X)⊗ Sn(X)? Well, I can form the composite:

Sp(X)⊗ Sn(X)1×(αX,X∆∗)ÐÐÐÐÐÐÐÐ→ Sp(X)⊗ Sp(X)⊗ Sn−p(X)

⟨−,−⟩⊗1ÐÐÐÐ→ Sn−p(X)

Using our explicit formula for α, we can write:

∩ ∶ β ⊗ σ ↦ β ⊗ (σ αp)⊗ (σ ωq)↦ (β(σ αp)) (σ ωq)

32. CAP PRODUCT AND “ČECH” COHOMOLOGY 95

We are evaluating the cochain on part of the chain, leaving a lower dimensionalchain left over.

This composite is a chain map, and so induces a map in homology:

∩ ∶Hp(X)⊗Hn(X)→Hn−p(X) .

Here are some properties of the cap product.

Lemma 31.4. (α ∪ β) ∩ x = α ∩ (β ∩ x) and 1 ∩ x = x.

Proof. Easy to check from the definition.

This makes H∗(X) into a module over H∗(X). These are not hard thingsto check. There’s a lot of structure, and the fact that H∗(X) forms an algebrais a good thing. Notice how the dimensions work. Long ago a bad choicewas made: we should index cohomology with negative numbers, so that thegrading in ∩ ∶ Hp(X) ⊗Hn(X) → Hn−p(X) makes sense. A cochain complexwith positive grading is the same as a chain complex with negative grading.

There are also two slant products. Maybe we won’t talk about them. Wewill check a few things about cap products, and then we’ll get into the machin-ery of Poincaré duality.

32 Cap product and “Čech” cohomology

We have a few more things to say about the cap product, and will then use itto give a statement of Poincaré duality.

Let R be a commutative ring of coefficients. The cap product

∩ ∶Hp(X)⊗Hn(X)→Hq(X) , p + q = n ,

comes from a chain level composite

Sp(X)⊗ Sn(X) 1⊗αÐÐ→ Sp(X)⊗ Sp(X)⊗ Sq(X)⟨−,−⟩⊗1ÐÐÐÐ→ R⊗ Sq(X) ≅ Sq(X) .

Using the Alexander-Whitney map this can be written as follows.

∩ ∶ β ⊗ σ ↦ β ⊗ (σ αp)⊗ (σ ωq)↦ (β(σ αp)) (σ ωq)

Proposition 32.1. The cap product enjoys the following properties.(1) (a ∪ b) ∩ x = a ∩ (b ∩ x) and 1 ∩ x = x: H∗(X) is a module for H∗(X).(2) Given a map f ∶X → Y , b ∈Hp(Y ), and x ∈Hn(X),

f∗(f∗(b) ∩ x) = b ∩ f∗(x) .

(3) Let ε ∶H∗(X)→ R be the augmentation. Then

ε(b ∩ x) = ⟨b, x⟩ .

(4) Cap and cup are adjoint:

⟨a ∩ b, x⟩ = ⟨a, b ∩ x⟩


Proof. (1) We proved this in the last lecture.(2) Let β be a cocycle representing b, and σ an n-simplex in X. Then

f∗(f∗(β) ∩ σ) = f∗((f∗(β)(σ αp)) ⋅ (σ ωq))= f∗(β(f σ αp) ⋅ (σ ω))= β(f σ αp) ⋅ f∗(σ ωq)= β(f σ αp) ⋅ (f σ ωq)= β ∩ f∗(σ)

This formula goes by many names: the “projection formula,” or “Frobeniusreciprocity.”(3) We get zero unless p = n. Again let σ ∈ Sinn(X), and compute:

ε(β ∩ σ) = ε(β(σ) ⋅ c0σ(n)) = β(σ)ε(c0σ(n)) = β(σ) = ⟨β,σ⟩ .

Here now is a statement of Poincaré duality. It deals with the homologicalstructure of compact topological manifolds. The statement will use the notionof orientability, which we will have more to say about. Since M is locallyEuclidean, excision guarantees that for any p ∈M there is an isomorphism

ja ∶Hn(M,M − a;R)←Hn(Rn,Rn − 0;R) = R .

An “orientation” is a choice of generators for these free R-modules that variescontinuously with a. With coefficients in F2, any manifold is uniquely oriented.

Theorem 32.2 (Poincaré duality). Let M be a topological n-manifold that iscompact and oriented with respect to a PID R. Then there is a unique class[M] ∈Hn(M ;R) that restricts to the orientation class in Hn(M,M − a;R) forevery a ∈M . It has the property that

− ∩ [M] ∶Hp(M ;R)→Hq(M ;R) , p + q = n ,

is an isomorphism for all p.

You might want to go back to Lecture 25 and verify thatRP3×RP3 satisfiesthis theorem.

Our proof of Poincaré duality will be by induction. In order to make theinduction go we will prove a substantially more general theorem, one thatinvolves relative homology and cohomology. So we begin by understandinghow the cup product behaves in relative homology.

32. CAP PRODUCT AND “ČECH” COHOMOLOGY 97

Suppose A ⊆X is a subspace. We have:

0

0

Sp(X)⊗ Sn(A)

1⊗i∗

i∗⊗1 // Sp(A)⊗ Sn(A) ∩ // Sq(A)

i∗

Sp(X)⊗ Sn(X) ∩ //

Sq(X)

Sp(X)⊗ Sn(X,A)

// Sq(X,A)

0 0

The left sequence is exact because 0 → Sn(A) → Sn(X) → Sn(X,A) → 0 splitsand tensoring with Sp(X) (which is not free!) therefore leaves it exact. Thesolid arrow diagram commutes precisely by the chain-level projection formula.There is therefore a uniquely defined map on cokernels.

This chain map yields the relative cap product

∩ ∶Hp(X)⊗Hn(X,A)→Hq(X,A)

It renders H∗(X,A) a module for H∗(X).I want to come back to an old question, about the significance of relative

homology. Suppose that K ⊂X is a subspace, and consider the relative homol-ogy H∗(X,X −K). Since the complement of X −K in X is K, these groupsshould be regarded as giving information about K. If I enlarge K, I makeX − K smaller, K ⊆ L ⊆ induces H∗(X,X − L) → H∗(X − K): the relativehomology is contravariant in the variable K (regarded as an element of theposet of subspaces of X, at least).

Excision gives insight into how H∗(X,X − K) depends on K. SupposeK ⊆ U ⊆X such that K ⊆ Int(U). To simplify things, let’s just suppose that Kis closed and U is open. Then X−U is closed, X−K is open, and X−U ⊆X−K,so excision asserts that the inclusion map

H∗(U,U −K)→H∗(X,X −K)

is an isomorphism.The cap product puts some structure on H∗(X,X − K): it’s a module

over H∗(X). But we can do better! We just decided that H∗(X,X −K) =H∗(U,U −K), so the H∗(X) action factors through an action by H(U), for anyopen set U containing K. How does this refined action change when I decreaseU?


Lemma 32.3. Let U ⊇ V ⊇K. Then:

Hp(U)⊗Hn(X,X −K)

i∗⊗1

∩

))Hq(X,X −K)

Hp(V )⊗Hn(X,X −K)

∩55

commutes.

Proof. Hint: use projection formula again.

Let UK be the set of open neighborhoods of K in X. It is partially orderedby reverse inclusion. This poset is directed, since an intersection of two opensis open. By the lemma, Hp ∶ UK →Ab is a directed system.

Definition 32.4. The Čech cohomology of K is

Hp(K) ∶= limÐ→U∈UK

Hp(U) .

I apologize for this bad notation; its apparent dependence on the way K issitting in X is not recorded.

Since tensor product commutes with direct limits, we now get a cap productpairing

∩ ∶ Hp(K)⊗Hn(X,X −K)→Hq(X,X −K)satifying the expected properties. This is the best you can do. It’s the naturalstructure that this relative homology has: H∗(X,X − K) is a module overH∗(K).

There are compatible restricton mapsHp(U)→Hp(K), so there is a naturalmap

H∗(K)→Hp(K) .This map is often an isomorphism. Suppose K ⊆ X satisfies the condition (a“regular neighborhood” condition) that for every open U ⊇ K, there exists anopen V such that U ⊇ V ⊇ K such that K V is a homotopy equivalence (oractually just a homology isomorphism).

Lemma 32.5. Under these conditions, H∗(K)→H∗(K) is an isomorphism.

Proof. We will check that the map to Hp(K) satisfies the conditions we estab-lished in Lecture 23 to be a direct limit.

So let x ∈ Hp(K). Let U be a neighborood of K in X such that Hp(U) →Hp(K) is an isomorphism. Then indeed x is in the image of Hp(U).

Then let U be a neighborhood of K and let x ∈ Hp(U) restrict to 0 inHp(K). Let V be a sub-neighborood such that Hp(V )→Hp(K) is an isomor-phism. Then x restricts to 0 in Hp(V ).

33. H∗ AS A COHOMOLOGY THEORY, AND THE FULLY RELATIVE∩ PRODUCT 99

On the other hand, here’s an example that distinguishes H∗ from H∗. Thisis a famous example – the “topologist’s sine curve.” The topologist’s sine curveis the subspace of R2 defined as follows. It is union of two subsets, A and B. Ais the graph of sin(π/x) where 0 < x ≤ 1. B is a continuous curve from (0,−1)to (1,0) and not meeting A. This is a counterexample for a lot of things; you’veprobably seen it in 18.901.

What is the singular homology of the topologist sine curve? Use Mayer-Vietoris! I can choose V to be some connected portion of the continuous curvefrom (0,−1) to (1,0), and U to contain the rest of the space in a way thatintersects V in two open intervals. Then V is contractible, and U is made upof two contractible connected components. (This space is not locally connected,and one of these path components is not closed.)

The Mayer-Vietoris sequence looks like

0→H1(X) ∂Ð→H0(U ∩ V )→H0(U)⊕H0(V )→H0(X)→ 0 .

The two path components of U ∩ V do not become connected in U , so ∂ = 0

and we find that ε ∶H∗(X) ≅Ð→H∗(∗) and hence H∗(X) ≅H∗(∗).How about H∗? Let X ⊂ U be an open neighborhood. The interval is

contained in some ε-neighborhood that’s contained in U . This implies thatthere exists a neighborhood X ⊆ V ⊆ U such that V ∼ S1. This implies that

limÐ→U∈UX

H∗(U) ≅H∗(S1)

by an argument we will detail later. So H∗(X) ≠H∗(X).

33 H∗ as a cohomology theory, and the fully relative ∩product

Let X be any space, and let K ⊆ X be a closed subspace. We’ve defined theČech cohomology of K as the direct limit of H∗(U) as U ranges over the posetUK of open neighborhoods of K. This often coincides with H∗(K) but will notbe the same in general. Nevertheless it behaves like a cohomology theory. Toexpand on this claim, we should begin by defining a relative version.

Suppose L ⊆ K is a pair of closed subsets of a space X. Let (U,V ) be a“neighborhood pair” for (K,L):

L ⊆ K⊆ ⊆

V ⊆ U

These again form a directed set UK,L, with partial order given by reverse in-clusion of pairs. Then define

Hp(K,L) = limÐ→(U,V )∈UK,L

Hp(U,V ) .


Theorem 33.1. Let (K,L) be a closed pair in X. There is a long exactsequence

⋯→ Hp(K,L)→ Hp(K)→ Hp(L) δÐ→ Hp+1(K,L)→ ⋯

that is natural in the pair.

Also, a form of excision holds:

Theorem 33.2 (Excision). Suppose A,B ⊆ X are closed. Then the inclusioninduces isomorphisms

Hp(A ∪B,A) ≅Ð→ Hp(B,A ∩B) .

So Čech cohomology is better suited to closed subsets than singular coho-mology is.

Čech cohomology appeared as the natural algebra acting on H∗(X,X −K),where K is a closed subspace of X:

∩ ∶ Hp(K)⊗Hn(X,X −K)→Hq(X,X −K) , p + q = n .

If we fix xK ∈Hn(X,X −K), then capping with xK gives a map

∩xK ∶ Hp(K)→Hq(X,X −K) , p + q = n .

We will be very interested in showing that this map is an isomorphism undercertain conditions. This is a kind of duality result, comparing cohomology andrelative homology and reversing the dimensions. We’ll try to show that such amap is an isomorphism by embedding it in a map of long exact sequences andusing the five-lemma.

For a start, let’s think about how these maps vary as I change K. So let Lbe a closed subset of K, so X −K ⊆X −L and I get a “restriction map”

i∗ ∶Hn(X,X −K)→Hn(X,X −L) .

Define xL as the image of xK . The diagram

Hp(K) //

−∩xK

Hp(L)

−∩xL

Hq(X,X −K) // Hq(X,X −L)

commutes by the projection formula. This embeds in a bigger diagram:

Theorem 33.3. There is a “fully relative” cap product

∩ ∶ Hp(K,L)⊗Hn(X,X −K)→Hq(X −L,X −K) , p + q = n ,


such that for any xK ∈Hn(X,X −K) the diagram

⋯ // Hp(K,L) //

−∩xK

Hp(K) //

−∩xK

Hp(L) δ //

−∩xL

Hp+1(K,L) //

−∩xK

⋯

⋯ // Hq(X −L,X −K) // Hq(X,X −K) // Hq(X,X −L) ∂ // Hq−1(X −L,X −K) // ⋯

commutes. Here xL is xK restricted to Hn(X,X −L).

What I have to do is define a cap product of the following form (bottomrow):

Hp(K)⊗Hn(X,X −K) ∩ // Hq(X,X −K)

Hp(K,L)⊗Hn(X,X −K)

OO

∩ // Hq(X −L,X −K)

(where p + q = n)Our map Hp(K) ⊗ Hn(X,X −K) → Hq(X,X −K) came from Sp(U) ⊗

Sn(U,U−K)→ Sq(U,U−K) where U ⊇K, defined via β⊗σ ↦ β(σαp)⋅(σωq).I’m hoping to get:

Sp(U)⊗ Sn(U,U −K) // Sq(U,U −K)

Sp(U,V )⊗ Sn(U −L)/Sn(U −K) //

OO

Sq(U −L)/Sq(U −K)

OO

where again we have inclusions (U,V open and K,L closed):

K _

L _

_?oo

U V_?oo

The bottom map Sp(U,V ) ⊗ Sn(U − L)/Sn(U −K) → Sq(U − L)/Sq(U −K)makes sense. We can evaluate a cochain that kills everything on V . This meansthat we can add in Sn(V ) to get Sp(U,V )⊗(Sn(U −L)+Sn(V ))/Sn(U −K)→Sq(U −L)/Sq(U −K) by sending β⊗τ ↦ 0 where τ ∶ ∆n → V . This means thatthe diagram:


Sp(U,V )⊗ (Sn(U −L) + Sn(V ))/Sn(U −K) //

OO


OO

commutes. It’s not that far off from where we want to go.


Now, (U − L) ∪ V = U . I have this covering of U by two open sets. InSn(U−L)+Sn(V ) we’re taking the sum of n-chains. We have a map S∗(U−L)+S∗(V )→ S∗(U). We have already worked through this – the locality principle!This tells us that S∗(U−L)+S∗(V )→ S∗(U) is a homotopy equivalence. Hencewe can extend our diagram:


Sp(U,V )⊗ (Sn(U −L) + Sn(V ))/Sn(U −K) //

≃

OO


OO

Sp(U,V )⊗ Sn(U)/Sn(U −K)

We want the homology of Sn(U)/Sn(U −K) to approximate Hn(X,X −K).

Claim 33.4. There is an isomorphism Hn(S∗(U)/S∗(U − K)) = Hn(U,U −K)→Hn(X,X −K).

Proof. This is exactly excision! Remember our recasting of excision in theprevious lecture.

This means that what we’ve constructed really is what we want! We nowhave our large lexseq:

⋯ // Hp(K,L) //

−∩xK

Hp(K) //

−∩xk

Hp(L) δ //

−∩xL

Hp+1(K,L) //

−∩xK

⋯

⋯ // Hq(X −L,X −K) // Hq(X,X −K) // Hq(X,X −L) ∂ // Hq−1(X −L,X −K) // ⋯

⋯ // Hq(U −L,U −K) //

≅, five-lemma

OO

Hq(U,U −K) //

≅, locality

OO

Hq(U,U −L) //

≅, locality

OO

⋯

As desired.The diagram:

Hp(L) δ //

−∩xL

Hp+q(K,L)

−∩xK

Hq(X,X −L) ∂ // Hq−1(X −L,X −K)

says that:(δb) ∩ xk = ∂(b ∩ xL)

It’s rather wonderful! You have a decreasing sequence below and an increasingone above.


I want to reformulate all of this in a more useful fashion, from Mayer-Vietoris. We had two different proofs, one from locality, and another one thatwe’ll remind you of:

⋯ // An //

Bn //

≅

Cn //

An−1//

⋯

⋯ // A′n

// B′n

// C ′n

// A′n−1

// ⋯

then you get a lexseq:

⋯→ Cn+1 → C ′n+1 ⊕An → A′

n

∂Ð→ Cn → ⋯

You can use this to prove Mayer-Vietoris – I will do this in a special case. (Thisis exactly what I did in a homework assignment1!) We have a ladder of lexseqs:

⋯ // Hq(X,X −A ∪B) //

Hq(X,X −A) //

Hq−1(X −A,X −A ∪B) //

≅, excision

⋯

⋯ // Hq(X,X −B) // Hq(X,X −A ∩B) // Hq−1(X −A ∩B,X −B) // ⋯

This means that (using the lexseq of the ladder) you have a lexseq:

⋯→Hq(X,X−A∪B)→Hq(X,X−A)⊕Hq(X,X−B)→Hq(X,X−A∩B)→Hq−1(X,X−A∪B)→ ⋯

This can be used to give a lexseq for Čech cohomology:

⋯→ Hp(A ∪B)→ Hp(A)⊕ Hp(B)→ Hp(A ∩B)→ Hp+q(A ∪B)→ ⋯

so that we’re going to get a commutative Mayer-Vietoris ladder:

Theorem 33.5. There’s a “Mayer-Vietoris” ladder:

→ Hp(A ∪B) //

−∩xA∪B

Hp(A)⊕ Hp(B) //

(−∩xA)⊕−∩xB

Hp(A ∩B) //

Hp+q(A ∪B) //

⋯

→Hq(X,X −A ∪B) // Hq(X,X −A)⊕Hq(X,X −B) // Hq(X,X −A ∩B) // Hq−1(X,X −A ∪B) // ⋯1Suppose A ⊆ X is a subspace of X. Then there is a lexseq in reduced homology ⋯ →

Hn(A) → Hn(X) → Hn(X,A) → Hn−1(A) → ⋯ that can be obtained by using the lexseq inhomology of the sexseq 0→ S∗(A)→ S∗(X)→ S∗(X,A)→ 0.

Now suppose X = A ∪B. Consider the ladder:

⋯ // Hn+1(A,A ∩B) //

Hn(A ∩B) //

Hn(A) //

Hn(A,A ∩B) //

⋯

⋯ // Hn+1(X,B) // Hn(B) // Hn(X) // Hn(X,B) // ⋯

The first and fourth maps as shown are isomorphisms because of excision. The lexseq fromthe ladder (see above) therefore yields the Mayer-Vietoris sequence ⋯ → Hn(A ∩ B) →Hn(B)⊕ Hn(A)→ Hn(X)→ Hn−1(A ∩B)→ ⋯.


where I have four cohomology classes xA∪B , xA, xB , xA∩B that commute in:

Hn(X,X −A)

))Hn(X,X −A ∩B)

55

))

Hn(X,X −A ∩B)

Hn(X,X −B)

55

This is the most complicated blackboard for the rest of the course. Alsoxymatrix is not compiling properly because the diagram is too big!

34 H∗ as a cohomology theory

Office hours: today, Hood in 4-390 from 1:30 to 3:30 and Miller in 4-478 from1-3 on Tuesday. Note that pset 6 is due Wednesday. Also, Wednesday we’llhave a lightning review of π1 and covering spaces.

We’re coming to the end of the course, and there are going to be oral exams.I have some questions that I’d like to ask you. They won’t be super advanced,detailed questions – they’ll be basic things. I’ll post a list of examples ofquestions. I won’t select questions from that list, that’s cruel and isn’t thepoint. The oral will be 40 minutes. It’ll be fun – better than a written exam.It’s much better than grading a written exam!

PLEASE DRAW A PICTURE WHEN READING THIS IF YOU DIDN’TCOME TO CLASS!

Cofinality

Let I be a directed set. Let A ∶ I → Ab be a functor. If I have a functorf ∶ K → I, then I get Af ∶ K →Ab, i.e., (Af)j = Af(j).

I can form limÐ→KAf and limÐ→I A. I claim you have a map limÐ→KAf → limÐ→I A.All I have to do is the following:

limÐ→J Af// limÐ→I A

Af(j)

inj

OO

So I have to give you maps Af(j) → limÐ→I A for various j. I know what to do,because I have inf(j) ∶ Af(j) → limÐ→I A. Are they compatible when I changej? Suppose I have j′ ≤ j. Then I get a map f(j′) → f(j), so I have a map

34. H∗ AS A COHOMOLOGY THEORY 105

Af(j′) → Af(j), and thus the maps are compatible. Hence I get:

limÐ→J Af// limÐ→I A

(Af)j = Af(j)

inf(j)88

inj

OO

Example 34.1. Suppose K ⊇ L be closed, then I get a map H∗(K) →H∗(L). Is this a homomorphism? Well, H∗(K) = limÐ→U∈UK H

∗(U) and H∗(L) =limÐ→V ∈ULH

∗(V ). This is an example of a I and K that I care about. Well,

UK ⊆ UL, and thus I get a map H∗(K)→ H∗(L), which is what I wanted.I can do something for relative cohomology. Suppose:

K _

L _

? _oo

K ′ L′? _oo

I get a homomorphism H∗(K,L)→ H∗(K ′, L′) because I have UK,L → UK′,L′ .

This isn’t exactly what we need:

Question 34.2. When does f ∶ K → I induce an isomorphism limÐ→J Af →limÐ→I A?

This is a lot like taking a sequence and a subsequence and asking whenthey have the same limit. There’s a cofinality condition in analysis, that has asimilar expression here.

Definition 34.3. f ∶ K → I is cofinal if for all i ∈ I, there exists j ∈ K suchthat i ≤ f(j).

Example 34.4. If f is surjective.

Lemma 34.5. If f is cofinal, then limÐ→J Af → limÐ→I A is an isomorphism.

Proof. Check that Af(j) → limÐ→I A satisfies the necessary and sufficient con-ditions:

1. For all a ∈ limÐ→I A, there exists j and aj ∈ Af(j) such that aj ↦ a. Weknow that there exists some i and ai ∈ A such that ai ↦ a. Pick j suchthat f(j) ≥ i, so we get a map ai → af(j), and by compatibility, we getaf(j) ↦ a.

2. The other condition is also just as easy.


This is a very convenient condition.

Example 34.6. I had a perverse way of constructingQ by using the divisibilitydirected system. A much simpler (linear!) directed system is Z

2Ð→ Z3Ð→ Z

4Ð→Z → ⋯. This has the same colimit as the divisibility directed system becausen∣n!, so we have a cofinal map between directed systems.

How about the direct limits in the Čech cohomology case?

Example 34.7. Do I have a map H∗(K,L)→ H∗(K)? Suppose:

K _

L _

? _oo

U V?_oo

Then Hp(K,L) = limÐ→(U,V )∈UK,LHp(U,V ) and Hp(K) = limÐ→U∈UK H

p(U). I

have a map of directed sets UK,L → UK by sending (U,V ) ↦ U . I didn’t haveto use cofinality. I want a long exact sequence, though, and I’m going to dothis by saying that it’s a directed limit of a long exact sequence. I’m going tohave to have all of these various Čech cohomologies as being the directed limitover the same indexing set.

I’d really like to say that Hp(K) = limÐ→U∈UK Hp(U) ≅ limÐ→(U,V )∈UK,L

Hp(U).Thus I need to show that UK,L → UK where (U,V ) ↦ U is cofinal. This iseasy, because if U ∈ UK , just pick (U,U), i.e., UK,L → UK is cofinal. Howabout UK,L → UL by (U,V ) ↦ V ; is it cofinal? Yes! For V ∈ UL, pick (X,V )!This means that ⋯Hp−1(L) → Hp(K,L) → Hp(K) → Hp(L) → Hp+1(K,L) islimÐ→UK,L (⋯→Hp(U,V )→ ⋯), and hence exact.

How about excision? I need this to get to Mayer-Vietoris!

Lemma 34.8. Assume X is normal and A,B are closed subsets. Then Hp(A∪B,B)→ Hp(A,A ∩B) is an isomorphism.

Proof. Well, Hp(A ∪ B,B) is limÐ→ over UA∪B,B and Hp(A,A ∩ B) is limÐ→ overUA,A∩B . Let W ⊇ A and Y ⊇ B are neighborhoods. I claim that UA × UB →UA∪B,B sending (W,Y )↦ (W ∪Y,Y ) and UA×UB → UA,A∩B sending (W,Y )↦(W,W ∩ Y ) are cofinal.

If I give you (U,V ) ∈ UA∪B,B , define (W,V ) ∈ UA × UB where W = U andY = V , so UA ×UB → UA∪B,B is surjective, hence cofinal. The latter is trickier.Let U ⊇ A and V ⊇ A ∩B. Here’s where normality comes into play. SeparateB − V from A. Let T ⊇ B − V . Shit. Shit!

Maybe I’ll leave this to you. I’ll put this on the board on Wednesday.Anyway, I’ll use normality to show that UA ×UB → UA,A∩B is cofinal, and thusthis verifies excision – so you actually have excision.

35. FINISH OFF THE PROOF OF Hp EXCISION, TOPOLOGICALMANIFOLDS, FUNDAMENTAL CLASSES 107

35 Finish off the proof of Hp excision, topologicalmanifolds, fundamental classes

The end of the proof

Let’s finish off the proof from last time. Suppose A,B are closed in normal X.Excision for Hp:

limÐ→(W,Y )∈UA×UBHp(W ∪ Y,Y )

≅, ordinary excision //

≅, cofinality/surjectivity

limÐ→UA×UB Hp(W,W ∩ Y )

≅, cofinality, see below

limÐ→(U,V )∈UA∪B,B

Hp(U,V ) // limÐ→(U,V )∈UA,A∩BHp(U,V )

Hp(A ∪B,B) // Hp(A,A ∩B)

UA × UB → UA,A∩B is cofinal since: start with (U,V ) ⊇ (A,A ∩B). Usingnormality, separate B∩(X −V ) ⊆ T and A ⊆ S. TakeW = U ∩S and Y = V ∪T .Then A ⊆W ⊆ U and A ∩B ⊆W ∩ Y = S ∩ V ⊆ V .

This means that Hp satisfies excision, hence Mayer-Vietoris. Let’s put thisin the drawer for now.

Topological manifolds + Poincaré duality

yayyyyyyyyyyyyy finally

Fundamental class and orientation local system

Definition 35.1. A topological manifold is a Hausdorff space M such that forevery x ∈M , there exists a neighborhood U ∋ x that is homeomorphic to someEuclidean space Rn. It’s called an n-manifold if all U are homeomorphic toRn for the same n.

Example 35.2. Rn, duh. ∅ is an n-manifold for every n. The sphere Sn.The Grassmannian Grk(Rn), introduced in the beginning of the course. Idon’t know exactly what the dimension of this is, but you can figure it out.Also, Vk(Rn), and surfaces.

These things are the most interesting things to look at.

Warning 35.3. We assume the following.

1. There exists a countable basis.

2. There exists a good cover, i.e., all nonempty intersections are Euclideanas well (always true for differentiable manifolds because you can takegeodesic neighborhoods, and in particular for the manifolds we listedabove).


This is the context in which duality works.

Definition 35.4. Let X be any space, and let a ∈X. The local homology of Xat a is the homology H∗(X,X − a). We’re always working over a commutativering.

For example, Hq(Rn,Rn − 0) =⎧⎪⎪⎨⎪⎪⎩

free of rank 1 q = n0 q ≠ n

. This means that

local homology is picking out the characteristic feature of Euclidean space.

Therefore we also have Hq(M,M −a) =⎧⎪⎪⎨⎪⎪⎩

free of rank 1 q = n0 q ≠ n

for n-manifolds.

Notation 35.5. Let ja ∶ (M,∅)→ (M,M − a) be the inclusion.

Definition 35.6. A fundamental class forM (an n-manifold) is [M] ∈Hn(M)such that for every a ∈M , the image of [M] under ja,∗ ∶Hn(M)→Hn(M,M −a) is a generator of Hn(M,M − a).

This is somehow trying to say that this class [M] covers the whole manifold.

Example 35.7. When does a space have a fundamental class?

R2 RP2 T 2

R = Z no! no! yes! you did this for homeworkR = Z/2Z no! yes! yes!

Something about orientability and compactness seem to be involved.

What do we have?

Definition 35.8. oM =∐a∈M Hn(M,M −a) as a set. This has a map p ∶ oM →M .

Construction 35.9. This can be topologized in Euclidean neighborhoods.Let U ≅ Rn be an Euclidean neighborhood of a. I can always arrange sothat a corresponds to 0. We have the open disk sitting inside the closed disk:Dn ⊆ Dn ⊆Rn that corresponds to some open V ⊆ V ⊆ U . Let x ∈ V . I have adiagram:

Hn(M,M − V )

Hn(U,U − V )excision of M−U

≅oo

≅

Hn(Rn,Rn −Dn)

≅, homotopy equivalence

Hn(M,M − x) Hn(U,U − x)oo Hn(Rn,Rn − 0)

Hence Hn(M,M−V ) ≅Hn(M,M−x). Thus I can collect points in oM togetherwhen they come from the same class in Hn(M,M − V ), so they form “sheets”.

I have a map V × Hn(M,M − V ) → oM ∣V = p−1(V ) by sending (x, c) ↦(jx)∗(c) ∈ Hn(M,M − x), and this map is bijective (that’s what comes fromexcision). This LHS has a nice topology by letting Hn(M,M −V ) be discrete.I’m topologizing oM as the weakest topology these generate.

36. FUNDAMENTAL CLASS 109

“Have I been sufficiently obscure enough? This is not supposed to be acomplicated point”. This oM →M is called the orientation local system, and isa covering space.

Definition 35.10. A continuous map p ∶ E → B is a covering space if:

1. p−1(b) is discrete for all b ∈ B.

2. For every b there’s a neighborhood V and a map p−1(V ) → p−1(b) suchthat p−1(V ) ≅Ð→ V × p−1(b) is a homeomorphism.

That’s exactly the way we topologized oM . There’s more structure thoughbecause Hn(M,M − V ) is an R-module!

Definition 35.11. A local system (of R-modules) p ∶ E → B is a covering spacetogether with structure maps E ×B E ∶= (e, e′)∣pe = pe′ +Ð→ E and z ∶ B → Esuch that:

E ×B E+ //

##

E

R ×Eoo

B

??

B

z

OO

making p−1(b) a R-module.

We have Hn(M) jxÐ→ Hn(M,M − x), which gives a section of oM . If Ihave a covering space p ∶ E → B, a section is a continuous map s ∶ B → Esuch that ps = 1B . Write Γ(E) to be the set of sections. If E is a local

system, this is an R-module. Hence Hn(M) jxÐ→ Hn(M,M − x) gives a mapj ∶ Hn(M) → Γ(oM). This is pretty cool because it’s telling you about thishigh-dimensional homology of M into something “discrete”.

Theorem 35.12. If M is compact then j ∶ Hn(M) → Γ(oM) is an isomor-phism, and Hq(M) = 0 for q > n.

This is case of Poincaré duality actually because Γ(oM) is somewhat likezero-dimensional cohomology. If this is trivial, like it is for a torus, so if themanifold is connected, then Γ(oM) is just R.

36 Fundamental class

Note that if M is a compact manifold, then Hq(M ;R) = 0 for q ≫ 0, and if Ris a PID, then for all q, Hq(M ;R) is finitely-generated. This follows from:

Claim 36.1. Suppose X admits an open cover Uini=1 such that all intersec-tions are either empty or contractible (this is what you get for a good cover ona manifold). Then Hq(X;R) = 0 for q ≥ n, and if R is a PID, then for all q,Hq(X;R) is finitely-generated.


Proof. Induct. Certainly true for n = 1. Let Y = ⋃n−1i=1 Ui, then this statement

is true by induction – and similarly for Y ∩ Un. Now use Mayer-Vietoris.You have ⋯ → Hq(Y ∩ Un) → Hq(Y ) ⊕Hq(Un) → Hq(X) → Hq−1(Y ∩ Un) →⋯. When q = n − 1, Hq(Y ∩ Un) could be nonzero, and so you might getsomething nontrivial (???). Also, you’ll get a sexseq by unsplicing the lexseq:0→Hq(Y )⊕Hq(Un)/something→Hq(X)→ submodule of Hq−1(Y ∩Un)→ 0,where you use R being a PID to conclude that submodule of Hq−1(Y ∩Un) isfinitely generated.

Let M be an n-manifold. We had a map j ∶ Hn(M) → Γ(M ; oM). HereΓ(M ; oM) is the collection of compatible elements of Hn(M,M −x) for x ∈M .This map j ∶ Hn(M) → Γ(M ; oM) sends c ↦ (x ↦ jxc) where jx ∶ Hn(M,∅) →Hn(M,M − x). I want to make two refinements.

You can’t expect j to be surjective, except maybe when M is compact.Here’s why. Let c ∈ Zn(M). It’s a sum of simplices, and each simplex is com-pact, and so the union of the images is compact, and hence there’s a compactsubset K ⊆ M such that c ∈ Zn(K). Now if I take x /∈ K, then the mapHn(K) → Hn(M) splits as Hn(K) → Hn(M − x) → Hn(M). In the relativehomology, Hn(M,M − x), the map Hn(K) → Hn(M) → Hn(M,M − x) sendsc to zero.

Definition 36.2. Let σ be a section of p ∶ E → B (local system). Then thesupport of σ is defined as supp(σ) = x ∈ B∣σ(x) ≠ 0. The collection of allsections with compact support is Γc(B;E), and it’s a submodule of Γ(B;E).

The first refinement is that j ∶ Hn(M) → Γ(M ; oM) lands in Γc(M ; oM),because homology is compactly supported.

The second refinement seems a little artificial but is part of the inductiveprocess. Let A ⊆ M be closed. Then you have a restriction map Hn(M,M −A) jxÐ→ Hn(M,M − x) for x ∈ A. Thus you get a map j ∶ Hn(M,M − A) →Γc(A; oM ∣A), the latter of which we’ll just denote Γc(A; oM).

Theorem 36.3. The map j ∶Hn(M,M −A)→ Γc(A; oM ∣A) is an isomorphismand Hq(M,M −A) = 0 for q > n. (If A =M then j ∶ Hn(M) → Γc(M ; oM) isan isomorphism.)

Proof. ForX =Rn and A =Dn. Well, oRn =Rn×Hn(Rn,Rn−0) is trivial (i.e.,a product projection), so Γ(Dn; oRn) = HomTop(Dn,Hn(Rn,Rn − 0)) whereHn(Rn,Rn − 0) is discrete, and this is therefore just a map from π0 into this,and thus Γ(Dn; oRn) = R (your coefficient). But also, Hn(Rn,Rn −Dn) ≅ R,so you have that j gives Hn(Rn,Rn −Dn)→ Γc(Dn; oRn ∣Dn).

Say that this is true forA,B,A∩B – we’ll prove this forA∪B. Obviously, useMayer-Vietoris. I have a restriction Γc(A ∪B; oM) → Γc(A; oM) ⊕ Γc(B; oM)that sits in an exact sequence 0 → Γc(A ∪ B; oM) inclusion, determined by A,BÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐ→Γc(A; oM)⊕Γc(B; oM)→ Γc(A∩B; oM). This is a gluing lemma. We also have

36. FUNDAMENTAL CLASS 111

a relative Mayer-VietorisHn(M,M−A∪B)→Hn(M,M−A)⊕Hn(M,M−B)→Hn(M,M −A ∩B), so we have:

0 // Γc(A ∪B; oM)inclusion A,B→A∪B// Γc(A; oM)⊕ Γc(B; oM) // Γc(A ∩B; oM)

Hn+1(M,M −A ∩B) = 00 // Hn(M,M −A ∪B) //

OO

Hn(M,M −A)⊕Hn(M,M −B) //

j∗ ≅

OO

Hn(M,M −A ∩B)

j∗ ≅

OO

This is a “local-to-global” argument. “I don’t feel like going through the point-set topology – the rest of the proof is just annoyance.” See Bredon’s book forthe conclusion of the proof.

Corollary 36.4. j ∶Hn(M)→ Γc(M ; oM) is an isomorphism.

Definition 36.5. An R-orientation for M is a section σ of Γ(M ; o×M) whereo×M is the covering space of M given by the generators (as R-modules) of thefibers of oM .

If M is compact, then j ∶ Hn(M) → Γ(M ; oM), and you get [M] ↔ σ.When does that exist?

Over Z: o×M →M is a double cover of M (over every element you have twopossible elements given by the two possible orientations (±1)). If M is an n-manifold and f ∶ N →M is a covering space, then N is also locally Euclidean.I have the orientation local system to get a pullback local system:

f∗oM = N ×M oM //

oM

N // M

Because N → M is a covering space, the fibers of f∗oM are the same as thefibers of oN , so actually, f∗oM ≅ oN . For example, suppose N = o×M . Whathappens if I consider:

oN = N ×M N //

N

N // M

But now, I have the identity N → N that sits compatibly as:

Nid //

id

N

N // M

And hence you get N → o×N , which is a section of o×N → N . The conclusionis that N = o×M is canonically oriented (even if M is not oriented!). If M isoriented, then the local system is trivial and you have the trivial double cover.

The overarching conclusion is: if M is an n-manifold, then:


1. Hq(M) = 0 for q > n.

2. If M is compact, then Hn(M) ≅Ð→ Γ(M,oM).

3. If M is connected and compact, then:

a) if M is oriented with respect to R, then Hn(M) ≅ Γ(M,oM) ≅ R.b) (I have no idea what was happening here, we didn’t reach to a

conclusion for a while.) ifM is not orientable, then o×M is nontrivial.If o×M has a section, then it’s trivial (and so is oM ) because if it hasa section σ ∶ M → o×M , define M × R× ≅Ð→ o×M by sending (x, r) ↦rσ(x) ∈ o×M (and the same thing M × R ≅Ð→ oM for the orientationlocal system itself). I don’t see an argument to conclude that if Mis nonorientable, then there aren’t any section of oM . In particular,if R = Z, then Hn(M ;Z) = 0. I’m going to leave this as a statementwithout proof, unless any of you can help me.

If a section σ(x) = 0 for some x, then σ = 0.

Remark 36.6. Prof. Miller talked with me about this after class. If I recallcorrectly, one way to think about this is as follows. If you have a local systemp ∶ E → B, this can be viewed as a representation of π1(B) → R×, and theΓ(B;E) = (Ex)π1(B) where π1(B) acts on the fibers by multiplication. Thus(Ex)π1(B) = Rπ1(B). If R = Z, then R× = ±1, so Rπ1(B) = r∣ar = r, a ∈π1(B), so that Zπ1(B) = 0. Hence there are no sections of oM , as desired.For a ring R, oM,R = oM,Z ⊗R. Something else for Z/2Z. A higher homotopytheoretic perspective is that if you have a fibration E → B, then E = PB×ΩBFwhere F is the fiber of the fibration, so that Γ(B;E) = MapΩB(PB,F ) = FhΩB .In the case of a covering space you recover what you have above since π0(ΩB) =π1(B).

37 Covering spaces and Poincaré duality

Miller’s office hours are tomorrow, from 1-3 in 2-478. The first half of thislecture was just explaining the remark above by using less technology.

On the website, there are notes on π1(X,∗). I’m assuming people haveseen this thing. Assume X is path-connected, and let ∗ ∈ X. There’s anothertechnical assumption: semi-locally simply connected (SLSC), which means thatfor every b ∈ X and neighborhood b ∈ U , there exists a smaller neighborhoodb ∈ V ⊆ U such that π1(V, b) → π1(X, b) is trivial. This is a very very weakcondition.

Theorem 37.1. Let X be a path-connected, SLSC space with ∗ ∈ X. Thenthere is an equivalence of categories between covering spaces over X and setswith an action of π1(X,∗). The way this functor goes is by sending p ∶ E →Xto p−1(∗), which has an action of π1(X,∗) in the obvious way by path-lifting.

37. COVERING SPACES AND POINCARÉ DUALITY 113

Example 37.2 (Stupidest possible case). Suppose id ∶ X → X is sent to ∗with the trivial action. This is the terminal covering space over X.

We’ve been interested in Γ(E;X), which is the same thing as MapX(X →X,E → X) ≅ Mapπ1(X)(∗,E∗) = (E∗)π1(B), the fixed points of the action. Wealso thought about the case of E being a local system of R-modules, and thesame functor gives an equivalence between local systems of R-modules andR[π1(X)]-modules, i.e., representations of π1(X).

Recall that oM is the orientation local system, but now over Z. Thus, overa general ring, oM,R = oM ⊗R. We were thinking about what happens with aclosed path-connected SLSC subset ∗ ∈ A ⊆M of an n-manifold M , and thenconsidering Γ(A,oM ⊗R), which we now see to be (oM ⊗R)π1(A,∗)

∗ . How manyoptions do we have here?

That is to say, this local system oM is the same thing as the free abeliangroup Hn(M,M − ∗) with an action of π1(X,∗). There aren’t many op-tions for this action. In other words, this is a homomorphism π1(M,∗) →Aut(Hn(M,M−∗)). I haven’t chosen a generator forHn(M,M−∗), and there’sonly two automorphisms, i.e., we get a homomorphism w1 ∶ π1(M,∗) → Z/2Z.This homomorphism is called the “first Stiefel-Whitney class”. 18.906 willdescribe all the Stiefel-Whitney classes. With R-coefficients, I get a mapπ1(M,∗) → Aut(Hn(M,M − ∗;R)) ≅ R×. This is a natural construction,so this homomorphism π1(M,∗) → R× factors through π1(M,∗) → Z/2Z.This lets us get a good handle on what the sections are: Γ(A; oM ⊗ R) =Hn(M,M − ∗;R)π1(X,∗), but our analysis shows that:

Γ(A; oM ⊗R) =Hn(M,M − ∗;R)π1(X,∗)

=⎧⎪⎪⎨⎪⎪⎩

Hn(M,M − ∗;R) ≅ R if w1 = 1, well-defined up to sign; the orientable case

ker(R 2Ð→ R) if w1 ≠ 1, and this is a canonical identification

where we get the latter thing because then a = −a, i.e., 2a = 0. In particular, ifR = Z/2Z, since AutZ/2Z(Z/2Z) = 1, you always have a unique orientation. If

R = Z/pZ,Z,Q, then ker(R 2Ð→ R) = 0.We had a general theorem:

Theorem 37.3. Hn(M,M−A;R) j,≅Ð→ Γc(A; oM⊗R) and Hq(M,M−A;R) = 0for q > n.

Corollary 37.4. If M is connected and A = M , and if M is not compact,then Hn(M ;R) = 0. If M is compact, then the work we just did shows that

Hn(M ;R) =⎧⎪⎪⎨⎪⎪⎩

R oriented

ker(R 2Ð→ R) nonorientable.

Poincaré duality, finally

Assume M is R-oriented. Let K ⊆ M be compact. Then Hn(M,M −K) ≅Ð→Γ(K; oM ⊗R). Picking an orientation picks an isomorphism Γ(K; oM ⊗R) ≅ R.


This gives some [M]K , which is called the fundamental class along K. IfK =M , then [M]M =∶ [M] ∈Hn(M ;R).

Suppose K ⊆ L are compact subsets. We now combine all of our resultsabove:

Theorem 37.5 (Fully relative Poincaré duality). If p+q = n, then Hp(K,L;R)∩[M]KÐÐÐÐ→

Hq(M −L,M −K;R) is an isomorphism.

Proof. “It’s, like, not hard at this point.” One thing we did was set up an LESfor H of a pair, which implies that we may assume that L = ∅. We want to

prove that Hp(K;R)∩[M]KÐÐÐÐ→Hq(M,M −K;R) is an isomorphism. Now there’s

a standard local-to-global process.In the local case, ifM =Rn andK =Dn, then this is saying that Hp(Dn;R) ≅

Hp(Dn;R)∩[Rn]DnÐÐÐÐÐ→ Hq(Rn,Rn −Dn;R) where the first isomorphism comes

from analysis we did earlier about Čech and ordinary cohomology coincid-ing. If p ≠ 0, then both sides are zero. When p = 0, we are asking that

H0(Dn;R)∩[Rn]DnÐÐÐÐÐ→ Hn(Rn,Rn −Dn;R). They’re both equal to R, and we

are just capping along [Rn]Dn , because we found that 1 ∩ [Rn]Dn = [Rn]Dn ,as desired.

We carefully set up the Mayer-Vietoris sequence ladder (Theorem 33.5) thatallows us to put this all together. “We’re not going to go through the detailsbecause there’s point set topology that I don’t like there.” Note that normalityis not needed for K,L compact because compact sets in Hausdorff spaces canalways be separated, normal or not. I just reversed the order in which thingsare usually taught in books.

We have time for one beginning application.

Corollary 37.6 (Relative Poincaré duality). Suppose K =M and M is com-

pact and R-oriented. Then Hp(M,L;R)∩[M]ÐÐÐ→ Hn−p(M − L,R) is an isomor-

phism.

Corollary 37.7 (Poincaré duality, corollary of corollary). Let M be compact

and R-oriented, then Hp(M ;R)∩[M]ÐÐÐ→Hn−p(M ;R) is an isomorphism.

Proof. Follows from the above corollary since Hp(M ;R) is literally equal toHp(M ;R)

That’s the most beautiful form of all. If you do have an L, you have thisladder, where all vertical maps are isomorphisms:

⋯ // Hp(M,L) //

−∩[M]

Hp(M) //

−∩[M]

Hp(L) δ //

−∩[M]L

Hp+1(M,L) //

−∩[M]

⋯

⋯ // Hq(L) // Hq(M) // Hq(M,M −L) ∂ // Hq−1(L) // ⋯

38. APPLICATIONS 115

This is a consistency statement for Poincaré duality. On Wednesday, we’llspecialize even further, and prove the Jordan curve theorem as well as studythe cohomology rings of things we haven’t worked through before.

38 Applications

Please check exam schedule! Also, a sample exam is posted. This is the payoffday. All this stuff about Poincaré duality has got to be good for something.Recall:

Theorem 38.1 (Fully relative duality). LetM be a R-oriented n-manifold. LetL ⊆K ⊆M be compact (M need not be compact). Then [M]K ∈Hn(M,M−K),and capping gives an isomorphism:

Hp(K,L;R)∩[M]k,≅ÐÐÐÐÐ→Hn−p(M −L,M −K;R)

Today we’ll think about the case L = ∅, so this is saying:

Hp(K;R)∩[M]k,≅ÐÐÐÐÐ→Hn−p(M,M −K;R)

Corollary 38.2. Hq(K;R) = 0 for q > n.

We can contrast this with singular (co)homology. Here’s an example:

Example 38.3 (Barratt-Milnor). A two-dimensional version K of the Hawai-ian earring, i.e., nested spheres all tangent to a point whose radii are going tozero. What they proved is that Hq(K;Q) is uncountable for every q > 1. Butif you look at the Čech cohomology, stuff vanishes.

That’s nice.How about an even more special subcase? Suppose M =Rn. The result is

called Alexander duality. This says:

Theorem 38.4 (Alexander duality). If ∅ ≠ K ⊆ Rn be compact. ThenHn−q(K;R) ≅Ð→ Hq−1(Rn −K;R)

Proof. We have the LES of a pair, which gives an isomorphism ∂ ∶Hq(Rn,Rn−K;R) ≅Ð→ Hq−1(Rn−K;R), so the composition ∂(−∩[M]K) is an isomorphismby Poincaré duality.

For most purposes, this is the most useful duality theorem.

Example 38.5 (Jordan curve theorem). q = 1 and R = Z. Then this is sayingthat Hn−1(K) ≅Ð→ H0(Rn −K). But H0(Rn −K) is free on #π0(Rn −K) − 1generators. If n = 2, for example, and K ≅ S1, then Hn−1(K) = Hn−1(K) ≅Hn−1(S1), so H1(S1) ≅ H0(R2 −K). Hence there are two components in thecomplement of K. This could also be the topologist’s sine curve as well. Thisis the Jordan curve theorem.


Consider the UCT, which states that there’s a sexseq 0→ Ext1Z(Hq−1(X),Z)→

Hq(X) → Hom(Hq(X),Z) → 0 that splits, but not naturally. First, note thatHom(Hq(X),Z) is always torsion-free. If I assume that Hq−1(X) is finitelygenerated, then Ext1

Z(Hq−1(X),Z) is a finite abelian group, but in particularit’s torsion.

The UCT is making the decomposition of Hq(X) into its torsion-free andtorsion parts. I can divide by torsion, so that Hq(X)/tors ≅ Hom(Hq(X),Z).But there’s also an isomorphism Hom(Hq(X)/tors,Z) → Hom(Hq(X),Z) be-cause Z is torsion-free. Therefore I get an isomorphism α ∶ Hq(X)/tors →Hom(Hq(X)/tors,Z). I.e.:

0 // Ext1Z(Hq−1(X),Z) // Hq(X) //

Hom(Hq(X),Z) // 0

Hq(X)/tors

≅55

α// Hom(Hq(X)/tors/Z)

≅

OO

Or I could say it like this: the Kronecker pairing can be quotiented by torsion,and you get an induced map Hq(X)/tors⊗Hq(X)/tors→ Z is a perfect pairing,which means that the adjoint map Hq(X)/tors

≅Ð→ Hom(Hq(X)/tors,Z). Let’scombine this with Poincaré duality.

Let X = M be a compact oriented n-manifold. Then Hn−q(X)−∩[M],≅ÐÐÐÐÐ→

Hq(M), and so we get a perfect pairing Hq(X)/tors ⊗ Hn−q(X)/tors → Z.And what is that pairing? It’s the cup product! We have:

Hq(M)⊗Hn−q(M) //

1⊗(−∩[M])

Z

Hq(M)⊗Hq(M)⟨,⟩

77

And, well:⟨a, b ∩ [M]⟩ = ⟨a ∪ b, [M]⟩

Thus the map Hq(M) ⊗ Hn−q(M) → Z is a ⊗ b ↦ ⟨a ∪ b, [M]⟩, and it’s aperfect pairing. This is a purely cohomological version, and is the most usefulstatement.

Example 38.6. Suppose M = CP2 = D0 ∪ D2 ∪ D4, and its homology isZ0Z0Z, and so its cohomology is the same. Let a ∈H2(CP2). Then we haveH2(CP2)⊗H2(CP2)→ Z, and so a∪a is a generator of H4(CP2), and hencespecifies an orientation for CP2. The conclusion is that H∗(CP2) = Z[a]/(a3)where ∣a∣ = 2.

How about CP3? It just adds a 6-cell, so its homology is Z0Z0Z0Z,and so its cohomology is the same. But then a3 = a ∪ a ∪ a is a generator ofH6(CP2), and etc. Thus in general, we have:

H∗(CPn) = Z[a]/(an+1)


These things are finite CW-complexes, so you find:

H∗(CP∞) = Z[a] (2.3)

Example 38.7. Suppose I look at maps f ∶ Sm → Sn. One of the mostinteresting things is that there are lots of non null-homotopic maps Sm → Sn

if m > 2. For example, η ∶ S3 → S2 that’s the attaching map for the 4-cell inCP2. This is called the Hopf fibration. It’s essential. Why is it nullhomotopic?If η was null homotopic, then CP2 ≃ S2 ∧ S4. That’s compatible with thecohomology in each dimension, but not into the cohomology ring! There’s amap S2 ∧ S4 → S2 that collapses S4, and the generator in H∗(S2) has a2 = 0,so a2 = 0 in H∗(S2 ∧ S4). But this is not compatible with our computationthat H∗(CP2) = Z[a]/(a3) where ∣a∣ = 2.

With coefficients in a field k, then the torsion is zero, so you find that ifM iscompact k-oriented, then if the characteristic of k = 2, there’s no condition forM to be oriented, and if the characteristic of k is not 2, then M is Z-oriented.Thus we get that Hq(M ;k)⊗k Hn−q(M ;k)→ k is a perfect pairing.

Example 38.8. Exactly the same argument as for complex projective spaceshows that:

H∗(RPn;F2) = F2[a]/(an+1)

where ∣a∣ = 1. So:H∗(RP∞;F2) = F2[a] (2.4)

where ∣a∣ = 1.

I’ll end with the following application.

Theorem 38.9. Suppose f ∶Rm+1 ⊇ Sm → Sn ⊆Rn+1 that is equivariant withrespect to the antipodal action, i.e., f(−x) = −f(x). Then m ≤ n.

So there are no equivariant maps from Sm → Sn if m > n!

Proof. Suppose I have a map like that: the map on spheres induces a mapf ∶RPm →RPn. We claim that H1(f) is an isomorphism. Let π ∶ Sn →RPn

denote the map. Let σ ∶ I → Sm be defined via σ(0) = v and σ(1) = −v. Sothis gives a 1-cycle σ ∶ I → Sm → RPm, and H1(RPn) = [πσ] is generated bythis thing. When I map this thing to RPn, we send πσ to a generator. Whatwe’ve actually proved, therefore, is that H1(RPm) ≅ H1(RPn). This is alsotrue with mod 2 coefficients, i.e., H1(f,F2) ≠ 0.

That means that H1(f ;F2) ≠ 0 by UCT. But what is this? This is amap H1(f ;F2) ∶ H∗(RPn;F2) → H∗(RPn;F2), i.e., a map F2[a]/(an+1) →F2[a] → (am+1). Thus a ↦ a. There’s not a lot of ways to do this if m > n.Thus what we’ve shown that m ≤ n.


This is the Borsuk-Ulam theorem from the ’20s, I think. This is an exampleof how you can use the cohomology ring structure for projective space.

Please check the website for details about your finals. I will ask you to signa form, to make sure that you don’t share the questions or that you haven’theard the questions beforehand. I have a fixed set of questions that’ll guidethe conversation.

Part II

18.906 – homotopy theory

119


Introduction

Here is an overview of this part of the book.

1. General homotopy theory. This includes category theory; becauseit started as a part of algebraic topology, we’ll speak freely about ithere. We’ll also cover the general theory of homotopy groups, long exactsequences, and obstruction theory.

2. Bundles. One of the major themes of this part of the book is the use ofbundles to understand spaces. This will include the theory of classifyingspaces; later, we will touch upon connections with cohomology.

3. Spectral sequences. It is impossible to describe everything about spec-tral sequences in the duration of a single course, so we will focus on aspecial (and important) example: the Serre spectral sequence. As a con-sequence, we will derive some homotopy-theoretic applications. For in-stance, we will relate homotopy and homology (via the Hurewicz theorem,Whitehead’s theorem, and “local” versions like Serre’s mod C theory).

4. Characteristic classes. This relates the geometric theory of bundlesto algebraic constructions like cohomology described earlier in the book.We will discuss many examples of characteristic classes, including theThom, Euler, Chern, and Stiefel-Whitney classes. This will allow us toapply a lot of the theory we built up to geometry.

Chapter 3

Homotopy groups

Insert an outline of the content of each lecture.

39 Limits, colimits, and adjunctions

Limits and colimits

We will freely use the theory developed in the first part of this book (see §3).Suppose I is a small category (so that it has a set of objects), and let C beanother category.

Definition 39.1. Let X ∶ I → C be a functor. A cone under X is a naturaltransformation η from X to a constant functor; explicitly, this means that forevery object i of I, we must have a map ηi ∶ Xi → Y , such that for everyf ∶ i→ j in I, the following diagram commutes:

Xi

f∗

ηi

Xj

ηj // Y.

A colimit of X is an initial cone (L, τi) under X; explicitly, this means thatfor all cones (Y, ηi) under X, there exists a unique natural transformationh ∶ L→ Y such that h τi = ηi.

As always for category theoretic concepts, some examples are in order.

Example 39.2. If I is a discrete category (i.e., only a set, with identity maps),the colimit of any functor I → C is the coproduct. This already illustrates animportant point about colimits: they need not exist in general (since, for exam-ple, coproducts need not exist in a general category). Examples of categories Cwhere the colimit of a functor I → C exists: if C is sets, or spaces, the colimit isthe disjoint union. If C =Ab, a candidate for the colimit would be the product:

123

124 CHAPTER 3. HOMOTOPY GROUPS

but this only works if I is finite; in general, the correct thing is to take the(possibly infinite) direct sum.

Example 39.3. Let I = N, considered as a category via its natural posetstructure; then a functor I → C is simply a linear system of objects and mor-phisms in C. As a specific example, suppose C =Ab, and consider the diagramX ∶ I → C defined by the system

Z2Ð→ Z

3Ð→ Z→ ⋯

The colimit of this diagram is Q, where the maps are:

Z2 //

1

Z3 //

1/2

Z4 //

1/3!

⋯

Q

Example 39.4. Let G be a group; we can view this as a category with oneobject, where the morphisms are the elements of the group (composition isgiven by the group structure). If C = Top is the category of topological spaces,a functor G→ C is simply a group action on a topological space X. The colimitof this functor is the orbit space of the G-action on X.

Example 39.5. Let I be the category whose objects and morphisms are de-termined by the following graph:

a

b c.

The colimit of a diagram I → C is called a pushout.If C = Top, again, a functor I → C is determined by a diagram of spaces:

A

f

g

B C.

The colimit of such a functor is just the pushout B ∪A C ∶= B ⊔ C/ ∼, wheref(a) ∼ g(a) for all a ∈ A. We have already seen this in action before: the sameconstruction appears in the process of attaching cells to CW-complexes.

If C is the category of groups, instead, the colimit of such a functor is thefree product quotiented out by a certain relation (the same as for topologicalspaces); this is called the amalgamated free product.

Example 39.6. Suppose I is the category defined by the following graph:

a b.

The colimit of a diagram I → C is called the coequalizer of the diagram.

39. LIMITS, COLIMITS, AND ADJUNCTIONS 125

One can also consider cones over a diagram X ∶ I → C: this is simply acone in the opposite category.

Definition 39.7. With notation as above, the limit of a diagram X ∶ I → C isa terminal object in cones over X.

For instance, products are limits, just like in Example 39.2. (This examplealso shows that abelian groups satisfy an interesting property: finite productsare the same as finite coproducts!)

Exercise 39.8. Revisit the examples provided above: what is the limit of eachdiagram? For instance, the limit of the diagram described in Example 39.4 isjust the fixed points!

Adjoint functors

Adjoint functors are very useful — and very natural — objects. We alreadyhave an example: let CI be the functor category Fun(I,C). (We’ve beenworking in this category this whole time!) Let’s make an additional assumptionon C, namely that all I-indexed colimits exist. All examples considered abovesatisfy this assumption.

There is a functor C → CI , given by sending any object to the constantfunctor taking that value. The process of taking the colimit of a diagramsupplies us with a functor CI → C. We can characterize this functor via aformula1:

C(colimi∈IXi, Y ) = CI(X, constY ),

where X is some functor from I to C. This formula is reminiscent of theadjunction operator in linear algebra, and is in fact our first example of anadjunction.

Definition 39.9. Let C,D be categories, with specified functors F ∶ C → Dand G ∶ D → C. An adjunction between F and G is an isomorphism:

D(FX,Y ) = C(X,GY ),

which is natural in X and Y . In this situation, we say that F is a left adjointof G and G is a right adjoint of X.

This notion was invented by Dan Kan, who worked in the MIT mathematicsdepartment until he passed away in 2013.

We’ve already seen an example above, but here is another one:

1There is an analogous formula for the limit of a diagram:

C(W, limi∈I

Xi) = CI(constW ,X).


Definition 39.10 (Free groups). There is a forgetful functor u ∶ Grp → Set.Any set X gives rise to a group FX, namely the free group on X elements.This is determined by a universal property: set maps X → uΓ are the same asgroup maps FX → Γ, where Γ is any group. This is exactly saying that thefree group functor the left adjoint to the forgetful functor u.

In general, “free objects” come from left adjoints to forgetful functors.

Definition 39.11. A category C is said to be cocomplete if all (small) colimitsexist in C. Similarly, one says that C is complete if all (small) limits exist in C.

The Yoneda lemma

One of the many important concepts in category theory is that an object isdetermined by the collection of all maps out of it. The Yoneda lemma is a wayof making this precise. An important reason to even bother thinking aboutobjects in this fashion comes from our discussion of colimits. Namely, how dowe even know that the notion is well-defined?

The colimit of an object is characterized by maps out of it; precisely:

C(colimj∈J Xj , Y ) = CJ (X, constY ).

The two sides are naturally isomorphic, but if the colimit exists, how do weknow that it is unique? This is solved by Yoneda lemma2:

Theorem 39.12 (Yoneda lemma). Consider the functor C(X,−) ∶ C → Set.Suppose G ∶ C → Set is another functor. It turns out that:

nt(C(X,−),G) ≃ G(X).

Proof. Let x ∈ G(X). Define a natural transformation that sends a mapf ∶ X → Y to f∗(x) ∈ G(Y ). On the other hand, we can send a naturaltransformation θ ∶ C(X,−) → G to θX(1X). Proving that these are inverses isleft as an exercise — largely in notation — to the reader.

In particular, if G = C(Y,−) — these are called corepresentable functors— then nt(C(X,−),C(Y,−)) ≃ C(Y,X). Simply put, natural isomorphismsC(X,−) → C(Y,−) are the same as isomorphisms Y → X. As a consequence,the object that a corepresentable functor corepresents is unique (at least up toisomorphism).

From the Yoneda lemma, we can obtain some pretty miraculous conclusions.For instance, functors with left and/or right adjoints are very well-behaved(the “constant functor” functor is an example where both adjoints exist), asthe following theorem tells us.

Theorem 39.13. Let F ∶ C → D be a functor. If F admits a right adjoint, itpreserves colimits. Dually, if F admits a left adjoint, it preserves limits.

2Sometimes “you-need-a-lemma”!

40. COMPACTLY GENERATED SPACES 127

Proof. We’ll prove the first statement, and leave the other as an (easy) exercise.Let F ∶ C → D be a functor that admits a right adjoint G, and let X ∶ I → C bea small I-indexed diagram in C. For any object Y of C, there is an isomorphism

Hom(colimIX,Y ) ≃ limI

Hom(X,Y ).

This follows easily from the definition of a colimit. Let Y be any object of D;then, we have:

D(F (colimIX), Y ) ≃ C(colimIX,G(Y ))≃ limIC(X,G(Y ))

≃ limID(F (X), Y )

≃ D(colimI F (X), Y ).

The Yoneda lemma now finishes the job.

40 Compactly generated spaces

A lot of homotopy theory is about loop spaces and mapping spaces. Standardtopology doesn’t do very well with mapping spaces, so we will narrate thestory of compactly generated spaces. One nice consequence of working withcompactly generated spaces is that the category is Cartesian-closed (a conceptto be defined below).

CGHW spaces

Some constructions commute for “categorical reasons”. For instance, limitscommute with limits. Here is an exercise to convince you of a special case ofthis.

Exercise 40.1. Let X be an object of a category C. The overcategory (orthe slice category) C/X has objects given by morphisms p ∶ Y → X in C, andmorphisms given by the obvious commutativity condition.

1. Assume that C has finite products. What is the left adjoint to the functorX × − ∶ C → C/X that sends Y to the object X × Y

pr1ÐÐ→X?

2. As a consequence of Theorem 39.13, we find thatX×− ∶ C → C/X preserveslimits. The composite C → C/X → C, however, probably does not.

• What is the limit of a diagram in C/X?• Let Y ∶ I → C be any diagram. Show that

limi∈IC/X (X × Yi) ≃X × lim

i∈ICYi.

What happens if I only has two objects and only identity mor-phisms?


However, colimits and limits need not commute! An example comes fromalgebra. The coproduct in the category of commutative rings is the ten-sor product (exercise!). But (limZ/pkZ) ⊗ Q ≃ Zp ⊗ Q ≃ Qp is clearly notlim (Z/pkZ⊗Q) ≃ lim 0 ≃ 0!

We also need not have an isomorphism betweenX×colimj∈J Yj and colimj∈J (X×Yj). One example comes a quotient map Y → Z: in general, the induced mapX × Y →X ×Z is not necessarily another quotient map. A theorem of White-head’s says that this problem is rectified if we assume that X is a compactHausdorff space. Unfortunately, a lot of interesting maps are built up frommore “elementary” maps by such a procedure, so we would like to repair thisproblem.

We cannot simply do this by restricting ourselves to compact Hausdorffspaces: that’s a pretty restrictive condition to place. Instead (motivated par-tially by the Yoneda lemma), we will look at topologies detected by maps fromcompact Hausdorff spaces.

Definition 40.2. Let X be a space. A subspace F ⊆X is said to be compactlyclosed if, for any map k ∶ K → X from a compact Hausdorff space K, thepreimage k−1(F ) ⊆K is closed.

It is clear that any closed subset is compactly closed, but there might becompactly closed sets which are not closed in the topology onX. This motivatesthe definition of a k-space:

Definition 40.3. A topological space X is said to be a k-space if every com-pactly closed set is closed.

The k comes either from “kompact” and/or Kelly, who was an early topol-ogist who worked on such foundational topics.

It’s clear that X is a k-space if and only if the following statement is true:a map X → Y is continuous if and only if, for every compact Hausdorff spaceK and map k ∶K →X, the composite K →X → Y is continuous. For instance,compact Hausdorff spaces are k-spaces. First countable (so metric spaces) andCW-complexes are also k-spaces.

In general, a topological space X need not be a k-space. However, it canbe “k-ified” to obtain another k-space denoted kX. The procedure is simple:endow X with the topology consisting of all compactly closed sets. The readershould check that this is indeed a topology on X; the resulting topologicalspace is denoted kX. This construction immediately implies, for instance, thatthe identity kX →X is continuous.

Let kTop be the category of k-spaces. This is a subcategory of the categoryof topological spaces, via a functor i ∶ kTop Top. The process of k-ificationgives a functor Top→ kTop, which has the property that:

kTop(X,kY ) = Top(iX,Y ).

Notice that this is another example of an adjunction! We can conclude fromthis that k(iX × iY ) =X ×kTop Y , where X and Y are k-spaces. One can alsocheck that kiX ≃X.

40. COMPACTLY GENERATED SPACES 129

The takeaway is that kTop has good categorical properties inherited fromTop: it is a complete and cocomplete category. As we will now explain, thiscategory has more categorical niceness, that does not exist in Top.

Mapping spaces

Let X and Y be topological spaces. The set Top(X,Y ) of continuous mapsfrom X to Y admits a topology, namely the compact-open topology. If Xand Y are k-spaces, we can make a slight modification: define a topology onkTop(X,Y ) generated by the sets

W (k ∶K →X, open U ⊆ Y ) ∶= f ∶X → Y ∶ f(k(K)) ⊆ U.

We write Y X for the k-ification of kTop(X,Y ).

Proposition 40.4. 1. The functor (kTop)op × kTop → kTop given by(X,Y )→ Y X is a functor of both variables.

2. e ∶ X × ZX → Z given by (x, f) ↦ f(x) and i ∶ Y → (X × Y )X given byy ↦ (x↦ (x, y)) are continuous.

Proof. The first statement is left as an exercise to the reader. For the secondstatement, see [Str09, Proposition 2.11].

As a consequence of this result, we can obtain a very nice adjunction. Definetwo maps:

• kTop(X × Y,Z)→ kTop(Y,ZX) via

(f ∶X × Y → Z)↦ (Y iÐ→ (X × Y )X → ZX).

• kTop(Y,ZX)→ kTop(X × Y,Z) via

(f ∶ Y → ZX)↦ (X × Y →X ×ZX eÐ→X).

By [Str09, Proposition 2.12], these two maps are continuous inverses, so thereis a natural homeomorphism

kTop(X × Y,Z) ≃ kTop(Y,ZX).

This motivates the definition of a Cartesian closed category.

Definition 40.5. A category C with finite products is said to be Cartesianclosed if, for any object X of C, the functor X × − ∶ C → C has a right adjoint.

Our discussion above proves that kTop is Cartesian closed, while this is notsatisfied by Top. As we will see below, this has very important ramificationsfor algebraic topology.

Exercise 40.6. Insert Exercise 2 from18.906.Insert Exercise 2 from18.906.


41 “Cartesian closed”, Hausdorff, Basepoints

Pushouts are colimits, so the quotient space X/A =X ∪A ∗ is an example of acolimit. Let Y be a topological space, and consider the functor Y × − ∶ Top →Top. Applying this to the pushout square, we find that (Y × X) ∪Y ×A ∗ ≃(Y × X)/(Y × A). As we discussed in §40, this product is not the same asY × (X/A)! There is a bijective map Y ×X/Y ×A→ Y × (X/A), but it is not,in general, a homeomorphism. From a categorical point of view (see Theorem39.13), the reason for this failure stems from Y × − not being a left adjoint.

The discussion in §40 implies that, when working with k-spaces, that functoris indeed a left adjoint (in fancy language, the category kTop is Cartesianclosed), which means that — in kTop — there is a homeomorphism Y ×X/Y ×A→ Y ×(X/A). This addresses the issues raised in §40. The ancients had comeup with a good definition of a topology — but k-spaces are better! Sometimes,though, we can be greedy and ask for even more: for instance, we can demandthat points be closed. This leads to a further refinement of k-spaces.I don’t like point-set topology, so I’ll return to editing this lecture at theend.

“Hausdorff”

Definition 41.1. A space is “weakly Hausdorff” if the image of every mapK →X from a compact Hausdorff space K is closed.

Another way to say this is that the map itself if closed. Clearly Hausdorffimplies weakly Hausdorff. Another thing this means is that every point in Xis closed (eg K = ∗).

Proposition 41.2. Let X be a k-space.

1. X is weakly Hausdorff iff ∆ ∶X →X×kX is closed. In algebraic geometrysuch a condition is called separated.

2. Let R ⊆ X ×X be an equivalence relation. If R is closed, then X/R isweakly Hausdorff.

Definition 41.3. A space is compactly generated if it’s a weakly Hausdorffk-space. The category of such spaces is called CG.

We have a pair of adjoint functors (i, k) ∶ Top→ kTop. It’s possible to de-fine a functor kTop→CG given by X ↦X/⋂all closed equivalence relations.It is easy to check that if Z is weakly Hausdorff, then ZX is weakly Hausdorff(where X is a k-space). What this implies is that CG is also Cartesian closed!

I’m getting a little tired of point set stuff. Let’s start talking about homo-topy and all that stuff today for a bit. You know what a homotopy is. I willnot worry about point-set topology anymore. So when I say Top, I probably

42. FIBER BUNDLES, FIBRATIONS, COFIBRATIONS 131

mean CG. A homotopy between f, g ∶X → Y is a map h ∶ I ×X → Y such thatthe following diagram commutes:

X

i0 ""

f

))I ×X h // Y

X

i1

<<

g

55

We write f ∼ g. We define [X,Y ] = Top(X,Y )/ ∼. Well, a map I ×X → Y isthe same as a map X → Y I but also I → Y X . The latter is my favorite! It’s apath of maps from f to g. So [X,Y ] = π0Y

X .To talk about higher homotopy groups and induct etc. we need to talk

about basepoints.

Basepoints

A pointed space is (X,∗) with ∗ ∈ X. This gives a category Top∗ where themorphisms respect the basepoint. This has products because (X,∗) × (Y,∗) =(X × Y, (∗,∗)). How about coproducts? It has coproducts as well. This is thewedge product, defined as X ⊔ Y /∗X ∼ ∗Y =∶X ∨ Y . This is \vee, not \wedge.Is this category also Cartesian closed?

Define the space of pointed maps ZX∗ ⊆ ZX topologized as a subspace. Doesthe functor Z ↦ ZX∗ have a left adjoint? Well Top(W,ZX) = Top(X ×W,Z).What about Top(W,ZX∗ )? This is f ∶ X ×W → Z ∶ f(∗,w) = ∗∀w ∈ W.That’s not quite what I wanted either! Thus Top∗(W,ZX∗ ) = f ∶X ×W → Z ∶f(∗,w) = ∗ = f(x,∗)∀x ∈X,w ∈W. These send both “axes” to the basepoint.Thus, Top∗(W,ZX∗ ) = Top∗(X ∧W,Z) where X ∧W =X ×W /X ∨W becauseX ∨W are the “axes”.

So Top∗ is not Cartesian closed, but admits something called the smashproduct3. What properties would you like? Here’s a good property: (X∧Y )∧Zand X ∧ (Y ∧Z) are bijective in pointed spaces. If you work in kTop or CG,then they are homeomorphic! It also has a unit.

Oh yeah, some more things about basepoints! So there’s a canonical for-getful functor i ∶ Top∗ → Top. Let’s see. If I have Top(X, iY ) = Top∗(??, Y )?This is X+ = X ⊔ ∗. Thus we have a left adjoint (−)+. It is clear that(X ⊔ Y )∗ =X+ ∨ Y+. The unit for the smash product is ∗+ = S0.

On Friday I’ll talk about fibrations and fiber bundles.

42 Fiber bundles, fibrations, cofibrations

Having set up the requisite technical background, we can finally launch our-selves from point-set topology to the world of homotopy theory.

3Remark by Sanath: this is like the tensor product.


Fiber bundles

Definition 42.1. A fiber4 bundle is a map p ∶ E → B, such that for everyb ∈ B, there exists:

• an open subset U ⊆ B that contains b, and

• a map p−1(U) → p−1(b) such that p−1(U) → U × p−1(b) is a homeomor-phism.

If p ∶ E → B is a fiber bundle, E is called the total space, B is called the basespace, p is called a projection, and F (sometimes denoted p−1(b)) is called thefiber over b.

In simpler terms: the preimage over every point in B looks like a product,i.e., the map p ∶ E → B is “locally trivial” in the base.

Here is an equivalent way of stating Definition 42.1: there is an open coverU (called the trivializing cover) of B, such that for every U ⊆ U , there is aspace F , and a homeomorphism p−1(U) ≃ U × F that is compatible with theprojections down to U . (So, for instance, a trivial example of a fiber bundle isjust the projection map B × F

pr1ÐÐ→ B.)Fiber bundles are naturally occurring objects. For instance, a covering

space E → B is a fiber bundle with discrete fibers.

Example 42.2 (The Hopf fibration). The Hopf fibration is an extremely im-portant example of a fiber bundle. Let S3 ⊂ C2 be the 3-sphere. There is amap S3 → CP1 ≃ S2 that is given by sending a vector v to the complex linethrough v and the origin. This is a non-nullhomotopic map, and is a fiberbundle whose fiber is S1.

Here is another way of thinking of the Hopf fibration. Recall that S3 =

SU(2); this contains as a subgroup the collection of matrices (λλ−1). This

subgroup is simply S1, which acts on S3 by translation; the orbit space is S2.

The Hopf fibration is a map between smooth manifolds. A theorem ofEhresmann’s says that it is not too hard to construct fiber bundles over smoothmanifolds:

Theorem 42.3 (Ehresmann). Suppose E and B are smooth manifolds, andlet p ∶ E → B be a smooth (i.e., C∞) map. Then p is a fiber bundle if:

1. p is a submersion, i.e., dp ∶ TeE → Tp(e)B is a surjection, and

2. p is proper, i.e., preimages of compact sets are compact.

The purpose of this part of the book is to understand fiber bundles throughalgebraic methods like cohomology and homotopy. This means that we willusually need a “niceness” condition on the fiber bundles that we will be study-ing; this condition is made precise in the following definition (see [May99]).

4Or “fibre”, if you’re British.

42. FIBER BUNDLES, FIBRATIONS, COFIBRATIONS 133

Definition 42.4. Let X be a space. An open cover U of X is said to benumerable if there exists a subordinate partition of unity, i.e., for each U ∈ U ,there is a function fU ∶X → [0,1] = I such that f−1((0,1]) = U , and any x ∈Xbelongs to only finitely many U ∈ U . The space X is said to be paracompact ifany open cover admits a numerable refinement.

This isn’t too restrictive for us algebraic topologists since CW-complexesare paracompact.

Definition 42.5. A fiber bundle is said to be numerable if it admits a numer-able trivializing cover.

Fibrations and path liftings

For our purposes, though, fiber bundles are still too narrow. Fibrations capturethe essence of fiber bundles, although it is not at all immediate from theirdefinition that this is the case!

Definition 42.6. A map p ∶ E → B is called a (Hurewicz5) fibration if it sat-isfies the homotopy lifting property (commonly abbreviated as HLP): supposeh ∶ I ×W → B is a homotopy; then there exists a lift6 (given by the dottedarrow) that makes the diagram commute:

Wf //

in0

E

p

I ×W

h//

h

<<

B,

(3.1)

At first sight, this seems like an extremely alarming definition, since theHLP has to be checked for all spaces, all maps, and all homotopies! The HLPis not impossible to check, though.

Exercise 42.7. Check that the projection pr1 ∶ B × F → B is a fibration.

Exercise 42.8. Check the following statements.

• Fibrations are closed under pullbacks. In other words, if p ∶ E → B is afibration and X → B is any map, then the induced map E ×B X → X isa fibration.

• Fibrations are closed under exponentiation and products. In other words,if p ∶ E → B is a fibration, then EA → BA is another fibration.

• Fibrations are closed under composition.

5Named after Witold Hurewicz, who was one of the first algebraic topologists at MIT.6Note that we place no restriction on the uniqueness of this lift.


Exercise 42.9. Let p ∶ E0 → B0 be a fibration, and let f ∶ B → B0 be a homo-topy equivalence. Prove that the induced map B ×B0 E0 → E0 is a homotopyequivalence. (Warning: this exercise has a lot of technical details! The end ofthis chapter describes an alternative7 solution to this exercise, when E0 andB ×B0 E0 are CW-complexes.)Don’t forget to do this!Don’t forget to do this!

There is a simple geometric interpretation of what it means for a map tobe a fibration, in terms of “path liftings”. To understand this description, wewill reformulate the diagram (3.1). Given that we are working in the categoryof CGWH spaces, one of the first things we can attempt to do is adjoint the I;this gives the following diagram.

Ep // B

W

f

OO

h

// BI

ev0

OO (3.2)

By the definition of the pullback of a diagram, the data of this diagram isequivalent to a map W → BI ×B E. Explicitly,

BI ×B E = (ω, e) ∈ BI ×E such that ω(0) = p(e).

Suppose the desired dotted map exists (i.e., p ∶ E → B satisfied the HLP).This would beget (again, by adjointness) a lifted homotopy h ∶W → EI . Sincewe already have a map8 p ∶ EI → BI ×B E given by ω ↦ (pω,ω(0)), theexistence of the lift h in the diagram (3.1) is equivalent to the existence of alift in the following diagram.

EI

p

W //

h

;;

BI ×B E

Obviously the universal example of a space W that makes the diagram (3.2)commute is BI ×B E itself. If p is a fibration, we can make the lift in thefollowing diagram.

EI

p

BI ×B E

λ

88

1 // BI ×B E

The map λ is called a lifting function. To understand why, suppose (ω, e) ∈BI ×B E, so that ω(0) = p(e). In this case, λ(ω, e) defines a path in E suchthat

p λ(ω, e) = ω, and λ(ω, e)(0) = e.7“Alternative” in the sense that the proof uses statements not covered yet in this book.8Clearly (pω)(0) = p(ω(0)), so this map is well-defined (i.e., the image lands in BI ×BE).

43. FIBRATIONS AND COFIBRATIONS 135

Taking a step back and assessing the situation, we find that the lifting functionλ starts with a path ω in B, and some point in E mapping down to ω(0), andproduces a “lifted” path in E which lives over ω. In other words, the map λis a path lifting: it’s a continuous way to lift paths in the base space B to thetotal space E.

The following result is a “consistency check”.

Theorem 42.10 (Dold). Let p ∶ E → B be a map. Assume there’s a numerablecover of B, say U , such that for every U ∈ U , the restriction p∣p−1(U) ∶ p−1U → Uis a fibration. (In other words, p is locally a fibration over the base). Then pitself is a fibration.

In particular, one consequence of this theorem is that every numerable fiberbundle is a fibration. Our discussion above tells us that numerable fiber bundlessatisfy the homotopy (and hence path) lifting property. This is great news, aswe will see shortly.

43 Fibrations and cofibrations

Comparing fibers over different points

Let p ∶ E → B be a fibration. Above, we saw that this implies that paths inB “lift” to paths in E. Let us consider a path ω ∶ I → B with ω(0) = a andω(1) = b. Denote by Fa the fiber over a. If the world plays fairly, the pathlifting property of fibrations should beget a (unique9) map Fa → Fb. The goalof this subsection is to construct such a map.

Consider the diagram:

Fa

in0

// E

p

I × Fa

h

66

pr1

// I ω// B.

This commutes since ω(0) = a. Utilizing the homotopy lifting property, thereis a dotted arrow that makes the entire diagram commute. If x ∈ Fa, the imageh(1, x) is in Fb, and h(0, x) = x. This supplies us with a map f ∶ Fa → Fb, givenby f(x) = h(1, x).

We’re now faced with a natural question: is f unique up to homotopy?Namely: if we have two homotopic paths ω0, ω1 with ω0(0) = ω1(0) = a, andω0(1) = ω1(1) = b, along with a given homotopy g ∶ I × I → B between ω0 andω1, such that f0, f1 ∶ Fa → Fb are the associated maps (defined by h0(1, x) andh1(1, x)), respectively, are f0 and f1 homotopic?

9At least up to homotopy.


We have a diagram of the form:

((∂I × I) ∪ (I × 0)) × Fa

in0

// E

p

I × I × Fa pr1

//

33

I × I g// B

To get a homotopy between f0 and f1, we need the dotted arrow to exist.It’s an easy exercise to recognize that our diagram is equivalent to the

following.

I × Fa≃ // ((∂I × I) ∪ (I × 0)) × Fa

in0

// E

p

I × I × Fa ≃

// I × I × Fa pr1

//

44

I × I g// B

Letting W = I × Fa in the definition of a fibration (Definition 42.6) thus givesus the desired lift, i.e., a homotopy f0 ≃ f1.

We can express the uniqueness (up to homotopy) of lifts of homotopic pathsin a functorial fashion. To do so, we must introduce the fundamental groupoidof a space.

Definition 43.1. Let X be a topological space. The fundamental groupoidΠ1(X) of X is a category (in fact, groupoid), whose objects are the pointsof X, and maps are homotopy classes of paths in X. The composition ofcompatible paths σ and ω is defined by:

(σ ⋅ ω)(t) =⎧⎪⎪⎨⎪⎪⎩

ω(2t) 0 ≤ t ≤ 1/2σ(2t − 1) 1/2 ≤ t ≤ 1.

The results of the previous sections can be succinctly summarized in thefollowing neat statement.

Proposition 43.2. Any fibration p ∶ E → B gives a functor Π1(B)→ Top.

This is the beginning of a beautiful story involving fibrations. (The inter-ested reader should look up “Grothendieck construction”.)

Cofibrations

Let i ∶ A → X be a map of spaces. If Y is another topological space, when isthe induced map Y X → Y A a fibration? This is asking for the map i to be“dual” to a fibration.

By the definition of a fibration, we want a lifting:

W //

in0

Y X

I ×W

;;

// Y A.

43. FIBRATIONS AND COFIBRATIONS 137

Adjointing over, we get:

A ×W i×1 //

1×in0

X ×W

A ×W × I //

++

X × I ×W

%%Y.

Again adjointing over, this diagram transforms to:

A //

X

A × I //

))

X × I

##YW .

This discussion motivates the following definition of a “cofibration”: as men-tioned above, this is “dual” to the notion of fibration.

Definition 43.3. A map i ∶ A → X of spaces is said to be a cofibration if itsatisfies the homotopy extension property (sometimes abbreviated as “HEP”):for any space Y , there is a dotted map in the following diagram that makes itcommute:

A //

X

A × I //

))

X × I

""Y.

Again, using the definition of a pushout, the universal example of such aspace Y is the pushout X ∪A (A× I). Equivalently, we are therefore asking forthe existence of a dotted arrow in the following diagram.

X ∪A (A × I) //

&&

X × I

Z,


for any Z. Using the universal property of a pushout, this is equivalent to theexistence of a dotted arrow in the following diagram.

X ∪A (A × I) //

((

X × I

X ∪A (A × I)

Z,

which is, in turn, equivalent to asking X ∪A (A × I) to be a retract of X × I.

Example 43.4. Sn−1 Dn is a cofibration.Properly draw out this fig-ure!Properly draw out this fig-ure!

Figure 3.1: Drawing by John Ni.

In particular, setting n = 1 in this example, 0,1 I is a cofibration.

Here are some properties of the class of cofibrations of CGWH spaces.

• It’s closed under cobase change: if A→X is a cofibration, and A→ B isany map, the pushout B →X ∪A B is also cofibration. (Exercise!)

• It’s closed under finite products. (This is surprising.)

• It’s closed under composition. (Exercise!)

• Any cofibration is a closed inclusion10.This is not obvious; shouldwe include a proof?This is not obvious; shouldwe include a proof?

10Note that the dual statement for fibrations would state: any fibration p ∶ E → B isa quotient map. This is definitely not true: fibrations do not have to be surjective! Forinstance, the trivial map ∅→ B is a fibration. (Fibrations are surjective on path componentsthough, because of path lifting.)

44. HOMOTOPY FIBERS 139

44 Homotopy fibers

An important, but easy, fact about fibrations is that the canonical map X → ∗from any spaceX is a fibration 11. This is because the dotted lift in the diagrambelow can be taken to the map (t,w)↦ f(w):

W

f // X

I ×W

;;

// ∗.

However:

Exercise 44.1. The inclusion ∗ X is not always a cofibration; if it is, saythat ∗ is a nondegenerate basepoint of X. Give an example of a compactlygenerated space X for which this is true.

If ∗ has a neighborhood in X that contracts to ∗, the inclusion ∗ X isa cofibration. Note that if ∗ is a nondegenerate basepoint, the canonical mapXA evÐ→ X is a fibration, where A is a pointed subspace of X (with basepointgiven by ∗). The fiber of ev is exactly the space of pointed maps A→X.

Remark 44.2. In Example 43.4, we saw that 0,1 I is a cofibration; thisimplies that the map Y I → Y × Y (given by ω ↦ (ω(0), ω(1))) is a fibration.

“Fibrant replacements”

The purpose of this subsection is to provide a proof of the following result,which says that every map can be “replaced” (up to homotopy) by a fibration.

Theorem 44.3. For any map f ∶ X → Y , there is a space T (f), along with afibration p ∶ T (f) → Y and a homotopy equivalence X

≃Ð→ T (f), such that thefollowing diagram commutes:

X≃ //

f ""

T (f)

p

Y.

Proof. Consider the map Y I(ev0ev1

)ÐÐÐ→ Y × Y . Let T (f) be the pullback of the

following diagram:T (f) //

Y I

(ev0ev1

)

X × Yf×1// Y × Y.

11Model category theorists get excited about this, because this says that all objects in theassociated model structure on topological spaces is fibrant.


So, as a set, we can write

T (f) = (x,ω) ∈X × Y I ∣f(x) = ω(0).

Let us check that the canonical map T (f) → Y , given by (x,ω) ↦ ω(1), isa fibration. The projection map pr ∶ X × Y → Y is a fibration, so it suffices toshow that the map T (f)→X ×Y is also a fibration. Since fibrations are closedunder pullbacks, we are reduced to checking that the map Y I → Y × Y is afibration; but this is exactly saying that the inclusion 0,1 I is a cofibration,which it is (Example 43.4).

To prove that X is homotopy equivalent to T (f), we need to produce a mapX → T (f). This is equivalent to giving maps X →X×Y and X → Y I that have

compatible images in Y × Y . The first map can be chosen to be X(1f)

ÐÐ→X × Y .Define the map X → Y I by sending x ∈ X to the constant loop at f(x). It isclear that both composites X → X × Y → Y × Y and X → Y I → Y × Y are thesame; this defines a map X → T (f), denoted g. As one can easily check, thecomposite X → T (f) pÐ→ Y is the map f ∶ X → Y that we started off with. Itremains to check that this map X

gÐ→ T (f) is a homotopy equivalence. We willconstruct a homotopy inverse to this map.

The composite X → T (f) → X × Y → X is the identity, so one candidatefor a homotopy inverse to g is the composite

T (f)→X × Y pr1ÐÐ→X.

To prove that this map is indeed a homotopy inverse to g, we need to considerthe composite T (f) → X

gÐ→ T (f), which sends (x,ω) ↦ x ↦ (x, cf(x)) where,recall, cf(x) is the constant path at x. We need to produce a homotopy betweenthis composite and the identity on T (f).

Let s ∈ I. Given ω ∈ Y I , define a new loop ωs by ωs(t) = ω(st). Forinstance, ω1 = ω, and ω0 = cω(0) — so, the loop ωs “sucks in” the point ω(1).This is precisely what we need to produce a homotopy between the compositeT (f)→X

gÐ→ T (f) and idT (f), since the only constraint on (x,ω) ∈ T (f) is onω(0). The following map provides the desired homotopy equivalence X ≃ T (f).

H ∶ I × T (f)→T (f)(s, (x,ω))↦(x,ωs).

Example 44.4 (Path-loop fibration). This is a silly, but important, example.If X = ∗, the space T (f) consists of paths ω in Y such that ω(0) = ∗. In otherwords, T (f) = Y I∗ ; this is called the (based) path space of Y , and is denoted byP (Y,∗), or simply by PY . The fiber of the fibration T (f) = PY → Y consistsof paths that begin and end at ∗, i.e., loops on Y based at ∗. This is denotedΩY , and is called the (based) loop space of Y . The resulting fibration PY → Yis called the path-loop fibration.

44. HOMOTOPY FIBERS 141

Exercise 44.5 (“Cofibrant replacements”). In this exercise, you will prove theanalogue of Theorem 44.3 for cofibrations. Let f ∶ X → Y be any map. Showthat f factors (functorially) as a composite X → M → Y , where X → M is acofibration and M → Y is a homotopy equivalence.

Solution 44.6. Define Mf via the pushout:

Xf //

in0

Y

g

I ×X // Mf.

Define r ∶Mf → Y via r(y) = y on Y and r(x, s) = f(x) on X×I. Then, clearly,rg = idY . There is a homotopy idMf ≃ gr given by the map h ∶Mf × I →Mf ,defined by the formulae

h(y, t) = y, and h((x, s), t) = (x, (1 − t)s).

We now have to check that X → Mf is a cofibration, i.e., that Mf × Iretracts onto Mf × 0 ∪X (X × I). This can be done by “pushing” Y × I toY × 0 and X × I × I down to X × I, while fixing X × 0.

It is easy to see that this factorization is functorial: if f ∶ X → Y is sent tog ∶W → Z via p ∶X →W and q ∶ Y → Z, then Mf →Mg can be defined as thedotted map in the following diagram (which exists, by the universal propertyof the pushout):

Xf //

in0

p

$$

Y

q

##W g

//

Z

X × I //

p×id $$

Mf

""W × I // Mg.

Fix the overlapping arrowshere, I don’t know how todo this...

Fix the overlapping arrowshere, I don’t know how todo this...Homotopy fibers

One way to define the fiber (over a basepoint) of a map f ∶ X → Y is via thepullback

f−1(∗) //

X

f

∗ // Y.


If g ∶W → X is another map such that the composite WgÐ→ X

fÐ→ Y is trivial,the map g factors through f−1(∗). In homotopy theory, maps are generallynot trivial “on the nose”; instead, we usually have a nullhomotopy of a map.Nullhomotopies of composite maps do not factor through this “strict” fiber;this leads to the notion of a homotopy fiber.

Definition 44.7 (Homotopy fiber). The homotopy fiber of a map f ∶ X → Yis the pullback:

F (f,∗) //

T (f) ≃ //

p

X

f||

∗ // Y

As a set, we have

F (f,∗) = (x,ω) ∈X × Y I ∣f(x) = ω(0), ω(1) = ∗. (3.3)

A nullhomotopic composite W →XfÐ→ Y factors as W → F (f,∗)→X

fÐ→ Y .

Warning 44.8. The ordinary fiber and the homotopy fiber of a map are gen-erally not the same! There is a canonical map p−1(∗) → F (p,∗), but it isgenerally not a homotopy equivalence.

Proposition 44.9. Suppose p ∶X → Y is a fibration. Then the canonical mapp−1(∗)→ F (p,∗) is a homotopy equivalence.

You will prove this in a series of exercises.

Exercise 44.10. Prove Proposition 44.9 by working through the followingstatements.

1. Let p ∶ E → B be a fibration. Suppose g ∶ X → B lifts across p up tohomotopy, i.e., there exists a map f ∶ X → E such that p f ≃ g. Provethat there exists a map f ′ ∶ X → E that is homotopic to f , such thatp f ′ = g (on the nose).

2. Show that if p ∶ E → B and p′ ∶ E′ → B are fibrations, and f ∶ E → E′

such that p′ f = p, the map f is a fiber homotopy equivalence: thereis a homotopy inverse g ∶ E′ → E such that g, and the two homotopiesfg ≃ idE′ and gf ≃ idE are all fiber preserving (e.g., p g = p′).

3. Conclude Proposition 44.9.

Before we proceed, recall that we constructed the homotopy fiber by re-placing f ∶X → Y by a fibration. In doing so, we implicitly made a choice: wecould have replaced the map ∗→ Y by a fibration. Are the resulting pullbacksthe same?

45. BARRATT-PUPPE SEQUENCE 143

By replacing ∗→ Y by a fibration (namely, the path-loop fibration), we endup with the following pullback diagram:

F ′(f,∗) //

P (Y,∗) ≃ //

∗

X

f// Y

As a set, we have

F ′(f,∗) = (x,ω) ∈X × Y I such that ω(0) = ∗ and ω(1) = f(x).

Our description of F (f,∗) in (3.3) is almost exactly the same — except that thedirections of the paths are reversed. Thus there’s a homeomorphism F ′(f,∗) ≃F (f,∗) given by reversing directions of paths.

Remark 44.11. One could also replace both f ∶ X → Y and ∗ → Y byfibrations, and the resulting pullback is also homeomorphic to F (f,∗). (Provethis, if the statement is not immediate.)

45 Barratt-Puppe sequence

Fiber sequences

Recall, from the previous section, that we have a pullback diagram:

F (f,∗) //

p

J

PY

p

≃

f−1(∗)

::

// Xf

// Y ∗oo

Consider a pointed map12 f ∶ X → Y (so that f(∗) = ∗). Then, we will writeFf for the homotopy fiber F (f,∗).

Since we’re exploring the homotopy fiber Ff , we can ask the following,seemingly silly, question: what is the fiber of the canonical map p ∶ Ff → X(over the basepoint of X)? This is precisely the space of loops in Y ! Since pis a fibration (recall that fibrations are closed under pullbacks), the homotopy

12Some people call such a map “based”, but this makes it sound like we’re doing chemistry,so we won’t use it.


fiber of p is also the “strict” fiber! Our expanded diagram is now:

ΩY = p−1(∗)

F (f,∗) //

p

PY

p

≃

f−1(∗)

88

// Xf

// Y ∗.oo

It’s easy to see that the composite FfpÐ→ X

fÐ→ Y sends (x,ω) ↦ f(x); this isa pointed nonconstant map. (Note that the basepoint we’re choosing for Ff isthe image of the basepoint in f−1(∗) under the canonical map f−1(∗) Ff .)

While the composite fp ∶ Ff → Y is not zero “on the nose”, it is nullhomo-topic, for instance via the homotopy h ∶ Ff × I → Y , defined by

h(t, (x,ω)) = ω(t).

Exercise 45.1. Let f ∶ X → Y and g ∶W → X be pointed maps. Establish ahomeomorphism between the space of pointed maps W

pÐ→ Ff such that fp = gand the space of pointed nullhomotopies of the composite fg.

This exercise proves that the homotopy fiber is the “kernel” in the homotopycategory of pointed spaces and pointed maps between them.

Define [W,X]∗ = π0(XW∗ ); this consists of the pointed homotopy classes

of maps W → X. We may view this as a pointed set, whose basepoint is theconstant map. FixingW , this is a contravariant functor inX, so there are maps[W,Ff]∗ → [W,X]∗ → [W,Y ]∗. This composite is not just nullhomotopic: itis “exact”! Since we are working with pointed sets, we need to describe whatexactness means in this context: the preimage of the basepoint in [W,Y ]∗ isexactly the image of [W,Ff]∗ → [W,X]∗. (This is exactly a reformulation of

Exercise 45.1.) We say that Ff →XfÐ→ Y is a fiber sequence.

Remark 45.2. Let f ∶ X → Y be a map of spaces, and suppose we have ahomotopy commutative diagram:

ΩY

Ωg

// Ff

// X

h

f // Y

g

ΩY ′ // Ff ′ // X ′

f ′// Y.

Then the dotted map exists, but it depends on the homotopy f ′h ≃ gf .

45. BARRATT-PUPPE SEQUENCE 145

Iterating fiber sequences

Let f ∶ X → Y be a pointed map, as before. As observed above, we have acomposite map Ff

pÐ→XfÐ→ Y , and the strict fiber (homotopy equivalent to the

homotopy fiber) of p is ΩY . This begets a map i(f) ∶ ΩY → Ff ; iterating theprocedure of taking fibers gives:

⋯ // Fp3p4 // Fp2

p3 // Fp1p2 // Ff

p1 // Xf // Y

ΩFp0

≃

OO

i(p2)

;;

// ΩX //

≃

OO

i(p1)

<<

ΩY

≃

OO

i(f)

==

All the pi in the above diagram are fibrations. Each of the dotted maps in theabove diagram can be filled in up to homotopy. The most obvious guess forwhat these dotted maps are is simply ΩX

ΩfÐ→ ΩY . But that is the wrong map!

The right map turns out to be ΩXΩfÐ→ ΩY :

Lemma 45.3. The following diagram commutes to homotopy:

Fp

ΩXΩf

//

i(p)<<

ΩY ;

OO

here, Ωf is the diagonal in the following diagram:

ΩX− //

Ωf

""Ωf

ΩX

Ωf

ΩY −

// ΩY,

where the map − ∶ ΩX → ΩX sends ω ↦ ω.

Proof. typeset the following imagefor the proof... it’s goingto be impossible to writethis up in symbols.

typeset the following imagefor the proof... it’s goingto be impossible to writethis up in symbols.


Figure 3.2: A proof of this lemma.

By the above lemma, we can extend our diagram to:

⋯ // Fp4// Fp3

// Fp2// Fp1

p2 // Ffp1 // X

f // Y

⋯ // ΩFp1 Ωp2//

≃

OO

ΩFf

≃

OO

i(p2)

<<

Ωp // ΩX Ωf //

≃

OO

i(p1)

<<

ΩY

≃

OO

i(f)

==

Ω2X

≃

OO

Ω2f

// Ω2Y

≃

OO

Ωi(p0)

;;

We have a special name for the sequence of spaces sneaking along the bottomof this diagram:

⋯→ Ω2X → Ω2Y → ΩFf → ΩX → ΩY → Ff →XfÐ→ Y ;

this is called the Barratt-Puppe sequence. Applying [W,−]∗ to the Barratt-Puppe sequence of a map f ∶X → Y gives a long exact sequence.

46. RELATIVE HOMOTOPY GROUPS 147

The most important case of this long exact sequence comes from settingW = S0 = ±1; in this case, we get terms like π0(ΩnX). We can identifyπ0(ΩnX) with [Sn,X]∗: to see this for n = 2, recall that Ω2X = (ΩX)S

1

;because (S1)∧n = Sn (see below for a proof of this fact), we find that

(ΩX)S1

≃ (XS1

∗ )S1

∗ =XS1∧S1

∗ =XS2

∗ , (3.4)

as desired.The space ΩX is a group in the homotopy category; this implies that

π0ΩX = π1X is a group! For n > 1, we know that

πn(X) = [(Dn, Sn−1), (X,∗)] = [(In, ∂In), (X,∗)].

Exercise 45.4. Prove that πn(X) is an abelian group for n > 2.

Applying π0 to the Barratt-Puppe sequence (see Equation 3.4) thereforegives a long exact sequence (of groups when the homotopy groups are in degreesgreater than 0, and of pointed sets in degree 0):

⋯→ π2X → π2Y → π1Ff → π1X → π1Y → π0Ff → π0X → π0X.

46 Relative homotopy groups

Spheres and homotopy groups

The functor Ω (sending a space to its based loop space) admits a left adjoint.To see this, recall that ΩX =XS1

∗ , so that

Top∗(W,ΩX) = Top∗(S1 ∧W,X).

Definition 46.1. The reduced suspension ΣW is S1 ∧W .

If A ⊆X, then

X/A ∧ Y /B = (X × Y )/((A × Y ) ∪A×B (X ×B)).

Since S1 = I/∂I, this tells us that ΣX = S1∧X can be identified with I×X/(∂I×X ∪ I × ∗): in other words, we collapse the top and bottom of a cylinder to apoint, as well as the line along a basepoint.

The same argument says that ΣnX (defined inductively as Σ(Σn−1X)) isthe left adjoint of the n-fold loop space functor X ↦ ΩnX. In other words,ΣnX = (S1)∧n ∧X. We claim that S1 ∧ Sn ≃ Sn+1. To see this, note that

S1 ∧ Sn = I/∂I ∧ In ∧ ∂In = (I × In)/(∂I × In ∪ I × ∂In).

The denominator is exactly ∂In+1, so S1 ∧Sn ≃ Sn+k. It’s now easy to see thatSk ∧ Sn ≃ Sk+n.

Definition 46.2. The nth homotopy group of X is πnX = π0(ΩnX).

This is, as we noted in the previous section, [S0,ΩnX]∗ = [Sn,X]∗ =[(In, ∂In), (X,∗)].


The homotopy category

Define the homotopy category of spaces Ho(Top) to be the category whoseobjects are spaces, and whose hom-sets are given by taking π0 of the mappingspace. To check that this is indeed a category, we need to check that if f0, f1 ∶X → Y and g ∶ Y → Z, then gf0 ≃ gf1 — but this is clear. Similarly, we’dneed to check that f0h ≃ f1h for any h ∶ W → X. We can also think aboutthe homotopy category of pointed spaces (and pointed homotopies) Ho(Top∗);this is the category we have been spending most of our time in. Both Ho(Top)and Ho(Top∗) have products and coproducts, but very few other limits orcolimits. From a category-theoretic standpoint, these are absolutely terrible.

Let W be a pointed space. We would like the assignment X ↦XW∗ to be a

homotopy functor. It clearly defines a functor Top∗ → Top∗, so this desire isequivalent to providing a dotted arrow in the following diagram:

Top∗

X↦XW∗ // Top∗

Ho(Top∗) // Ho(Top∗).

Before we can prove this, we will check that a homotopy f0 ∼ f1 ∶X → Y is thesame as a map I+ ∧X → Y . There is a nullhomotopy if the basepoint of I isone of the endpoints, so a homotopy is the same as a map I ×X/I × ∗ → Y .The source is just I+ ∧X, as desired.

A homotopy f0 ≃ f1 ∶ X → Y begets a map (I+ ∧ X)W → YW∗ . For theassignment X ↦ XW

∗ to be a homotopy functor, we need a natural transfor-mation I+ ∧XW

∗ → YW∗ , so this map is not quite what’s necessary. Instead, wecan attempt to construct a map I+ ∧XW

∗ → (I+ ∧X)W∗ .We can construct a general map A ∧ XW

∗ → (A ∧ X)W∗ : there is a mapA ∧XW

∗ → AW∗ ∧XW∗ , given by sending a↦ ca; then the exponential law gives

a homotopy AW∗ ∧XW∗ → (A ∧X)W∗ . This, in turn, gives a map I+ ∧XW

∗ →(I+ ∧X)W∗ → YW∗ , thus making X ↦XW

∗ a homotopy functor.Motivated by our discussion of homotopy fibers, we can study composites

which “behave” like short exact sequences.

Definition 46.3. A fiber sequence in Ho(Top∗) is a composite X → Y → Z

that is isomorphic, in Ho(Top∗), to some composite FfpÐ→ E

fÐ→ B; in otherwords, there exist (possibly zig-zags of) maps that are homotopy equivalences,that make the following diagram commute:

X //

Y //

Z

Ff p

// Ef// B.

Let us remark here that if A′ ∼Ð→ A is a homotopy equivalence, and A →B → C is a fiber sequence, so is the composite A′ ∼Ð→ A→ B → C.

46. RELATIVE HOMOTOPY GROUPS 149

Exercise 46.4. Prove the following statements.

• Ω takes fiber sequences to fiber sequences.

• ΩFf ≃ FΩf . Check this!

We’ve seen examples of fiber sequences in our elaborate study of the Barratt-Puppe sequence.

Example 46.5. Recall our diagram:

⋯ // Fp4// Fp3

// Fp2// Fp1

p2 // Ffp1 // X

f // Y

⋯ // ΩFp1 Ωp2//

≃

OO

ΩFf

≃

OO

i(p2)

<<

Ωp // ΩX Ωf //

≃

OO

i(p1)

<<

ΩY

≃

OO

i(f)

==

Ω2X

≃

OO

Ωf// ΩY

≃

OO

Ωi(f)

;;

The composite Ff → XfÐ→ Y is canonically a fiber sequence. The above

diagram shows that ΩY → FpÐ→ X is another fiber sequence: it is isomorphic

to Fp → F → X in Ho(Top∗). Similarly, the composite ΩXΩfÐ→ ΩY → F is

another fiber sequence; this implies that ΩXΩfÐ→ ΩY → F is also an example of

a fiber sequence (because these two fiber sequences differ by an automorphismof ΩX)

Applying Ω again, we get ΩFΩpÐ→ ΩX

ΩfÐ→ ΩY . Since this is a loopingof a fiber sequence, and taking loops takes fiber sequences to fiber sequences(Exercise 46.4), this is another fiber sequence. Looping again gives another

fiber sequence Ω2YΩiÐ→ ΩF

ΩpÐ→ ΩX. (For the category-theoretically–mindedfolks, this is an unstable version of a triangulated category.)

The long exact sequence of a fiber sequence

As discussed at the end of §45, applying π0 = [S0,−]∗ to the Barratt-Puppesequence associated to a map f ∶X → Y gives a long exact sequence:

⋯ // π2Y

uuπ1F // π1X // π1Y

uuπ0F // π0X.


of pointed sets. The space Ω2X is an abelian group object in Ho(Top) (inother words, the multiplication on Ω2X is commutative up to homotopy). Thisimplies π1(X) is a group, and that πk(X) is abelian for k ≥ 2; hence, in ourdiagram above, all maps (except on π0) are group homomorphisms.

Consider the case when X → Y is the inclusion i ∶ AX of a subspace. Inthis case,

Fi = (a,ω) ∈ A ×XI∗ ∣ω(1) = a;

this is just the collection of all paths that begin at ∗ ∈ A and end in A. Thismotivates the definition of relative homotopy groups:

Definition 46.6. Define:

πn(X,A,∗) = πn(X,A) ∶= πn−1Fi = [(In, ∂In, (∂In×I)∪(In−1×0)), (X,A,∗)].

We have a sequence of inclusions

∂In × I ∪ In−1 × 0 ⊂ ∂In ⊂ In.

One can check that

πn−1Fi = [(In, ∂In, (∂In × I) ∪ (In−1 × 0)), (X,A,∗)].

This gives a long exact sequence on homotopy, analogous to the long exactsequence in relative homology:

⋯ // π2(X,A)

uuπ1A // π1X // π1(X,A)

uuπ0A // π0X

(3.5)

47 Action of π1, simple spaces, and the Hurewicztheorem

In the previous section, we constructed a long exact sequence of homotopygroups:

⋯ // π2(X,A)

uuπ1A // π1X // π1(X,A)

uuπ0A // π0X,

47. ACTION OF π1, SIMPLE SPACES, AND THE HUREWICZTHEOREM 151

which looks suspiciously similar to the long exact sequence in homology. Thegoal of this section is to describe a relationship between homotopy groups andhomology groups.

Before we proceed, we will need the following lemma.

Lemma 47.1 (Excision). If A ⊆X is a cofibration, there is an isomorphism

H∗(X,A) ≃Ð→ H∗(X/A).

Under this hypothesis,

X/A ≃Mapping cone of i ∶ A→X;

here, the mapping cone is the homotopy pushout in the following diagram:

Ai //

in1

X

CA // X ∪A CA,

where CA is the cone on A, defined by

CA = A × I/A × 0.

This lemma is dual to the statement that the homotopy fiber is homotopyequivalent to the strict fiber for fibrations.

Unfortunately, π∗(X,A) is definitely not π∗(X/A)! For instance, there is acofibration sequence

S1 →D2 → S2.

We know that π∗S1 is just Z in dimension 1, and is zero in other dimensions.On the other hand, we do not, and probably will never, know the homotopygroups of S2. (A theorem of Edgar Brown in [Bro57] says that these groupsare computable, but this is super-exponential.)

π1-action

There is more structure in the long exact sequence in homotopy groups that weconstructed last time, coming from an action of π1(X). There is an action ofπ1(X) on πn(X): if x, y are points in X, and ω ∶ I →X is a path with ω(0) = xand ω(1) = y, we have a map fω ∶ πn(X,x) → πn(X,y); this, in particular,implies that π1(X,∗) acts on πn(X,∗). When n = 1, the action π1(X) on itselfis by conjugation.

In fact, one can also see that π1(A) acts on πn(X,A,∗). It follows (byconstruction) that all maps in the long exact sequence of Equation (3.5) areequivariant for this action of π1(A). Moreover:

Proposition 47.2 (Peiffer identity). Let α,β ∈ π2(X,A). Then (∂α) ⋅ β =αβα−1.


Definition 47.3. A topological space X is said to be simply connected if it ispath connected, and π1(X,∗) = 1.

Let p ∶ E → B be a covering space with E and B connected. Then, the fibersare discrete, hence do not have any higher homotopy. Using the long exactsequence in homotopy groups, we learn that πn(E)→ πn(B) is an isomorphismfor n > 1, and that π1(E) is a subgroup of π1(B) that classifies the coveringspace. In general, we know from Exercise 47.7 that ΩB acts on the homotopyfiber Fp. Since Ff is discrete, this action factors through π0(ΩB) ≃ π1(B).

Definition 47.4. A space X is said to be n-connected if πi(X) = 0 for i ≤ n.

Note that this is a well-defined condition, although we did not specify thebasepoint: 0-connected implies path connected. Suppose E → B is a coveringspace, with the total space E being n-connected. Then, Hopf showed that thegroup π1(B) determines the homology Hi(B) in dimensions i < n.

Sometimes, there are interesting spaces which are not simply connected, forwhich the π1-action is nontrivial.

Example 47.5. Consider the space S1∨S2. The universal cover is justR, witha 2-sphere S2 stuck on at every integer point. This space is simply connected,so the Hurewicz theorem says that π2(E) ≃ H2(E). Since the real line iscontractible, we can collapse it to a point: this gives a countable bouquet of2-spheres. As a consequence, π2(E) ≃H2(E) =⊕∞

i=0 Z.There is an action of π1(S1∨S2) on E: the action does is shift the 2-spheres

on the integer points of R (on E) to the right by 1 (note that π1(S1∨S2) = Z).This tells us that π2(E) ≃ Z[π1(B)] as a Z[π1(B)]-module; this is the sameaction of π1(E) on π2(E). We should be horrified: S1 ∨ S2 is a very simple3-complex, but its homotopy is huge!

Simply-connectedness can sometimes be a restrictive condition; instead, tosimplify the long exact sequence, we define:

Definition 47.6. A topological space X is said to be simple if it is path-connected, and π1(X) acts trivially on πn(X) for n ≥ 1.

Note, in particular, that π1(X) is abelian for a simple space.Being simple is independent of the choice of basepoint. If ω ∶ x ↦ x′ is a

path in X, then ω♯ ∶ πn(X,x) → πn(X,x′) is a group isomorphism. There isa (trivial) action of π1(X,x) on πn(X,x), and another (potentially nontriv-ial) action of π1(X,x′) on πn(X,x′). Both actions are compatible: hence, ifπ1(X,x) acts trivially, so does π1(X,x′).

If X is path-connected, there is a map πn(X,∗)→ [Sn,X]. It is clear thatthis map is surjective, so one might expect a factorization:

πn(X,∗) // //

((

[Sn,X]

π1(X,∗)/πn(X,∗)

OO

47. ACTION OF π1, SIMPLE SPACES, AND THE HUREWICZTHEOREM 153

Exercise 47.7. Prove that π1(X,∗)/πn(X,∗) ≃ [Sn,X]. To do this, workthrough the following exercises.

Let f ∶ X → Y be a map of spaces, and let ∗ ∈ Y be a fixed basepointof Y . Denote by Ff the homotopy fiber of f ; this admits a natural fibrationp ∶ Ff → X, given by (x,σ) ↦ x. If ΩY denotes the (based) loop space of Y ,we get an action ΩY × Ff → Ff , given by

(ω, (x,σ))↦ (x,σ ⋅ ω),

where σ ⋅ ω is the concatenation of σ and ω, defined, as usual, by

σ ⋅ ω(t) =⎧⎪⎪⎨⎪⎪⎩

ω(2t) 0 ≤ t ≤ 1/2σ(2t − 1) 1/2 ≤ t ≤ 1.

(Note that when X is the point, this defines a “multiplication” ΩY ×ΩY → ΩY ;this is associative and unital up to homotopy.) On connected components, wetherefore get an action of π0ΩY ≃ π1Y on π0Ff .

There is a canonical map

Ff ×ΩY → Ff ×X Ff,

given by ((x,σ), ω) ↦ ((x,σ), (x,σ) ⋅ ω). Prove that this map is a homotopyequivalence (so that the action of ΩY on Ff is “free”), and conclude that twoelements in π0Ff map to the same element of π0X if and only if they are inthe same orbit under the action of π1Y .

LetX be path connected, with basepoint ∗ ∈X. Conclude that π1(X,∗)/πn(X,∗) ≃[Sn,X] by proving that the surjection πn(X,∗) → [Sn,X] can be identifiedwith the orbit projection for the action of π1(X,∗) on πn(X,∗).

If X is simple, then the quotient π1(X,∗)/πn(X,∗) is simply πn(X,∗), soExercise 47.7 implies that πn(X,∗) ≅ [Sn,X] — independently of the base-point; in other words, these groups are canonically the same, i.e., two pathsω,ω′ ∶ x→ y give the same map ω♯ = ω′♯ ∶ πn(X,x)→ πn(X,y).

Exercise 47.8. A H-space is a pointed space X, along with a pointed mapµ ∶ X × X → X, such that the maps x ↦ µ(x,∗) and x ↦ µ(∗, x) are bothpointed homotopic to the identity. In this exercise, you will prove that pathconnected H-spaces are simple.

Denote by C the category of pairs (G,H), where G is a group that acts onthe group H (on the left); the morphisms in C are pairs of homomorphismswhich are compatible with the group actions. This category has finite products.Explain what it means for an object of C to have a “unital multiplication”, andprove that any object (G,H) of C with a unital multiplication has G and Habelian, and that the G-action on H is trivial. Conclude from this that pathconnected H-spaces are simple.


Hurewicz theorem

Definition 47.9. Let X be a path-connected space. The Hurewicz map h ∶πn(X,∗)→Hn(X)is defined as follows: an element in πn(X,∗) is representedby α ∶ Sn →X; pick a generator σ ∈Hn(Sn), and send

α ↦ α∗(σ) ∈Hn(X).

We will see below that h is in fact a homomorphism.This is easy in dimension 0: a point is a 0-cycle! In fact, we have an

isomorphism H0(X) ≃ Z[π0(X)]. (This isomorphism is an example of theHurewicz theorem.)

When n = 1, we have h ∶ π1(X,∗) → H1(X). Since H1(X) is abelian, thisfactors as π1(X,∗) → π1(X,∗)ab → H1(X). The Hurewicz theorem says thatthe map π1(X,∗)ab → H1(X) is an isomorphism. We will not prove this here;see [Hat15, Theorem 2A.1] for a proof.

The Hurewicz theorem generalizes these results to higher dimensions:

Theorem 47.10 (Hurewicz). Suppose X is a space for which πi(X) = 0 fori < n, where n ≥ 2. Then the Hurewicz map h ∶ πn(X) → Hn(X) is an isomor-phism.

Before the word “isomorphism” can make sense, we need to prove that h is ahomomorphism. Let α,β ∶ Sn →X be pointed maps. The product αβ ∈ πn(X)is the composite:

αβ ∶ Sn δ, pinching along the equatorÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐ→ Sn ∨ Sn β∨αÐÐ→X ∨X ∇Ð→X,

where ∇ ∶X ∨X →X is the fold map, defined by:

X

1

##X ∨X ∇ // X

X

OO

1

;;

To show that h is a homomorphism, it suffices to prove that for two mapsα,β ∶ Sn → X, the induced maps on homology satisfy (α + β)∗ = α∗ + β∗ —then,

h(α + β) = (α + β)∗(σ) = α∗(σ) + β∗(σ) = h(α) + h(β).

To prove this, we will use the pinch map δ ∶ Sn → Sn ∨ Sn, and the quotientmaps q1, q2 ∶ Sn ∨ Sn → Sn; the induced map Hn(Sn) → Hn(Sn) ⊕Hn(Sn) isgiven by the diagonal map a↦ (a, a). It follows from the equalities

(f ∨ g)ι1 = f, (f ∨ g)ι2 = g,

48. EXAMPLES OF CW-COMPLEXES 155

where ι1, ι2 ∶ Sn Sn∨Sn are the inclusions of the two wedge summands, thatthe map (f ∨ g)∗((ι1)∗ + (ι2)∗) sends (x,0) to f∗(x), and (0, x) to g∗(x). Inparticular,

(x,x)↦ f∗(x) + g∗(x),so the composite Hn(Sn) → Hn(X) sends x ↦ (x,x) ↦ f∗(x) + g∗(x). Thiscomposite is just (f + g)∗(x), since the composite (f ∨ g)δ induces the map(f + g)∗ on homology.

It is possible to give an elementary proof of the Hurewicz theorem, but wewon’t do that here: instead, we will prove this as a consequence of the Serrespectral sequence.

Example 47.11. Since πi(Sn) = 0 for i < n, the Hurewicz theorem tells usthat πn(Sn) ≃Hn(Sn) ≃ Z.

Example 47.12. Recall the Hopf fibration S1 → S3 ηÐ→ S2. The long exactsequence on homotopy groups tells us that πi(S3) ≃Ð→ πi(S2) for i > 2, wherethe map is given by α ↦ ηα. As we saw above, π3(S3) = Z, so π3(S2) ≃ Z,generated by η.

One can show that π4n−1(S2n) ⊗Q ≃ Q. A theorem of Serre’s says that,other than πn(Sn), these are the only non-torsion homotopy groups of spheres.

48 Examples of CW-complexes

Bringing you up-to-speed on CW-complexes

Definition 48.1. A relative CW-complex is a pair (X,A), together with afiltration

A =X−1 ⊆X0 ⊆X1 ⊆ ⋯ ⊆X,such that for all n, the space Xn sits in a pushout square:

∐α∈Σn Sn−1 //

attaching maps

∐α∈ΣnDn

characteristic maps

Xn−1

// Xn,

and X = limÐ→Xn.

If A = ∅, this is just the definition of a CW-complex. In this case, X is alsocompactly generated. (This is one of the reasons for defining compactly gen-erated spaces.) Often, X will be a CW-complex, and A will be a subcomplex.If A is Hausdorff, then so is X.

If X and Y are both CW-complexes, define

(X ×k Y )n = ⋃i+j=n

Xi × Yj ;

this gives a CW-structure on the product X ×kY . Any closed smooth manifoldadmits a CW-structure.


Example 48.2 (Complex projective space). The complex projective n-spaceCPn is a CW-complex, with skeleta CP0 ⊆ CP1 ⊆ ⋯ ⊆ CPn. Indeed, any

complex line through the origin meets the hemisphere defined by⎛⎜⎝

z0

⋮zn

⎞⎟⎠

with

∣∣z∣∣ = 1, I(zn) = 0, and R(zn) ≥ 0. Such a line meets this hemisphere (whichis just D2n) at one point — unless it’s on the equator; this gives the desiredpushout diagram:

S2n−1 //

D2n

CPn−1 // CPn.

Example 48.3 (Grassmannians). Let V = Rn or Cn or Hn, for some fixedn. Define the Grassmannian Grk(Rn) to be the collection of k-dimensionalsubspaces of V . This is equivalent to specifying a k × n rank k matrix.

For instance, Gr2(R4) is, as a set, the disjoint union of:

( 11) ,( 1 ∗

1) ,(1 ∗ ∗

1) ,( 1 ∗

1 ∗) ,(1 ∗ ∗

1 ∗) ,(1 ∗ ∗1 ∗ ∗) .

Motivated by this, define:

Definition 48.4. The j-skeleton of Gr(V ) is

skjGrk(V ) = A ∶ row echelon representation with at most j free entries.

For a proof that this is indeed a CW-structure, see [MS74, §6].

The top-dimensional cell tells us that

dim Grk(Rn) = k(n − k).

The complex Grassmannian has cells in only even dimensions. We know thehomology of Grassmannians: Poincaré duality is visible if we count the numberof cells. (Consider, for instance, in Gr2(R4)).

49 Relative Hurewicz and J. H. C. Whitehead

Here is an “alternative definition” of connectedness:

Definition 49.1. Let n ≥ 0. The space X is said to be (n − 1)-connected if,for all 0 ≤ k ≤ n, any map f ∶ Sk−1 →X extends:

Sk−1

// X

Dk

∃

==

49. RELATIVE HUREWICZ AND J. H. C. WHITEHEAD 157

When n = 0, we know that S−1 = ∅, and D0 = ∗. Thus being (−1)-connectedis equivalent to being nonempty. When n = 1, this is equivalent to path con-nectedness. You can check that this is exactly the same as what we said before,using homotopy groups.

As is usual in homotopy theory, there is a relative version of this definition.

Definition 49.2. Let n ≥ 0. Say that a pair (X,A) is n-connected if, for all0 ≤ k ≤ n, any map f ∶ (Dk, Sk−1)→ (X,A) extends:

(Dk, Sk−1)f //

(X,A)

(A,A)

99

up to homotopy. In other words, there is a homotopy between f and a mapwith image in A, such that f ∣Sk−1 remains unchanged.

0-connectedness implies that A meets every path component of X. Equiv-alently:

Definition 49.3. (X,A) is n-connected if:

• when n = 0, the map π0(A)→ π0(X) surjects.

• when n > 0, the canonical map π0(A) ≃Ð→ π0(X) is an isomorphism, andfor all a ∈ A, the group πk(X,A,a) vanishes for 1 ≤ k ≤ n. (Equivalently,π0(A) ≃Ð→ π0(X) and πk(A,a)→ πk(X,A) is an isomorphism for 1 ≤ k < nand is onto for k = n.)

The relative Hurewicz theorem

Assume that π0(A) = ∗ = π0(X), and pick a ∈ A. Then, we have a comparisonof long exact sequences, arising from the classical (i.e., non-relative) Hurewiczmap:

⋯ // π1(A) //

h

π1(X) //

h

π1(X,A) //

h

π0(A) //

h

π0(X)

h

⋯ // H1(A) // H1(X) // H1(X,A) // H0(A) // H0(X) // H0(X,A)

To define the relative Hurewicz map, let α ∈ πn(X,A), so that α ∶ (Dn, Sn−1)→(X,A); pick a generator ofHn(Dn, Sn−1), and send it to an element ofHn(X,A)via the induced map α∗ ∶Hn(Dn, Sn−1)→Hn(X,A).

BecauseHn(X,A) is abelian, the group π1(A) acts trivially onHn(X,A); inother words, h(ω(α)) = h(α). Consequently, the relative Hurewicz map factorsthrough the group π†

n(X,A), defined to be the quotient of πn(X,A) by thenormal subgroup generated by (ωα)α−1, where ω ∈ π1(A) and α ∈ πn(X,A).This begets a map π†

n(X,A)→Hn(X,A).


Theorem 49.4 (Relative Hurewicz). Let n ≥ 1, and assume (X,A) is n-connected. Then Hk(X,A) = 0 for 0 ≤ k ≤ n, and the map π†

n+1(X,A) →Hn+1(X,A) constructed above is an isomorphism.

We will prove this later using the Serre spectral sequence.

The Whitehead theorems

J. H. C. Whitehead was a rather interesting character. He raised pigs.Whitehead was interested in determining when a continuous map f ∶X → Y

that is an isomorphism in homology or homotopy is a homotopy equivalence.

Definition 49.5. Let f ∶X → Y and n ≥ 0. Say that f is a n-equivalence13 if,for every ∗ ∈ Y , the homotopy fiber F (f,∗) is (n − 1)-connected.

For instance, f being a 0-equivalence simply means that π0(X) surjectsonto π0(Y ) via f . For n > 0, this says that f ∶ π0(X) → π0(Y ) is a bijection,and that for every ∗ ∈X:

πk(X,∗)→ πk(Y, f(∗)) is⎧⎪⎪⎨⎪⎪⎩

an isomorphism 1 ≤ k < nonto k = n.

Using the “mapping cylinder” construction (see Exercise 44.5), we can alwaysassume f ∶ X → Y is a cofibration; in particular, that X Y is a closedinclusion. Then, f ∶ X → Y is an n-equivalence if and only if (Y,X) is n-connected.

Theorem 49.6 (Whitehead). Suppose n ≥ 0, and f ∶ X → Y is n-connected.Then:

Hk(X) fÐ→Hk(Y ) is⎧⎪⎪⎨⎪⎪⎩

an isomorphism 1 ≤ k < nonto k = n.

Proof. When n = 0, because π0(X) → π0(Y ) is surjective, we learn thatH0(X) ≃ Z[π0(X)] → Z[π0(Y )] ≃ H0(Y ) is surjective. To conclude, usethe relative Hurewicz theorem. (Note that the relative Hurewicz dealt withπ†n(X,A), but the map πn(X,A)→ π†

n(X,A) is surjective.)

The case n =∞ is special.

Definition 49.7. f is a weak equivalence (or an ∞-equivalence, to makeit sound more impressive) if it’s an n-equivalence for all n, i.e., it’s a π∗-isomorphism.

Putting everything together, we obtain:

Corollary 49.8. A weak equivalence induces an isomorphism in integral ho-mology.

13Some sources sometimes use “n-connected”.

50. CELLULAR APPROXIMATION, CELLULAR HOMOLOGY,OBSTRUCTION THEORY 159

How about the converse?If H0(X) → H0(Y ) surjects, then the map π0(X) → π0(Y ) also surjects.

Now, assume X and Y path connected, and that H1(X) surjects onto H1(Y ).We would like to conclude that π1(X) → π1(Y ) surjects. Unfortunately, thisis hard, because H1(X) is the abelianization of π1(X). To forge onward, wewill simply give up, and assume that π1(X)→ π1(Y ) is surjective.

Suppose H2(X) → H2(Y ) surjects, and that f∗ ∶ H1(X) ≃Ð→ H1(Y ). Weknow that H2(Y,X) = 0. On the level of the Hurewicz maps, we are still stuck,because we only obtain information about π†

2. Let us assume that π1(X) istrivial14. Under this assumption, we find that π1(Y ) = 0. This implies π2(Y,X)is trivial. Arguing similarly, we can go up the ladder.

Theorem 49.9 (Whitehead). Let n ≥ 2, and assume that π1(X) = 0 = π1(Y ).Suppose f ∶X → Y such that:

Hk(X)→Hk(Y ) is⎧⎪⎪⎨⎪⎪⎩

an isomorphism 1 ≤ k < nonto k = n;

then f is an n-equivalence.

Setting n =∞, we obtain:

Corollary 49.10. Let X and Y be simply-connected. If f induces an isomor-phism in homology, then f is a weak equivalence.

This is incredibly useful, since homology is actually computable! To wrapup the story, we will state the following result, which we will prove in a latersection.

Theorem 49.11. Let Y be a CW-complex. Then a weak equivalence f ∶X → Yis in fact a homotopy equivalence.

50 Cellular approximation, cellular homology,obstruction theory

In previous sections, we saw that homotopy groups play well with (mapsbetween) CW-complexes. Here, we will study maps between CW-complexesthemselves, and prove that they are, in some sense, “cellular” themselves.

Cellular approximation

Definition 50.1. Let X and Y be CW-complexes, and let A ⊆ X be a sub-complex. Suppose f ∶X → Y is a continuous map. We say that f ∣A is skeletal15

if f(Σn) ⊆ Yn.14This is a pretty radical assumption; for the following argument to work, it would techni-

cally be enough to ask that π1(X) acts trivially on π2(Y,X): but this is basically impossibleto check.

15Some would say cellular.


Note that a skeletal map might not take cells in A to cells in Y , but it takesn-skeleta to n-skeleta.

Theorem 50.2 (Cellular approximation). In the setup of Definition 50.1, themap f is homotopic to some other continuous map f ′ ∶ X → Y , relative to A,such that f ′ is skeletal on all of X.

To prove this, we need the following lemma.

Lemma 50.3 (Key lemma). Any map (Dn, Sn−1)→ (Y,Yn−1) factors as:

(Dn, Sn−1) //

''

(Y,Yn−1)

(Yn, Yn−1)

OO

“Proof.” Since Dn is compact, we know that f(Dn) must lie in some finitesubcomplex K of Y . The map Dn → K might hit some top-dimensional cellem ⊆K, which does not have anything attached to it; hence, we can homotopethis map to miss a point, so that it contracts onto a lower-dimensional cell.Iterating this process gives the desired result.

Using this lemma, we can conclude the cellular approximation theorem.

“Proof” of Theorem 50.2. We will construct the homotopy f ≃ f ′ one cell at atime. Note that we can replace the space A by the subspace to which we haveextended the homotopy.

Consider a single cell attachment A→ A ∪Dm; then, we have

A

skeletal

// A ∪Dm

may not be skeletal

Y

Using the “compression lemma” from above, the rightmost map factors (up tohomotopy) as the composite A ∪Dm → Ym → Y . Unfortunately, we have notextended this map to the whole of X, although we could do this if we knewthat the inclusion of a subcomplex is a cofibration. But this is true: there isa cofibration Sn−1 → Dn, and so any pushout of these maps is a cofibration!This allows us to extend; we now win by iterating this procedure.

As a corollary, we find:

Exercise 50.4. The pair (X,Xn) is n-connected.

50. CELLULAR APPROXIMATION, CELLULAR HOMOLOGY,OBSTRUCTION THEORY 161

Cellular homology

Let (X,A) be a relative CW-complex with A ⊆ Xn−1 ⊆ Xn ⊆ ⋯ ⊆ X. In theprevious part that H∗(Xn,Xn−1) ≃ H∗(Xn/Xn−1). More generally, if B → Yprovide a link!provide a link!is a cofibration, there is an isomorphism (see [Bre93, p. 433]):

H∗(Y,B) ≃ H∗(Y /B).

Since Xn/Xn−1 = ⋁α∈Σn Snα, we find that

H∗(Xn,Xn−1) ≃ Z[Σn] = Cn.

The composite Sn−1 →Xn−1 →Xn−1/Xn−2 is called a relative attaching map.There is a boundary map d ∶ Cn → Cn−1, defined by

d ∶ Cn =Hn(Xn,Xn−1)∂Ð→Hn−1(Xn−1)→Hn−1(Xn−1,Xn−2) = Cn−1.

Exercise 50.5. Check that d2 = 0.

Using the resulting chain complex, denoted C∗(X,A), one can prove thatthere is an isomorphism

Hn(X,A) ≃Hn(C∗(X,A)).

(In the previous part, we proved this for CW-pairs, but not for relative CW- provide a link!provide a link!complexes.) The incredibly useful cellular approximation theorem thereforetells us that the effect of maps on homology can be computed.

Of course, the same story runs for cohomology: one gets a chain complexwhich, in dimension n, is given by

Cn(X,A;π) = Hom(Cn(X,A), π) = Map(Σn, π),

where π is any abelian group.

Obstruction theory

Using the tools developed above, we can attempt to answer some concrete, anduseful, questions.

Question 50.6. Let f ∶ A→ Y be a map from a space A to Y . Suppose (X,A)is a relative CW-complex. When can we find an extension in the diagrambelow?

X

A?

OO

f// Y


The lower level obstructions can be worked out easily:

A

// X0

// X1

vv∅ ≠ Y

Thus, for instance, if two points in X0 are connected in X1, we only have tocheck that they are also connected in Y .

For n ≥ 2, we can form the diagram:

∐α∈Σn Sn−1α

f // _

Xn−1g //

Y

∐ΣnDnα

// Xn

==

The desired extension exists if the composite Sn−1α

fαÐ→ Xn−1 → Y is nullhomo-topic.

Clearly, g fα ∈ [Sn−1, Y ]. To simplify the discussion, let us assume thatY is simple; then, Exercise 47.7 says that [Sn−1, Y ] = πn−1(Y ). This proce-dure begets a map Σn

θÐ→ πn−1(Y ), which is a n-cochain, i.e., an element ofCn(X,A;πn−1(Y )). It is clear that θ = 0 if and only if the map g extends toXn → Y .

Proposition 50.7. θ is a cocycle in Cn(X,A;πn−1(Y )), called the “obstructioncocycle”.

Proof. θ gives a map Hn(Xn,Xn−1) → πn−1(Y ). We would like to show thatthe composite

Hn+1(Xn+1,Xn)∂Ð→Hn(Xn)→Hn(Xn,Xn−1)

θÐ→ πn−1(Y )

is trivial. We have the long exact sequence in homotopy of a pair (see Equation(3.5)):

πn+1(Xn+1,Xn)

// // Hn+1(Xn+1,Xn)

∂

πn(Xn) //

Hn(Xn)

πn(Xn,Xn−1) // //

∂

Hn(Xn,Xn−1)

θ

πn−1(Xn−1) g∗

// πn−1(Y )

This diagram commutes, so θ is indeed a cocycle.

51. CONCLUSIONS FROM OBSTRUCTION THEORY 163

Our discussion above allows us to conclude:

Theorem 50.8. Let (X,A) be a relative CW-complex and Y a simple space.Let g ∶Xn−1 → Y be a map from the (n−1)-skeleton of X. Then g∣Xn−2 extendsto Xn if and only if [θ(g)] ∈Hn(X,A;πn−1(Y )) is zero.

Corollary 50.9. If Hn(X,A;πn−1(Y )) = 0 for all n > 2, then any map A→ Yextends to a map X → Y (up to homotopy16); in other words, there is a dottedlift in the following diagram:

A //

Y

X

>>

For instance, every mapA→ Y factors through the cone ifHn(CA,A;πn−1(Y )) ≃Hn−1(A;πn−1(Y )) = 0.

51 Conclusions from obstruction theory

The main result of obstruction theory, as discussed in the previous section, isthe following.

Theorem 51.1 (Obstruction theory). Let (X,A) be a relative CW-complex,and Y a simple space. The map [X,Y ]→ [A,Y ] is:

1. is onto if Hn(X,A;πn−1(Y )) = 0 for all n ≥ 2.

2. is one-to-one if Hn(X,A;πn(Y )) = 0 for all n ≥ 1.

Remark 51.2. The first statement implies the second. Indeed, suppose wehave two maps g0, g1 ∶X → Y and a homotopy h ∶ g0∣A ≃ g0∣A. Assume the firststatement. Consider the relative CW-complex (X × I,A× I ∪X ×∂I). Because(X,A) is a relative CW-complex, the map AX is a cofibration; this impliesthat the map A × I ∪X × ∂I →X × I is also a cofibration.

Hn(X × I,A × I ∪X × ∂I;π) ≃ Hn(X × I/(A × I ∪X × ∂I);π)=Hn(ΣX/A;π) ≃ Hn−1(X/A;π).

We proved the following statement in the previous section.

Proposition 51.3. Suppose g ∶Xn−1 → Y is a map from the (n−1)-skeleton ofX to Y . Then g∣Xn−2 extends to Xn → Y iff [θ(g)] = 0 in Hn(X,A;πn−1(Y )).

An immediate consequence is the following.

Theorem 51.4 (CW-approximation). Any space admits a weak equivalencefrom a CW-complex.

16In fact, this condition is unnecessary, since the inclusion of a subcomplex is a cofibration.


This tells us that studying CW-complexes is not very restrictive, if we workup to weak equivalence.

It is easy to see that if W is a CW-complex and f ∶ X → Y is a weakequivalence, then [W,X] ≃Ð→ [W,Y ]. We can now finally conclude the result ofTheorem 49.11:

Corollary 51.5. Let X and Y be CW-complexes. Then a weak equivalencef ∶X → Y is a homotopy equivalence.

Postnikov and Whitehead towers

Let X be path connected. There is a space X≤n, and a map X → X≤n suchthat πi(X≥n) = 0 for i > n, and πi(X) ≃Ð→ πi(X≤n) for i ≤ n. This pair (X,X≤n)is essentially unique up to homotopy; the space X≤n is called the nth Postnikovsection of X. Since Postnikov sections have “simpler” homotopy groups, we cantry to understand X by studying each of its Postnikov sections individually,and then gluing all the data together.

Suppose A is some abelian group. We saw, in the first partthat there is aprovide a linkprovide a linkspace M(A,n) with homology given by:

Hi(M(A,n)) =⎧⎪⎪⎨⎪⎪⎩

A i = n0 i ≠ n.

This space was constructed from a free resolution 0 → F1 → F0 → A → 0 of A.We can construct a map ⋁Sn → ⋁Sn which realizes the first two maps; coningthis off gets M(A,n). By Hurewicz, we have:

πi(M(A,n)) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

0 i < nA i = n?? i > n

It follows that, when we look at the nth Postnikov section ofM(A,n), we have:

πi(M(A,n)≤n) =⎧⎪⎪⎨⎪⎪⎩

A i = n0 i ≠ n.

In some sense, therefore, this Postnikov section is a “designer homotopy type”.It deserves a special name: M(A,n)≤n is called an Eilenberg-MacLane space,and is denoted K(A,n). By the fiber sequence ΩX → PX → X with PX ≃ ∗,we find that ΩK(π,n) ≃K(π,n− 1). Eilenberg-MacLane spaces are unique upto homotopy.

Note that n = 1, A does not have to be abelian, but you can still constructK(A,1). This is called the classifying space of G; such spaces will be discussedin more detail in the next chapter. Examples are in abundance: if Σ is a closedsurface that is not S2 or R2, then Σ ≃ K(π1(Σ),1). Perhaps simpler is theidentification S1 ≃K(Z,1).

51. CONCLUSIONS FROM OBSTRUCTION THEORY 165

Example 51.6. We can identify K(Z,2) as CP∞. To see this, observe thatwe have a fiber sequence S1 → S2n+1 → CPn. The long exact sequence inhomotopy tells us that the homotopy groups of CPn are the same as thehomotopy groups of S1, until π∗S2n+1 starts to interfere. As n grows, weobtain a fibration S1 → S∞ → CP∞. Since S∞ is weakly contractible (it hasno nonzero homotopy groups), we get the desired result.

Example 51.7. Similarly, we can identify K(Z/2Z,1) as RP∞.

Since π1(K(A,n)) = 0 for n > 1, it follows that K(A,n) is automatically asimple space. This means that

[Sk,K(A,n)] = πk(K(A,n)) =Hn(Sk,A).

In fact, a more general result is true:

Theorem 51.8 (Brown representability). If X is a CW-complex, then [X,K(A,n)] =Hn(X;A).

We will not prove this here, but one can show this simply by showing thatthe functor [−,K(A,n)] satisfies the Eilenberg-Steenrod axioms. Somehow,these Eilenberg-MacLane spaces K(A,n) completely capture cohomology indimension n.

IfX is a CW-complex, then we may assume thatX≤n is also a CW-complex.(Otherwise, we can use cellular approximation and then kill homotopy groups.)Let us assume that X is path connected; then X≤1 = K(π1(X),1). We maythen form a (commuting) tower:

⋮

⋯oo

X≤3

K(π3(X),3)oo

X≤2

K(π2(X),2)oo

X //

>>

FF

II

X≤1 K(π1(X),1),

since K(πn(X), n) → X≤n → X≤n−1 is a fiber sequence. This decomposition ofX is called the Postnikov tower of X.


Denote by X>n the fiber of the map X → X≤n (for instance, X>1 is theuniversal cover of X); then, we have

⋯ //

⋯ // ⋮

⋯oo

X>3//

X // X≤3

K(π3(X),3)oo

X>2//

X // X≤2

K(π2(X),2)oo

X>1//

X // X≤1

K(π1(X),1)

X X // ∗

The leftmost tower is called the Whitehead tower of X, named after GeorgeWhitehead.

I can take the fiber of X>1 →X, and I get K(π1(X),0); more generally, thefiber of X>n →X>n−1 is K(πn(X), n − 1). This yields the following diagram:

⋯ ⋮

⋮ ⋮

⋯

K(π3(X),2) // X>3//

X // X≤3

K(π3(X),3)oo

K(π2(X),1) // X>2//

X // X≤2

K(π2(X),2)oo

K(π1(X),0) // X>1//

X // X≤1

K(π1(X),1)

X X // ∗

We can construct Eilenberg-MacLane spaces as cellular complexes by at-taching cells to the sphere to kill its higher homotopy groups. The complexityof homotopy groups, though, shows us that attaching cells to compute thecohomology of Eilenberg-MacLane spaces is not feasible.

Chapter 4

Vector bundles

52 Vector bundles, principal bundles

Let X be a topological space. A point in X can be viewed as a map ∗ → X;this is a cross section of the canonical map X → ∗. Motivated by this, we willdefine a vector space over B to be a space E → B over B with the followingextra data:

• a multiplication µ ∶ E ×B E → E, compatible with the maps down to B;

• a “zero” section s ∶ B → E such that the composite BsÐ→ E → B is the

identity;

• an inverse χ ∶ E → E, compatible with the map down to B; and

• an action of R:

R ×E

ppr2

''

(B ×R) ×B E //

E

p

yyB

Because R is a field, the last piece of data shows that p−1(b) is a R-vectorspace for any point b ∈ B.

Example 52.1. A rather silly example of a vector space over B is the projec-tion B × V → B where V is a (real) vector space, which we will always assumeto be finite-dimensional.

Example 52.2. Consider the map

R ×R(s,t)↦(s,st)ÐÐÐÐÐÐ→R ×R,

over R (the structure maps are given by projecting onto the first factor). Itis an isomorphism on all fibers, but is zero everywhere else. The kernel is

167

168 CHAPTER 4. VECTOR BUNDLES

therefore 0 everywhere, except over the point 0 ∈ R. This the “skyscraper”vector bundle over B.

Sheaf theory accommodates examples like this.One can only go so far you can go with this simplistic notion of a “vector

space” over B. Most interesting and naturally arising examples have a littlemore structure, which is exemplified in the following definition.

Definition 52.3. A vector bundle over B is a vector space over B that islocally trivial (in the sense of Definition 42.1).

Remark 52.4. We will always assume that the space B admits a numerableopen cover (see Definition 42.4) which trivializes the vector bundle. Moreover,the dimension of the fiber will always be finite.

If p ∶ E → B is a vector bundle, then E is called the total space, p is calledthe projection map, and B is called the base space. We will always use a Greekletter like ξ or ζ to denote a vector bundle, and E(ξ) → B(ξ) denotes theactual projection map from the total space to the base space. The phrase “ξis a vector bundle over B” will also be shortened to ξ ↓ B.

Example 52.5. 1. Following Example 52.1, one example of a vector bundleis the trivial bundle B ×Rn → B, denoted by nε.

2. In contrast to this silly example, one gets extremely interesting examplesfrom the Grassmannians Grk(Rn), Grk(Cn), and Grk(Hn). For simplic-ity, let K denote R,C, or H. Over Grk(Kn) lies the tautological bundleγ. This is a sub-bundle of nε (i.e., the fiber over any point x ∈ Grk(Kn)is a subspace of the fiber of nε over x). The total space of γ is defined as:

E(γ) = (V,x) ∈ Grk(Kn) ×Kn ∶ x ∈ V

This projection map down to Grk(Kn) is the literal projection map

(V,x)↦ V.

Exercise 52.6. Prove that γ, as defined above, is locally trivial; so γdefines a vector bundle over Grk(Kn).

For instance, when k = 1, we have Gr1(Rn) = RPn−1. In this case, γ isone-dimensional (i.e., the fibers are all of dimension 1); this is called aline bundle. In fact, it is the “canonical line bundle” over RPn−1.

3. LetM be a smooth manifold. Define τM to be the tangent bundle TM →M over M . For example, if M = Sn−1, then

TSn−1 = (x, v) ∈ Sn−1 ×Rn ∶ v ⋅ x = 0.

52. VECTOR BUNDLES, PRINCIPAL BUNDLES 169

Constructions with vector bundles

One cannot take the kernels of a map of vector bundles; but just about anythingwhich can be done for vector spaces can also be done for vector bundles:

1. Pullbacks are legal: if p′ ∶ E′ → B′, then the leftmost map in the diagrambelow is also a vector bundle.

E //

E′

p′

B

f// B′

For instance, if B = ∗, the pullback is just the fiber of E′ over the point∗ → B′. If ξ is the bundle E′ → B′, we denote the pullback E → B asf∗ξ.

2. If p ∶ E → B and p′ ∶ E′ → B′, then we can take the product E ×E′ p×p′

ÐÐ→B ×B′.

3. If B = B′, we can form the pullback:

E ⊕E′ //

E ×E′

B

∆// B ×B

The bundle E ⊕E′ is called the Whitney sum. For instance, it is an easyexercise to see that

nε = ε⊕⋯⊕ ε.

4. If E,E′ → B are two vector bundles over B, we can form another vectorbundle E ⊗R E′ → B by taking the fiberwise tensor product. Likewise,taking the fiberwise Hom begets a vector bundle HomR(E,E′)→ B.

Example 52.7. Recall from Example 52.5(2) that the tautological bundle γlives over RPn−1; we will write L = E(γ). The tangent bundle τRPn−1 alsolives over RPn−1. As this is the first explicit pair of vector bundles over thesame space, it is natural to wonder what is the relationship between these twobundles.

At first glance, one might guess that τRPn−1 = γ⊥; but this is false! Instead,

τRPn−1 = Hom(γ, γ⊥).

To see this, note that we have a 2-fold covering map Sn−1 →RPn−1; therefore,Tx(RPn−1) is a quotient of T (Sn) by the map sending (x, v)↦ (−x,−v), wherev ∈ Tx(Sn). Therefore,

TxRPn−1 = (x, v) ∈ Sn−1 ×Rn ∶ v ⋅ x = 0/((x, v) ∼ (−x,−v)).


This is exactly the fiber of Hom(γ, γ⊥) over x ∈RPn−1, since the line throughx can be mapped to the line through ±v.

Exercise 52.8. Prove that if γ is the tautological vector bundle over Grk(Kn),for K =R,C,H, then

τGrk(Kn) = Hom(γ, γ⊥).

Metrics and splitting exact sequences

A metric on a vector bundle is a continuous choice of inner products on fibers.

Lemma 52.9. Any vector bundle ξ over X admits a metric.

Intuitively speaking, this is true because if g, g′ are both inner products onV , then tg + (1 − t)g′ is another. Said differently, the space of metrics forms areal affine space.

Proof. Pick a trivializing open cover of X, and a subordinate partition of unity.This means that we have a map φU ∶ U → [0,1], such that the preimage of thecomplement of 0 is U . Moreover,

∑x∈U

φU(x) = 1.

Over each one of these trivial pieces, pick a metric gU on E∣U . Let

g ∶=∑U

φUgU ;

this is the desired metric on ξ.

We remark that, in general, one cannot pick metrics for vector bundles. Forinstance, this is the case for vector bundles which arise in algebraic geometry.

Definition 52.10. Suppose E,E′ → B are vector bundles over B. An iso-morphism is a map α ∶ E → E′ over B that is a linear isomorphism on eachfiber.

In particular, the map α admits an inverse (over B).

Corollary 52.11. Any exact1 sequence 0 → E′ → E → E′′ → 0 of vectorbundles (over the same base) splits.

Proof sketch. Pick a metric for E. Consider the composite

E′⊥ ⊆ E → E′′.

This is an isomorphism: the dimensions of the fibers are the same. It followsthat

E ≅ E′ ⊕E′⊥ ≅ E′ ⊕E′′,

as desired.

Note that this splitting is not natural.1This is the obvious definition.

53. PRINCIPAL BUNDLES, ASSOCIATED BUNDLES 171

53 Principal bundles, associated bundles

I-invariance

We will denote by Vect(B) the set of isomorphism classes of vector bundlesover B. (Justify the use of the word “set”!)

Consider a vector bundle ξ ↓ B. If f ∶ B′ → B, taking the pullback gives avector bundle denoted f∗ξ. This operation descends to a map f∗ ∶ Vect(B) →Vect(B′); we therefore obtain a functor Vect ∶ Topop → Set. One might expectthis functor to give some interesting invariants of topological spaces.

Theorem 53.1. Let I = ∆1. Then Vect is I-invariant. In other words, theprojection X × I →X induces an isomorphism Vect(X)→ Vect(X × I).

One important corollary of this result is:

Corollary 53.2. Vect is a homotopy functor.

Proof. Consider two homotopic maps f, g ∶ B → B′, so there exists a homotopyH ∶ B′ × I → B. If ξ ↓ B, we need to prove that f∗0 ξ ≃ f∗1 ξ. This is far fromobvious.

Consider the following diagram.

B′ × I H //

pr

B

B′

The leftmost map is an isomorphism under Vect, by Theorem 53.1. Let η ↓ B bea vector bundle such that pr∗η ≃ f∗ξ. For any t ∈ I, define a map ∈t∶ B′ → B′×Isends x↦ (x, t). We then have isomorphisms:

f∗t ξ ≃∈∗t f∗ξ ≃∈∗t pr∗η ≃ (pr ∈t)∗η ≃ η,

as desired.

It is easy to see that Vect(X)→ Vect(X×I) is injective. In the next lecture,we will prove surjectivity, allowing us to conclude Theorem 53.1.

Principal bundles

Definition 53.3. Let G be a topological group2. A principal G-bundle is aright action of G on P such that:

• G acts freely.

• The orbit projection P → P /G is a fiber bundle.

2We will only care about discrete groups and Lie groups.


These are not unfamiliar objects, as the next example shows.

Example 53.4. Suppose G is discrete. Then the fibers of the orbit projec-tion P → P /G are all discrete. Therefore, the condition that P → P /G is afiber bundle is simply that it’s a covering projection (the action is “properlydiscontinuous”).

As a special case, let X be a space with universal cover X ↓ X. Thenπ1(X) acts freely on X, and X ↓X is the orbit projection. It follows from ourdiscussion above that this is a principal bundle. Explicit examples include theprincipal Z/2-bundle Sn−1 ↓ RPn−1, and the Hopf fibration S2n−1 ↓ CPn−1,whcih is a principle S1-bundle.

By looking at the universal cover, we can classify covering spaces of X.Remember how that goes: if F is a set with left π1(X)-action, the dotted mapin the diagram below is the desired covering space.

X × F //

ppr1

X × F / ∼

q

yyX

Here, we say that (y, gz) ∼ (yg, z), for elements y ∈ X, z ∈ F , and g ∈ π1(X).Fix y0 ∈ X over ∗ ∈ X. Then it is easy to see that F

∼Ð→ q−1(∗), via themap z ↦ (y0, z). This is all neatly summarized in the following theorem frompoint-set topology.

Theorem 53.5 (Covering space theory). There is an equivalence of categories:

Left π1(X)-sets ≃Ð→ Covering spaces of X,

with inverse functor given by taking the fiber over the basepoint and lifting aloop in X to get a map from the fiber to itself.

Example 53.4 shows that covering spaces are special examples of principalbundles. The above theorem therefore motivates finding a more general picture.

Construction 53.6. Let P ↓ B is a principal G-bundle. If F is a left G-space,we can define a new fiber bundle, exactly as above:

P × F //

P × F / ∼

q

yyB

This is called an associated bundle, and is denoted P ×G F .

53. PRINCIPAL BUNDLES, ASSOCIATED BUNDLES 173

We must still justify that the resulting space over B is indeed a new fiberbundle with fiber F . Let x ∈ B, and let y ∈ P over x. As above, we have a mapF → q−1(∗) via the map z ↦ [y, z]. We claim that this is a homeomorphism.Indeed, define a map q−1(∗)→ F via

[y′, z′] = [y, gz′]↦ gz′,

where y′ = yg for some g (which is necessarily unique).

Exercise 53.7. Check that these two maps are inverse homeomorphisms.

Definition 53.8. A vector bundle ξ ↓ B is said to be an n-plane bundle if thedimensions of all the fibers are n.

Let ξ ↓ B be an n-plane bundle. Construct a principal GLn(R)-bundleP (ξ) by defining

P (ξ)b = bases for E(ξ)b = Iso(Rn,E(ξ)b).

To define the topology, note that (topologically) we have

P (B ×Rn) = B × Iso(Rn,Rn),

where Iso(Rn,Rn) = GLn(R) is given the usual topology as a subspace of Rn2

.There is a right action of GLn(R) on P (ξ) ↓ B, given by precomposition.

It is easy to see that this action is free and simply transitive. One thereforehas a principal action of GLn(R) on P (ξ). The bundle P (ξ) is called theprincipalization of ξ.

Given the principalization P (ξ), we can recover the total space E(ξ). Con-sider the associated bundle P (ξ)×GLn(R)R

n with fiber F =Rn, with GLn(R)acting on Rn from the left. Because this is a linear action, P (ξ) ×GLn(R) R

n

is a vector bundle. One can show that

P (ξ) ×GLn(R) Rn ≃ E(ξ).

Fix a topological group G. Define BunG(B) as the set of isomorphismclasses of G-bundles over B. An isomorphism is a G-equivariant homeo-morphism over the base. Again, arguing as above, this begets a functorBunG ∶ Top → Set. The above discussion gives a natural isomorphism offunctors:

BunGLn(R)(B) ≃ Vect(B).

The I-invariance theorem will therefore follow immediately from:

Theorem 53.9. BunG is I-invariant.

Remark 53.10. Principal bundles allow a description of “geometric structureson ξ”. Suppose, for instance, that we have a metric on ξ. Instead of looking at


all ordered bases, we can attempt to understand all ordered orthonormal basesin each fiber. This give the frame bundle

Fr(B) = ordered orthonormal bases of E(ξ)b;

these are isometric isomorphisms Rn → E(ξ)b. Again, there is an action ofthe orthogonal group on Fr(B): in fact, this begets a principal O(n)-bundle.Such examples are in abundance: consistent orientations give an SO(n)-bundle.Trivializations of the vector bundle also give principal bundles. This is called“reduction of the structure group”.

One useful fact about principal G-bundles (which should not be too sur-prising) is the following statement.

Theorem 53.11. Every morphism of principal G-bundles is an isomorphism.

Proof. Let p ∶ P → B and p′ ∶ P ′ → B be two principal G-bundles over B, andlet f ∶ P → P ′ be a morphism of principal G-bundles. For surjectivity of f , lety ∈ P ′. Consider x ∈ P such that p(x) = p′(y). Since p(x) = p′f(x) we concludethat y = f(x)g for some g ∈ G. But f(x)g = f(xg), so xg maps to y, as desired.To see that f is injective, suppose f(x) = f(y). Now p(x) = p′f(x) = p(y), sothere is some g ∈ G such that xg = y. But f(y) = f(xg) = f(x)g, so g = 1, asdesired. We will leave the continuity of f−1 as an exercise to the reader.

Theorem 53.11 says that if we view BunG(B) as a category where the mor-phisms are given by morphisms of principal G-bundles, then it is a groupoid.

54 I-invariance of BunG, and G-CW-complexes

Let G be a topological group. We need to show that the functor BunG ∶Topop → Set is I-invariant, i.e., the projection X × I prÐ→X induces an isomor-phism BunG(X) ≃Ð→ BunG(X × I). Injectivity is easy: the composite X

in0Ð→X × I prÐ→ X gives you a splitting BunG(X)

pr∗ÐÐ→ BunG(X × I) in0Ð→ BunG(X)whose composite is the identity.

The rest of this lecture is devoted to proving surjectivity. We will provethis when X is a CW-complex (Husemoller does the general case; see [Hus94,§4.9]). We begin with a small digression.

G-CW-complexes

We would like to define CW-complexes with an action of the group G. Thenaïve definition (of a space with an action of the group G) will not be sufficient;rather, we will require that each cell have an action of G.

In other words, we will build G-CW-complexes out of “G-cells”. This issupposed to be something of the form Dn×H/G, where H is a closed subgroupof G. Here, the space H/G is the orbit space, viewed as a right G-space. Theboundary of the G-cell Dn ×H/G is just ∂Dn ×H/G. More precisely:

54. I-INVARIANCE OF BunG, AND G-CW-COMPLEXES 175

Definition 54.1. A G-CW-complex is a (right) G-space X with a filtration0 =X−1 ⊆X0 ⊆ ⋯ ⊆X such that for all n, there exists a pushout square:

∐∂Dnα ×Hα/G //

∐Dnα ×Hα/G

Xn−1

// Xn,

and X has the direct limit topology.

Notice that a CW-complex is a G-CW-complex for the trivial group G.

Theorem 54.2. If G is a compact Lie group and M a compact smooth G-manifold, then M admits a G-CW-structure.

This is the analogue of the classical result that a compact smooth manifoldis homotopy equivalent to a CW-complex, but it is much harder to prove theequivariant statement.

Note that if G acts principally (Definition 53.3) on P , then every G-CW-structure on P is “free”, i.e., Hα = 0.

1. If X is a G-CW-complex, then X/G inherits a CW-structure whose n-skeleton is given by (X/G)n =Xn/G.

2. If P → X is a principal G-bundle, then a CW-structure on X lifts to aG-CW-structure on P .

Proof of I-invariance

Recall that our goal is to prove that every G-bundle over X × I is pulled backfrom some vector bundle over X.

As a baby case of Theorem 53.1 we will prove that if X is contractible, thenany principal G-bundle over X is trivial, i.e., P ≃X ×G as G-bundles.

Let us first prove the following: if P ↓ X has a section, then it’s trivial.Indeed, suppose we have a section s ∶ X → P . Since P has an action of thegroup on it, we may extend this to a map X×G→ P by sending (x, g)↦ gs(x).As this is a map of G-bundles over X, it is an isomorphism by Theorem 53.11,as desired.

To prove the statement about triviality of any principal G-bundle over acontractible space, it therefore suffices to construct a section for any principalG-bundle. Consider the constant map X → P . Then the following diagramcommutes up to homotopy, and hence (by Exercise 44.10(1)) there is an actualsection of P →X, as desired.

P

X

const

>>

// X


For the general case, we will assume X is a CW-complex. For notationalconvenience, let us write Y = X × I. We will use descending induction toconstruct the desired principal G-bundle over X.

To do this, we will filter Y by subcomplexes. Let Y0 =X × 0; in general, wedefine

Yn =X × 0 ∪Xn−1 × I.

It follows that we may construct Yn out of Yn−1 via a pushout:

∐α∈Σn−1(∂Dn−1 × I ∪Dn−1

α × 0) //

∐α∈Σn−1fα×1I∪φα×0

∐α(Dn−1α × I)

Yn−1

// Yn,

where the maps fα and φα are defined as:

∂Dn−1α

fα //

Xn−2

Dn−1α φα

// Xn−1

In other words, the fα are the attaching maps and the φα are the characteristicmaps.

Consider a principal G-bundle PpÐ→ Y = X × I. Define Pn = p−1(Yn); then

we can build Pn from Pn−1 in a similar way:

∐α(∂Dn−1α × I ∪Dn−1

α × 0) ×G //

∐α(Dn−1α × I) ×G

Pn−1

// Pn

Note that this isn’t quite a G-CW-structure. Recall that we are attempting tofill in a dotted map:

P //

P0

Y pr

// Y0 =X

I’m constructing this inductively– we have Pn−1 → P0. So I want to definefinish this...finish this...∐α(Dn−1

α × I) × G → P0 that’s equivariant. That’s the same thing as a map∐α(Dn−1

α ×I)→ P0 that’s compatible with the map from ∐(∂Dn−1α ×I ∪Dn−1

α ×

55. CLASSIFYING SPACES: THE GRASSMANN MODEL 177

0). Namely, I want to fill in:

∐α(∂Dn−1α × I ∪Dn−1

α × 0) //

∐α(Dn−1α × I)

∐α(∂Dn−1

α × I ∪Dn−1α × 0) ×G //

∐α(Dn−1α × I) ×G

Pn−1

induction

,,

// Pn

''P0

X

(4.1)

Now, I know that (Dn−1 × I, ∂Dn−1 × I ∪Dn−1 × 0) ≃ (Dn−1 × I,Dn−1 × 0). Sowhat I have is:

Dn−1 × 0

induction // P0

Dn−1 × I

φpr//

;;

X

So the dotted map exists, since P0 →X is a fibration!OK, so note that I haven’t checked that the outer diagram in Equation 4.1

commutes, because otherwise we wouldn’t get Pn → P0.

Exercise 54.3. Check my question above.Turns out this is easy, because you have a factorization:

Dn−1 × 0

// Pn−1induction// P0

Dn−1 × I

φpr//

55

X

Oh my god, look what time it is! Oh well, at least we got the proof done.

55 Classifying spaces: the Grassmann model

We will now shift our focus somewhat and talk about classifying spaces forprincipal bundles and for vector bundles. We will do this in two ways: the firstwill be via the Grassmann model and the second via simplicial methods.

Lemma 55.1. Over a compact Hausdorff space, any n-plane bundle embedsin a trivial bundle.


Proof. Let U be a trivializing open cover of the base B; since B is compact, wemay assume that U is finite with k elements. There is no issue with numerabil-ity, so there is a subordinate partition of unity φi. Consider an n-plane bundleE → B. By trivialization, there is a fiberwise isomorphism p−1(Ui)

fiÐ→ Rn

where the Ui ∈ U . A map to a trivial bundle is the same thing as a bundle mapin the following diagram:

E //

RN

B // ∗

We therefore define E → (Rn)k via

e↦ (φi(p(e))fi(e))i=1,⋯,k.

This is a fiberwise linear embedding, generally called a “Gauss map”. Indeed,observe that this map has no kernel on every fiber, so it is an embedding.

The trivial bundle has a metric on it, so choosing the orthogonal comple-ment of the embedding of Lemma 55.1, we obtain:

Corollary 55.2. Over a compact Hausdorff space, any n-plane bundle has acomplement (i.e. a ξ⊥ such that ξ ⊕ ξ⊥ is trivial).

Another way to say this is that if B is a compact Hausdorff space with ann-plane bundle ξ, there is a map f ∶ X → Grn(Rkn); this is exactly the Gaussmap. It has the property that taking the pullback f∗γn of the tautologousbundle over Grn(Rkn) gives back ξ.

In general, we do not have control over the number k. There is an easy fixto this problem: consider the tautologous bundle γn over Grn(R∞), definedas the union of Grn(Rm) and given the limit topology. This is a CW-complexof finite type (i.e. finitely many cells in each dimension). Note that Grn(Rm)are not the m-skeleta of Grn(R∞)!

The space Grn(R∞) is “more universal”:

Lemma 55.3. Any (numerable) n-plane bundle is pulled back from γn ↓ Grn(R∞)via the Gauss map.

Lemma 55.3 is a little bit tricky, since the covering can be wildly uncount-able; but this is remedied by the following bit of point-set topology.

Lemma 55.4. Let U be a numerable cover of X. Then there’s another nu-merable cover U ′ such that:

1. the number of open sets in U ′ is countable, and

2. each element of U ′ is a disjoint union of elements of U .

If U is a trivializing cover, then U ′ is also a trivializing cover.

55. CLASSIFYING SPACES: THE GRASSMANN MODEL 179

Proof. See [Hus94, Proposition 3.5.4].

It is now an exercise to deduce Lemma 55.3. The main result of this sectionis the following.

Theorem 55.5. The map [X,Grn(R∞)]→ Vectn(X) defined by [f]↦ [f∗γn]is bijective, where [f] is the homotopy class of f and [f∗γn] is the isomorphismclass of the bundle f∗γn.

This is why Grn(R∞) is also called the classifying space for n-plane bun-dles. The Grassmannian provides a very explicit geometric description for theclassifying space of n-plane bundles. There is a more abstract way to producea classifying space for principal G-bundles, which we will describe in the nextsection; the Grassmannian is the special case when G = GLn(R).

Proof. We have already shown surjectivity, so it remains to prove injectivity.Suppose f0, f1 ∶ X → Grn(R∞) such that f∗0 γ

n and f∗1 γn are isomorphic over

X. We need to construct a homotopy f0 ≃ f1. For ease of notation, let usidentify f∗0 γ

n and f∗1 γn with each other; call it ξ ∶ E ↓X.The maps fi are the same thing as Gauss maps gi ∶ E → R∞, i.e., maps

which are fiberwise linear embeddings. The homotopy f0 ≃ f1 is created bysaying that we have a homotopy from g0 to g1 through Gauss maps, i.e., throughother fiberwise linear embeddings.

In fact, we will prove a much stronger statement: any two Gauss mapsg0, g1 ∶ E →R∞ are homotopic through Gauss maps. This is very far from trueif I didn’t have a R∞ on the RHS there.

Let us attempt (and fail!) to construct an affine homotopy between g0 andg1. Consider the map tg0 + (1 − t)g1 for 0 ≤ t ≤ 1. In order for these mapsto define a homotopy via Gauss maps, we need the following statement to betrue: for all t, if tg0(v) + (1 − t)g1(v) = 0 ∈R∞, then v = 0. In other words, weneed tg0 + (1 − t)g1 to be injective. Of course, this is not guaranteed from theinjectivity of g0 and g1!

Instead, we will construct a composite of affine homotopies between g0 andg1 using the fact that R∞ is an infinite-dimensional Euclidean space. Considerthe following two linear isometries:

R∞ = ⟨e0, e1,⋯⟩

α

ei↦e2i

wwβ

ei↦e2i+1

''R∞ R∞

Then, we have four Gauss maps: g0, α g0, β g1, and g1. There are affinehomotopies through Gauss maps:

g0 ≃ α g0 ≃ β g1 ≃ g1.

We will only show that there is an affine homotopy through Gauss maps g0 ≃α g0; the others are left as an exercise. Let t and v be such that tg0(v)+ (1−


t)αg0(v) = 0. Since g0 and αg0 are Gauss maps, we may suppose that 0 < t < 1.Since αg0(v)i has only even coordinates, it follows by definition of the mapα that g0(v) only had nonzero coordinates only in dimensions congruent to 0mod 4. Repeating this argument proves the desired result.

56 Simplicial sets

In order to discuss the simplicial model for classifying spaces of G-bundles, wewill embark on a long digression on simplicial sets (which will last for threesections). We begin with a brief review of some of the theory of simplicialobjects (see also Part 1).

Review

We denote by [n] the set 0,1,⋯, n, viewed as a totally ordered set. Definea category ∆ whose objects are the sets [n] for n ≥ 0, with morphisms orderpreserving maps. There are maps di ∶ [n]→ [n + 1] given by omitting i (calledcoface maps) and codegeneracy maps si ∶ [n] → [n − 1] that’s the surjectionwhich repeats i. As discussed in Exercise ??, any order-preserving map can bewritten as the composite of these maps, and there are famous relations thatthese things satisfy. They generate the category ∆.

There is a functor ∆ ∶ ∆→ Top defined by sending [n]↦∆n, the standardn-simplex. To see that this is a functor, we need to show that maps φ ∶ [n] →[m] induce maps ∆n → ∆m. The vertices of ∆n are indexed by elements of[n], so we may just extend φ as an affine map to a map ∆n →∆m.

Let X be a space. The set of singular n-simplices Top(∆n,X) defines thesingular simplicial set Sin ∶ ∆op → Set.

Definition 56.1. Let C be a category. Denote by sC the category of simplicialobjects in C, i.e., the category Fun(∆op,C). We write Xn =X([n]), called then-simplices.

Explicitly, this gives an object Xn ∈ C for every n ≥ 0, as well as mapsdi ∶Xn+1 →Xn and si ∶Xn−1 →Xn given by the face and degeneracy maps.

Example 56.2. Suppose C is a small category, for instance, a group. Noticethat [n] is a small category, with:

[n](i, j) =⎧⎪⎪⎨⎪⎪⎩

≤ if i ≤ j∅ else.

We are therefore entitled to think about Fun([n],C). This begets a simplicialset NC, called the nerve of C, whose n-simplices are (NC)n = Fun([n],C).Explicitly, an n-simplex in the nerve is (n + 1)-objects in C (possibly withrepetitions) and a chain of n composable morphisms. The face maps are givenby composition (or truncation, at the end of the chain of morphisms). Thedegeneracy maps just compose with the identity at that vertex.

For example, if G is a group regarded as a category, then (NG)n = Gn.

56. SIMPLICIAL SETS 181

Realization

The functor Sin transported us from spaces to simplicial sets. Milnor describeda way to go the other way.

Let X be a simplicial set. We define the realization ∣X ∣ as follows:

∣X ∣ = (∐n≥0

∆n ×Xn) / ∼,

where ∼ is the equivalence relation defined as:

∆m ×Xm ∋ (v, φ∗x) ∼ (φ∗v, x) ∈ ∆n ×X

for all maps φ ∶ [m]→ [n] where v ∈ ∆m and x ∈Xn.

Example 56.3. The equivalence relation is telling us to glue together simplicesas dictated by the simplicial structure on X. To see this in action, let us lookat φ∗ = di ∶ Xn+1 → Xn and φ∗ = di ∶ ∆n → ∆n+1. In this case, the equivalencerelation then says that (v, dix) ∈ ∆n×Xn is equivalent to (div, x) ∈ ∆n+1×Xn+1.In other words: the n-simplex indexed by dix is identified with the ith face ofthe (n + 1)-simplex indexed by x.

There’s a similar picture for the degeneracies si, where the equivalencerelation dictates that every element of the form (v, six) is already representedby a simplex of lower dimension.

Example 56.4. Let n ≥ 0, and consider the simplicial set Hom∆(−, [n]). Thisis called the “simplicial n-simplex”, and is commonly denoted ∆n for goodreason: we have a homeomorphism ∣∆n∣ ≃ ∆n. It is a good exercise to provethis using the explicit definition.

For any simplicial set X, the realization ∣X ∣ is naturally a CW-complex,with

skn∣X ∣ = (∐k≤n

∆k ×Xk) / ∼ .

The face maps give the attaching maps; for more details, see [GJ99, PropositionI.2.3]. This is a very combinatorial way to produce CW-complexes.

The geometric realization functor and the singular simplicial set give twofunctors going back and forth between spaces and simplicial sets. It is naturalto ask: do they form an adjoint pair? The answer is yes:

sSet Top

∣−∣

Sin

⊣

For instance, let X be a space. There is a continuous map ∆n × Sinn(X)→Xgiven by (v, σ) ↦ σ(v). The equivalence relation defining ∣Sin(X)∣ says that


the map factors through the dotted map in the following diagram:

∣Sin(X)∣ // X

∐∆n × Sinn(X)

88OOOO

The resulting map is the counit of the adjunction.Likewise, we can write down the unit of the adjunction: if K ∈ sSet, the

map K → Sin∣K ∣ sends x ∈Kn to the map ∆n → ∣K ∣ defined via v ↦ [(v, x)].This is the beginning of a long philosophy in semi-classical homotopy theory,

of taking any homotopy-theoretic question and reformulating it in simplicialsets. For instance, one can define homotopy groups in simplicial sets. For moredetails, see [GJ99].

We will close this section with a definition that we will discuss in the nextsection. Let C be a category. From our discussion above, we conclude that therealization ∣NC∣ of its nerve is a CW-complex, called the classifying space BCof C; the relation to the notion of classifying space introduced in §55 will beelucidated upon in a later section.

57 Properties of the classifying space

One important result in the story of geometric realization introduced in thelast section is the following theorem of Milnor’s.

Theorem 57.1 (Milnor). Let X be a space. The map ∣Sin(X)∣→X is a weakequivalence.

As a consequence, this begets a functorial CW-approximation to X. Un-forunately, it’s rather large.

In the last section, we saw that ∣−∣ was a left adjoint. Therefore, it preservescolimits (Theorem 39.13). Surprisingly, it also preserves products:

Exercise 57.2 (Hard). Let X and Y be simplicial sets. Their product X × Yis defined to be the simplicial set such that (X × Y )n = Xn × Yn. Under thisnotion of product, there is a homeomorphism

∣X × Y ∣ ≃Ð→ ∣X ∣ × ∣Y ∣.

It is important that this product is taken in the category of k-spaces.

Last time, we defined the classifying space BC of C to be ∣NC∣.

Theorem 57.3. The natural map B(C×D) ≃Ð→ BC×BD is a homeomorphism3.

3Recall that if C and D are categories, the product C ×D is the category whose objectsare pairs of objects of C and D, and whose morphisms are pairs of morphisms in C and D.

57. PROPERTIES OF THE CLASSIFYING SPACE 183

Proof. It is clear that N(C × D) ≃ NC × ND. Since BC = ∣NC∣, the desiredresult follows from Exercise 57.2.

In light of Theorem 57.3, it is natural to ask how natural transformationsbehave under the classifying space functor. To discuss this, we need somecategorical preliminaries.

The category Cat is Cartesian closed (Definition 40.5). Indeed, the rightadjoint to the product is given by the functor D ↦ Fun(C,D), as can be directlyverified.

Consider the category [1]. This is particularly simple: a functor [1]→ C isjust an arrow in C. It follows that a functor [1]→ DC is a natural transformationbetween two functors f0 and f1 from C to D. By our discussion above, this isthe same as a functor C × [1]→ D.

By Theorem 57.3, we have a homeomorphism B([1]× C) ≃ B[1]×BC. Onecan show that B[1] = ∆1, so a natural transformation between f0 and f1 begetsa map ∆1 ×BC → BD between Bf0 and Bf1. Concretely:

Lemma 57.4. A natural transformation θ ∶ f0 → f1 where f0, f1 ∶ C → Dinduces a homotopy Bf0 ∼ Bf1 ∶ BC → BD.

An interesting comment is in order. The notion of a homotopy is “re-versible”, but that is definitely not true for natural transformations! The func-tor B therefore “forgets the polarity in Cat”.

Lemma 57.4 is quite powerful: consider an adjunction L ⊣ R where L ∶ C →D; then we have natural transformations given by the unit 1C → RL and thecounit LR → 1D. By Lemma 57.4 we get a homotopy equivalence between BCand BD. In other words, two categories that are related by any adjoint pairare homotopy equivalent.

A special case of the above discussion yields a rather surprising result.Consider the category [0]. Let D be another category such that there is anadjoint pair L ⊣ R where L ∶ [0]→ D. Then L determines an object ∗ of D. Letd be any object of D. We have the counit LR(d)→ d; but LR(d) = ∗, so thereis a unique morphism ∗ → X. (To see uniqueness, note that the adjunctionL ⊣ R gives an identification D(∗,X) = C(0,0) = 0.) In other words, such acategory D is simply a category with an initial object.

Arguing similarly, any category D with adjunction L ⊣ R where L ∶ D → [0]is simply a category with a terminal object. From our discussion above, weconclude that if D is any category with a terminal (or initial) object, then BDis contractible.


58 Classifying spaces of groups

The constructions of the previous sections can be summarized in a single dia-gram:

Catnerve // sSet

∣−∣

Gp?

OO

B// Top

The bottom functor is defined as the composite along the outer edge of thediagram. The space BG for a group G is called the classifying space of G. Atthis point, it is far from clear what BG is classifying. The goal of the next fewsections is to demystify this definition.

Lemma 58.1. Let G be a group, and g ∈ G. Let cg ∶ G → G via x ↦ gxg−1.Then the map Bcg ∶ BG→ BG is homotopic to the identity.

Proof. The homomorphism cg is a functor from G to itself. It suffices to provethat there is a natural transformation θ from the identity to cg. This is rathereasy to define: it sends the only object to the only object: we define θ∗ ∶ ∗→ ∗to be the map given by ∗ gÐ→ ∗ specified by g ∈ HomG(∗,∗) = G. In order forθ to be a natural transformation, we need the following diagram to commute,which it obviously does:

∗1

g // ∗

gxg−1

∗ g

// ∗.

Groups are famous for acting on objects. Viewing groups as categoriesallows for an abstract definition a group action on a set: it is a functor G→ Set.More generally, if C is a category, an action of C is a functor C XÐ→ Set. Wewrite Xc =X(c) for an object c of C.

Definition 58.2. The “translation” category XC has objects given by

ob(XC) =∐c∈C

Xc,

and morphisms defined via HomXC(x ∈Xc, y ∈Xd) = f ∶ c→ d ∶ f∗(x) = y.

There is a projection XC → C. (For those in the know: this is a special caseof the Grothendieck construction.)

Example 58.3. The group G acts on itself by left translation. We will writeG for this G-set. The translation category GG has objects as G, and mapsx→ y are elements yx−1. This category is “unicursal”, in the sense that there is

58. CLASSIFYING SPACES OF GROUPS 185

exactly one map from one object to another object. Every object is thereforeinitial and terminal, so the classifying space of this category is trivial by thediscussion at the end of §57. We will denote by EG the classifying spaceB(GG). The map GG→ G begets a canonical map EG→ BG.

The G also acts on itself by right translation. Because of associativity, theright and left actions commute with each other. It follows that the right actionis equivariant with respect to the left action, so we get a right action of G onEG.

Claim 58.4. This action of G on EG is a principal action, and the orbitprojection is EG→ BG.

To prove this, let us contemplate the set N(GG)n. An element is a chainof composable morphisms. In this case, it is actually just a sequence of n + 1elements in G, i.e., N(GG)n = Gn+1. The right action of G is simply thediagonal action. We claim that this is a free action. More precisely:

Lemma 58.5 (Shearing). If G is a group and X is a G-set, and if X ×∆G hasthe diagonal G-action and X × G has G acting on the second factor by righttranslation, then X ×∆ G ≃X ×G as G-sets.

Proof. Define a bijection X ×∆ G ↦ X × G via (x, g) ↦ (xg−1, g). This mapis equivariant since (x, g) ⋅ h = (xh, gh), while (xg−1, g) ⋅ h = (xg−1, gh). Theelement (xh, gh) is sent to (xh(gh)−1, gh), as desired. The inverse mapX×G→X ×∆ G is given by (x, g)↦ (xg, g).

We know that G acts freely on N(GG)n, soo a nonidentity group elementis always going to send a simplex to another simplex. It follows that G actsfreely on EG.

To prove the claim, we need to understand the orbit space. The shearinglemma shows that quotienting out by the action of G simply cancels out onecopy of G from the product N(GG) = Gn. In symbols:

N(GG)/G ≃ Gn ≃ (NG)n.

Of course, it remains to check the compatibility with the face and degeneracymaps. We will not do this here; but one can verify that everything works out:the realization is just BG!

We need to be careful: the arguments above establish that EG/G ≃ BGwhen G is a finite group. The case when G is a topological group is morecomplicated. To describe this generalization, we need a preliminary categoricaldefinition.

Let C be a category, with objects C0 and morphisms C1. Then we havemaps C1 ×C0 C1

composeÐÐÐÐ→ C1 and two maps (source and target) C1 → C0, andthe identity C0 → C1. One can specify the same data in any category D withpullbacks. Our interest will be in the case D = Top; in this case, we call C a“category in Top”.


Let G be a topological group acting on a space X. We can again define XG,although it is now a category in Top. Explicitly, (XG)0 = X and (XG)1 =G×X as spaces. The nerve of a topological category begets a simplicial space.In general, we will have

(NC)n = C1 ×C0 C1 ×⋯ ×C0 C1.

The geometric realization functor works in exactly the same way, so the re-alization of a simplicial space gets a topological space. The above discussionpasses through with some mild topological conditions on G (namely, if G is anabsolute neighborhood retract of a Lie group); we conclude:

Theorem 58.6. Let G be an absolute neighborhood retract of a Lie group.Then EG is contractible, and G acts from the right principally. Moreover, themap EG→ BG is the orbit projection.

A generalization of this result is:

Exercise 58.7. Let X be a G-set. Show that

EG ×GX ≃ B(XG).

59 Classifying spaces and bundles

Let π ∶ Y → X be a map of spaces. This defines a “descent category” C(π)whose objects are the points of Y , whose morphisms are points of Y ×X Y ,and whose structure morphisms are the obvious maps. Let cX denote thecategory whose objects and morphisms are both given by points of X, so thatthe nerve NcX is the constant simplicial object with value X. There is afunctor C(π)→ cX specified by the map π.

Let U be a cover of X. Let C(U) denote the descent category associatedto the obvious map ε ∶ ∐U∈U U → X. It is easy to see that ε ∶ BC(U) ≃X if U is numerable. The morphism determined by x ∈ U ∩ V is denotedxU,V . Suppose p ∶ P → X is a principal G-bundle. Then U trivializes p ifthere are homeomorphisms tU ∶ p−1(U) ≃Ð→ U × G over U . Specifying suchhomeomorphisms is the same as a trivialization of the pullback bundle ε∗P .

This, in turn, is the same as a functor θP ∶ C(U) → G. To see this, wenote that the G-equivariant composite tV t−1

U ∶ (U ∩ V ) × G → (U ∩ V ) × Gis determined by the value of (x,1) ∈ (U ∩ V ) × G. The map U ∩ V → G isdenoted fU,V . Then, the functor θP ∶ C(U)→ G sends every object of C(U) tothe point, and xU,V to fU,V (x).

On classifying spaces, we therefore get a map X≃←Ð BC(U) θPÐ→ BG, where

the map on the left is given by ε.

Exercise 59.1. Prove that θ∗PEG ≃ ε∗P .

59. CLASSIFYING SPACES AND BUNDLES 187

This suggests that BG is a classifying space for principal G-bundles (inthe sense of §55). To make this precise, we need to prove that two principalG-bundles are isomorphic if and only if the associated maps X → BG arehomotopic.

To prove this, we will need to vary the open cover. Say that V refines U iffor any V ∈ U , there exists U ∈ U such that V ⊆ U . A refinement is a functionp ∶ V → U such that V ⊆ p(V ). A refinement p defines a map∐V ∈V V →∐U∈U U ,denoted ρ.

As both ∐V ∈V V and ∐U∈U U cover X, we get a map C(V) → C(U) overcX. Taking classifying spaces begets a diagram:

BC(V) //

%%

BC(U)

X

Let t be trivialization of P for the open cover U . The construction describedabove begets a functor BC(U)→ BG, so we get a trivialization s for V. This isa homeomorphism sV ∶ p−1(V )→ V ×G which fits into the following diagram:

p−1(V ) sV∼

//

V ×G

p−1(ρ(V ))

tρ(V )

∼ // ρ(V ) ×G

By construction, there is a large commutative diagram:

BC(V) //

∼

%%

++BC(U)

∼

// BG

X.

This justifies dropping the symbol U in the notation for the map θP .Consider two principal G-bundles over X:

P≃ //

Q

X,

and suppose I have trivializations (U , t) of P and (W, s) of Q. Let V be a


common refinement, so that there is a diagram:

C(U)θUP

!!C(V)

;;

##

θVP,,

θVQ

22 G

C(W)θWQ

==

Included in the diagram is a mysterious natural transformation β ∶ θVP → θVQ,whose construction is left as an exercise to the reader. Its existence combinedShould we describe this?

It’s rather technical...Should we describe this?It’s rather technical... with Lemma 57.4 implies that the two maps θP , θQ ∶ BC(V) ≃ X → BG are

homotopic, as desired.

Topological properties of BG

Before proceeding, let us summarize the constructions discussed so far. Let Gbe some topological group (assumed to be an absolute neighborhood retract ofa Lie group). We constructed EG, which is a contractible space with G actingfreely on the right (this works for any topological group). There is an orbitprojection EG→ BG, which is a principal G-bundle under our assumption onG. The space BG is universal, in the sense that there is a bijection

BunG(X) ≃←Ð [X,BG]

given by f ↦ [f∗EG].Let E be a space such that G acts on E from the left. If P → B is any

principal G-bundle, then P × E → P ×G E is another principal G-bundle. Inthe case P = EG, it follows that if E is a contractible space on which G acts,then the quotient EG×GE is a model for BG. Recall that EG is contractible.Therefore, if E is a contractible space on which G acts freely, then the quotientG/E is a model for BG. Of course, one can run the same argument in the casethat G acts on E from the right. Although the construction with simplicialsets provided us with a very concrete description of the classifying space of agroup G, we could have chosen any principal action on a contractible space inorder to obtain a model for BG.

Suppose X is a pointed path connected space. Remember that X has acontractible path space PX = XI

∗ . The canonical map PX → X is a fibration,with fiber ΩX.

Consider the case when X = BG. Then, we can compare the above fibration

59. CLASSIFYING SPACES AND BUNDLES 189

with the fiber bundle EG→ BG:

G //

ΩBG

∗ ≃ EG

// PBG ≃ ∗

BG BG

The map EG → BG is nullhomotopic; a choice of a nullhomotopy is exactlya lift into the path space. Therefore, the dotted map EG → PBG exists inthe above diagram. As EG and PBG are both contractible, we conclude thatΩBG is weakly equivalent to G. In fact, this weak equivalence is a H-map,i.e., it commutes up to homotopy with the multiplication on both sides.

Remark 59.2 (Milnor). If X is a countable CW-complex, then ΩX is nota CW-complex, but it is homotopy equivalent (not just weakly equivalent) toone. Moreover, ΩX is weakly equivalent to a topological group GX such thatBGX ≃X.

Examples

We claim that BU(n) ≃ Grn(C∞). To see this, let Vn(C∞) is the contractiblespace of complex n-frames in C∞, i.e., isometric embeddings of Cn into C∞.The Lie group U(n) acts principally on Vn(C∞) by precomposition, and thequotient Vn(C∞)/U(n) is exactly the Grassmannian Grn(C∞). As Grn(C∞) isthe quotient of a principal action of U(n) on a contractible space, our discussionin the previous section implies the desired claim.

Let G be a compact Lie group (eg finite).

Theorem 59.3 (Peter-Weyl). There exists an embedding G U(n) for somen.

Since U(n) acts principally on Vn(C∞), it follows G also acts principallyon Vn(C∞). Therefore Vn(C∞)/G is a model for BG. It is not necessarily thatthis the most economic description of BG.

For instance, in the case of the symmetric group Σn, we have a much nicergeometric description of the classifying space. Let Confn(Rk) denote embed-dings of 1,⋯, n → Rk (ordered distinct n-tuples). This space is definitelynot contractible! However, the classifying space Confn(R∞) is contractible.The symmetric group obviously acts freely on this (for finite groups, a princi-pal action is the same as a free action). It follows that BΣn is the space ofunordered configurations of n distinct points in R∞. Using Cayley’s theoremfrom classical group theory, we find that if G is finite, a model for BG is thequotient Confn(R∞)/G.


We conclude this chapter with a construction of Eilenberg-Maclane spacesvia classifying spaces. If A is a topological abelian group, then the multiplica-tion µ ∶ A×A→ A is a homomorphism. Applying the classifying space functorbegets a map m ∶ BA ×BA → BA. If G is a finite group, then BA = K(A,1).The map m above gives a topological abelian group model for K(A,1). Thereis nothing preventing us from iterating this construction: the space B2A sitsin a fibration

BA→ EBA ≃ ∗→ B2A.

It follows from the long exact sequence in homotopy that the homotopy groupsof B2A are the same as that of BA, but shifted up by one. Repeating thisprocedure multiple times gives us an explicit model for K(A,n):

BnA =K(A,n).

Chapter 5

Spectral sequences

Spectral sequences are one of those things for which anybody who isanybody must suffer through. Once you’ve done that, it’s like linearalgebra. You stop thinking so much about the ‘inner workings’ later.

– Haynes Miller

60 The spectral sequence of a filtered complex

Our goal will be to describe a method for computing the homology of a chaincomplex. We will approach this problem by assuming that our chain complex isequipped with a filtration; then we will discuss how to compute the associatedgraded of an induced filtration on the homology, given the homology of theassociated graded of the filtration on our chain complex.

We will start off with a definition.

Definition 60.1. A filtered chain complex is a chain complex C∗ along witha sequence of subcomplexes FsC∗ such that the group Cn has a filtration by

F0Cn ⊂ F1Cn ⊆ ⋯,

such that ⋃FsCn = Cn.

The differential on C∗ begets the structure of a chain complex on the as-sociated graded grsCn = FsCn/Fs−1Cn; in other words, the differential on C∗respects the filtration, hence begets a differential d ∶ grsCn → grsCn−1.

The canonical example of a filtered chain complex to keep in mind is thehomology of a filtered space (such as a CW-complex). Let X be a filteredspace, i.e., a space equipped with a filtration X0 ⊆X1 ⊆ ⋯ such that ⋃Xn =X.We then have a filtration of the chain complex C∗(X) by the subcomplexesC∗(Xn).

For ease of notation, let us write

E0s,t = grsCs+t = FsCs+t/Fs−1Cs+t,

191

192 CHAPTER 5. SPECTRAL SEQUENCES

so the differential on C∗ gives a differential d0 ∶ E0s,t → E0

s,t−1. A first approxi-mation to the homology of C∗ might therefore be the homology Hs+t(grsC∗).We will denote this group by E1

s,t. This is the homology of the associatedgraded of the filtration F∗C∗.

We can get an even better approximation to H∗C∗ by noticing that thereis a differential even on E1

s,t. By construction, there is a short exact sequenceof chain complexes

0→ Fs−1C∗ → FsC∗ → grsC∗ → 0,

so we get a long exact sequence in homology. The differential on E1s,t is the

composite of the boundary map in this long exact sequence with the naturalmap H∗(Fs−1C∗)→H∗(grs−1C∗); more precisely, it is the composite

d1 ∶ E1s,t =Hs+t(grsC∗)

∂Ð→Hs+t−1(Fs−1C∗)→Hs+t−1(grs−1C∗) = E1s−1,t.

It is easy to check that (d1)2 = 0.This construction is already familiar from cellular chains: in this case, E1

s,t

is exactly Hs+t(Xs,Xs−1), which is exactly the cellular s-chains when t = 0(and is 0 if t ≠ 0). The d1 differential is constructed in exactly the same wayas the differential on cellular chains.

In light of this, we define E2s,t to be the homology of the chain complex

(E1∗,∗, d

1); explicitly, we let

E2s,t = ker(d1 ∶ E1

s,t → E1s−1,t)/ im(d1 ∶ E1

s+1,t → E1s,t).

Does this also have a differential d2? The answer is yes. We will inductivelydefine Ers,t via a similar formula: if Er−1

∗,∗ and the differential dr−1 ∶ Er−1s,t →

Er−1s−r+1,t+r−2 are both defined, we set

Ers,t = ker(dr−1 ∶ Er−1s,t → Er−1

s−r+1,t+r−2)/ im(dr−1 ∶ Er−1s+r−1,t−r+2 → Er−1

s,t ).

The differential dr ∶ Ers,t → Ers−r,t+r−1 is defined as follows. Let [x] ∈ Ers,t berepresented by an element of x ∈ E1

s,t, i.e., an element ofHs+t(grsC∗). As above,the boundary map induces natural maps ∂ ∶ Hs+t(grsC∗) → Hs+t−1(Fs−1C∗)and ∂ ∶ Hs+t−1(Fs−rC∗) → Hs+t−1(grs−rC∗). The element ∂x ∈ Hs+t−1(Fs−1C∗)in fact lifts to an element of Hs+t−1(Fs−rC∗). The image of this element under∂ inside Hs+t−1(grs−rC∗) = E1

s−r,t+r−1 begets a class in Ers−r,t+r−1; this is thedesired differential.

Exercise 60.2. Fill in the missing details in this construction of dr, and showthat (dr)2 = 0.

We have proven most of the statements in the following theorem.

Theorem-Definition 60.3. Let F∗C be a filtered complex. Then there existnatural

60. THE SPECTRAL SEQUENCE OF A FILTERED COMPLEX 193

1. bigraded groups (Ers,t)s≥0,t∈Z for any r ≥ 0, and

2. differentials dr ∶ Ers,t → Ers−r,t+r−1 for any r ≥ 0.

such that Er+1s,t is the homology of (Er∗,∗, dr), and (E0, d0) and (E1, d1) are

as above. If F∗C is bounded below, then this spectral sequence converges togr∗H∗(C), in the sense that there is an isomorphism:

E∞s,t ≃ grsHs+t(C). (5.1)

This is called a homology spectral sequence. One should think of each Er∗,∗as a “page”, with lattice points Ers,t. We still need to describe the symbols usedin the formula (5.1).

There is a filtration FsHn(C) ∶= im(Hn(FsC)→Hn(C)), and grsH∗(C) isthe associated graded of this filtration. Taking formula (5.1) literally, we onlyobtain information about the associated graded of the homology of C∗. Overvector spaces, this is sufficient to determine the homology of C∗, but in general,one needs to solve an extension problem.

To define the notation E∞ used above, let us assume that the filtration F∗Cis bounded below (so F−1C = 0). It follows that E0

s,t = FsCs+t/Fs−1Cs+t = 0for s < 0, so the spectral sequence of Theorem-Definition 60.3 is a “right halfplane” spectral sequence. It follows that in our example, the differentials fromthe group in position (s, t) must have vanishing ds+1 differential.

In turn, this implies that there is a surjection Es+1s,t → Es+2

s,t . This continues:we get surjections

Es+1s,t → Es+2

s,t → Es+3s,t → ⋯,

and the direct limit of this directed system is defined to be E∞s,t.

For instance, in the case of cellular chains, we argued above that E1s,t =

Hs+t(Xs,Xs−1), so that E1s,t = 0 if t ≠ 0, and the d1 differential is just the

differential in the cellular chain complex. It follows that E2s,t = Hcell

s (X) ift = 0, and is 0 if t ≠ 0. All higher differentials are therefore zero (becauseeither the target or the source is zero!), so Ers,t = E2

s,t for every r ≥ 2. Inparticular E∞

s,t =Hcells (X) when t = 0, and is 0 if t ≠ 0. There are no extension

problems either: the filtration on X is bounded below, so Theorem-Definition60.3 implies that grsHs+0(X) =Hs(X) ≃Hcell

s (X) = E∞s,t.

In a very precise sense, the datum of the spectral sequence of a filteredcomplex F∗C∗ determines the homology of C∗:

Corollary 60.4. Let CfÐ→ D be a map of filtered complexes. Assume that

the filtration on C and D are bounded below and exhaustive. Assume alsothat Er(f) is an isomorphism for some r. Then f∗ ∶ H∗(C) → H∗(D) is anisomorphism.

Proof. The map Er(f) is an isomorphism which is also also a chain map,i.e., it is compatible with the differential dr. It follows that Er+1(f) is anisomorphism. By induction, we conclude that E∞

s,t(f) is an isomorphism for


all s, t. Theorem-Definitino 60.3 implies that the map grs(f∗) ∶ grsH∗(C) →grsH(D) is an isomorphism.

We argue by induction using the short exact sequence:

0→ FsH∗(C)→ Fs+1H∗(C)→ grs+1H∗(C)→ 0.

We have gr0Hn(C) = F0Hn(C) = im(Hn(F0C) → Hn(C)), so the base casefollows from the five lemma. In general, f induces an isomorphism an iso-morphism on the groups on the left (by the inductive hypothesis) and right(by the above discussion), so it follows that Fsf∗ is an isomorphism by thefive lemma. Since the filtration F∗C∗ was exhaustive, it follows that f∗ is anisomorphism.

Serre spectral sequence

In this book, we will give two constructions of the Serre spectral sequence. Thesecond will appear later. Fix a fibration E

pÐ→ B, with B a CW-complex. Weobtain a filtration on E by taking the preimage of the s-skeleton of B, i.e.,Es = p−1sksB. It follows that there is a filtration on S∗(E) given by

FsS∗(E) = im(S∗(p−1sks(B))→ S∗E).

This filtration is bounded below and exhaustive. The resulting spectral se-quence of Theorem-Definition 60.3 is the Serre spectral sequence.

Let us be more explicit. We have a pushout square:

Es−1//

Es

Bs−1

// Bs

∐α∈Σs Ss−1α

//

OO

∐α∈ΣsDsα

OO

Let Fα be the preimage of the center of α cell. In particular, we have a pushout:

Es−1// Es

∐α∈Σs Ss−1α × Fα //

OO

∐α∈ΣsDsα × Fα

OO

We know that

E1s,t =Hs+t(Es,Es−1) = ⊕

α∈ΣsHs+t(Ds

α × Fα, Ss−1α × Fα).

We can suggestively view this as⊕α∈ΣsHs+1((Dsα, S

s−1α )×Fα). By the Künneth

formula (at least, if our coefficients are in a field), this is exactly⊕α∈ΣsHt(Fα).

61. EXACT COUPLES 195

In analogy with our discussion above regarding the spectral sequence comingfrom the cellular chain complex, one would like to think of this as “Cs(B;Ht(Fα))”.Sadly, there are many things wrong with writing this.

For instance, suppose B isn’t connected. The fibers Fα could have com-pletely different homotopy types, so the symbol Cs(B;Ht(Fα)) does not makeany sense. Even if B was path-connected, there would still be no canonicalway to identify the fibers over different points. Instead, we obtain a functorHt(p−1(−)) ∶ Π1(B)→Ab, i.e., a “local coefficient system” on B. So, the rightthing to say is “E2

s,t =Hs(B;Ht(fiber))”.To define precisely what Hs(B;Ht(fiber)) means, let us pick a basepoint

in B, and build the universal cover B → B. This has an action of π1(B,∗), sowe obtain an action of π1(B,∗) on the chain complex S∗(B). Said differently,S∗(B) is a chain complex of right modules over Z[π1(B)]. If B is connected,a local coefficient system on B is the same thing as a (left) action of π1(B) onHt(p−1(∗)). Then, we define a chain complex:

S∗(B;Ht(p−1(∗))) = S∗(B)⊗Z[π1(B)] Ht(p−1(∗));

the differential is induced by the Z[π1(B)]-equivariant differential on S∗(B).Our discussion above implies that the homology of this chain complex is theE2-page.

We will always be in the case where that local system is trivial, so thatH∗(B;H∗(p−1(∗))) is just H∗(B;H∗(p−1(∗))). For instance, this is the case ifπ1(B) acts trivially on the fiber. In particular, this is the case if B is simplyconnected.

61 Exact couples

Let us begin with a conceptual discussion of exact couples. As a special case,we will recover the construction of the spectral sequence associated to a filteredchain complex (Theorem-Definition 60.3).

Definition 61.1. An exact couple is a diagram of (possilby (bi)graded) abeliangroups

Ai // A

j

E

k

__

which is exact at each joint.

As jkjk = 0, the map EjkÐ→ E is a differential, denoted d. An exact couple


determines a “derived couple”:

A′ i′=i∣im i // A′

j′

~~E′

k′`` (5.2)

where A′ = im(i) and E′ = H∗(E,d). Iterating this procedure, we get exactsequences

Arir // Ar

jr

Er

kr

aa

where the next exact couple is the derived couple of the preceding exact couple.It remains to define the maps in the above diagram. Define j′(ia) = ja. A

priori, it is not clear that this well-defined. For one, we need [ja] ∈ E′; for this,we must check that dja = 0, but d = jk, and jkja = 0 so this follows. We alsoneed to check that j′ is well-defined modulo boundaries. To see this, supposeia = 0. We then need to know that ja is a boundary. But if ia = 0, then a = kefor some e, so ja = jke = de, as desired.

Define k′ ∶ H(E,d) → im i via k′([e]) ↦ ke. As before, we need to checkthat this is well-defined. For instance, we have to check that ke ∈ im i. Sincede = 0 and d = jk, we learn that jke = 0. Thus ke is killed by j, and therefore,by exactness, is in the image of i. We also need to check that k′ is independentof the choice of representative of the homology class. Say e = de′. Thenkd = kde′ = kjke′ = 0.

Exercise 61.2. Check that these maps indeed make diagram (5.2) into anexact couple.

It follows that we obtain a spectral sequence, in the sense of Theorem-Definition 60.3.

Exercise 61.3. By construction,

Ar = im(ir ∣A) = irA.

Show, by induction, that

Er = k−1(irA)j(ker ir)

and thatir(a) = ia, jr(ira) = [ja], kr(e) = ke.

Intuitively: an element of E1 will survive to Er if its image in A1 can bepulled back under ir−1. The differential dr is obtained by the homology classof the pushforward of this preimage via j to E1.


Remark 61.4. In general, the groups in consideration will be bigraded. It isclear by construction that deg(i′) = deg(i), deg(k′) = deg(k), and deg(j′) =deg(j) − deg(i). It follows by an easy inductive argument that

deg(dr) = deg(j) + deg(k) − (r − 1)deg(i).

The canonical example of an exact couple is that of a filtered complex;the resulting spectral sequence is precisely the spectral sequence of Theorem-Definition 60.3. If C∗ is a filtered chain complex, we let As,t =Hs+t(FsC∗), andE1s,t = Es,t =Hs+t(grsC∗). The exact couple is precisely that which arises from

the long exact sequence in homology associated to the short exact sequence ofchain complexes

0→ Fs−1C∗ → FsC∗ → grsC∗ → 0.

Note that in this case, the exact couple is one of bigraded groups, so Remark61.4 dictates the bidegrees of the differentials.

We will conclude this section with a brief discussion of the convergence ofthe spectral sequence constructed above. Assume that i ∶ A → A satisfies theproperty that

ker(i) ∩⋂ irA = 0.

Let A be the colimit of the directed system

AiÐ→ A

iÐ→ A→ ⋯

There is a natural filtration on A. Let I denote the image of the map A → A;the kernel of this map is ⋃ker(ir). The groups irI give an exhaustive filtrationof A, and the quotients irI/ir+1I are all isomorphic to I/iI (since i is anisomorphism on A). Then we have an isomorphism

E∞ ≃ I/iI. (5.3)

Indeed, we know from Exercise 61.3 that

E∞ ≃ k−1 (⋂ irA)j (⋃ker ir)

;

by our assumption on i, this is

ker(k)j (⋃ker ir)

≃ j(A)j (⋃ker ir)

.

But there is an isomorphism A/iA → j(A) which clearly sends iA + ⋃ker ir

to j (⋃ker ir). By our discussion above, A/⋃ker ir ≃ I, and iA/⋃ker ir ≃ iI.Modding out by iI on both sides, we get (5.3).

62 The homology of ΩSn, and the Serre exact sequence

The goal of this section is to describe a computation of the homology of ΩSn

via the Serre spectral sequence, as well as describe a “degenerate” case of theSerre spectral sequence.


The homology of ΩSn

Let us first consider the case n = 1. The space ΩS1 is the base of a fibrationΩS1 → PS1 → S1. Comparing this to the fibration Z → R → S1, we findthat ΩS1 ≃ Z. Equivalently, this follows from the discussion in §59 and theobservation that S1 ≃K(Z,1).

Having settled that case, let us now consider the case n > 1. Again, thereis a fibration ΩSn → PSn → Sn. In general, if F → E → B is a fibration andthe space F has torsion-free homology, we can (via the universal coefficientstheorem) rewrite the E2-page:

E2s,t =Hs(B;Ht(F )) ≃Hs(B)⊗Ht(F ).

Since Sn has torsion-free homology, the Serre spectral sequence (see §60) runs:

E2s,t =Hs(Sn)⊗Ht(ΩSn)⇒H∗(PSn) = Z.

Since Hs(Sn) is concentrated in degrees 0 and n, we learn that E2-page isconcentrated in columns s = 0, n. For instance, if n = 4, then the E2-page(without the differentials drawn in) looks like:

H∗(S4)

H∗(

ΩS

4)

0 1 2 3 4 5

0

0

0

0

0

0

H0(ΩSn)

H1(ΩSn)

H2(ΩSn)

H3(ΩSn)

H4(ΩSn)

H5(ΩSn)

H0(ΩSn)

H1(ΩSn)

H2(ΩSn)

H3(ΩSn)

H4(ΩSn)

H5(ΩSn)

We know that H0(ΩSn) = Z. Since the target has homology concentrated indegree 0, we know that E2

n,0 has to be killed. The only possibility is that it ishit by a differential, or that it supports a nonzero differential.

There are not very many possibilities for differentials in this spectral se-quence. In fact, up until the En-page, there are no differentials (either thetarget or source of the differential is zero), so E2 ≃ E3 ≃ ⋯ ≃ En. On the En-page, there is only one possibility for a differential: dn ∶ E2

n,0 → En0,n−1. This


differential has to be a monomorphism because if it had anything in its kernel,that will be left over in the position. In our example above (with n = 4), wehave

H∗(S4)

H∗(

ΩS

4)

0 1 2 3 4 5

0

0

0

0

0

0

H0(ΩSn)

H1(ΩSn)

H2(ΩSn)

H3(ΩSn)

H4(ΩSn)

H5(ΩSn)

H0(ΩSn)

H1(ΩSn)

H2(ΩSn)

H3(ΩSn)

H4(ΩSn)

H5(ΩSn)

d4d4

d4d4

d4d4

However, we still do not know the group En0,n−1. If it is bigger than Z, thendn is not surjective. There can be no other differentials on the Er-page forr ≥ n + 1 (because of sparsity), so the dn differential is our last hope in killingeverything in degree (0, n − 1). This means that dn is an epimorphism. Wefind that En0,n−1 =Hn−1(ΩSn) ≃ Z, and that dn is an isomorphism.

We have now discovered that Hn−1(ΩSn) ≃ Z — but there is a lot moreleft in the E2-page! For instance, we still have a Z in Enn,n−1. BecauseH∗(PSn) is concentrated in degree 0, this, too, must die! We are in exactlythe same situation as before, so the same arguments show that the differentialdn ∶ Enn,n−1 → En0,2(n−1) has to be an isomorphism. Iterating this argument, wefind:

Hq(ΩSn) ≃⎧⎪⎪⎨⎪⎪⎩

Z if (n − 1)∣q ≥ 0

0 else

This is a great example of how useful spectral sequences can be.

Remark 62.1. The loops ΩX is an associative H-space. Thus, as is the casefor any H-space, the homology H∗(ΩX;R) is a graded associative algebra.Recall that the suspension functor Σ is the left adjoint to the loops functor Ω, sothere is a unit map A→ ΩΣA. This in turn begets a map H∗(A)→H∗(ΩΣA).

Recall that the universal tensor algebra Tens(H∗(A)) is the free associativealgebra on H∗(A). Explicitly:

Tens(H∗(A)) =⊕n≥0

H∗(A)⊗n.


In particular, by the universal property of Tens(H∗(A)), we get a map α ∶Tens(H∗(A))→H∗(ΩΣA).

Theorem 62.2 (Bott-Samelson). The map α is an isomorphism if R is a PIDand H∗(A) is torsion-free.

For instance, if A = Sn−1 then ΩSn = ΩΣA. Theorem 62.2 then shows that

H∗(ΩSn) = Tens(H∗(Sn−1)) = ⟨1, x, x2, x3,⋯⟩,

where ∣x∣ = n − 1. It is a mistake to call this “polynomial”, since if n is even, xis an odd class (in particular, x squares to zero by the Koszul sign rule).

Theorem 62.2 suggests thinking of ΩΣA as the “free associative algebra” onA. Let us make this idea more precise.

Remark 62.3. The space ΩA is homotopy equivalent to a topological monoidΩMA, called the Moore loops on A. This means that ΩMA has a strict unitand is strictly associative (i.e., not just up to homotopy). Concretely,

ΩMA ∶= (`, ω) ∶ ` ∈R≥0, ω ∶ [0, `]→ A,ω(0) = ∗ = ω(`),

topologized as a subspace of the product. There is an identity class 1 ∈ ΩMA,given by 1 = (0, c∗) where c∗ is the constant loop at the basepoint ∗. Theaddition on this space is just given by concatenatation. In particular, thelengths get added; this overcomes the obstruction to ΩA not being strictlyassociative, so the Moore loops ΩMA are indeed strictly associative. If thebasepoint is nondegenerate, it is not hard to see that the inclusion ΩA ΩMAis a homotopy equivalence.

Given the space A, we can form the free monoid FreeMon(A). The elementsof this space are just formal sequences of elements of A (with topology comingfrom the product topology), and the multiplication is given by juxtaposition.Let us adjoin the element 1 = ∗. As with all free constructions, there is a mapA → FreeMon(A) which is universal in the sense that any map A → M to amonoid factors through FreeMon(A).

The unit A→ ΩΣA is a map from A to a monoid, so we get a monoid mapβ ∶ FreeMon(A)→ ΩΣA.

Theorem 62.4 (James). The map β ∶ FreeMon(A) → ΩΣA is a weak equiva-lence if A is path-connected.

The free monoid looks very much like the tensor product, as the followingtheorem of James shows.

Theorem 62.5 (James). Let J(A) = FreeMon(A). There is a splitting:

ΣJ(A) ≃w Σ(⋁n≥0

A∧n) .


Applying homology to the splitting of Theorem 62.5 shows that:

H∗(J(A)) ≃⊕n≥0

H∗(A∧n).

Assume that our coefficients are in a PID, and that H∗(A) is torsion-free;then this is just ⊕n≥0 H∗(A)⊗n. In particular, we recover our computation ofH∗(ΩSn) from these general facts.

The Serre exact sequence

Suppose π ∶ E → B is a fibration over a path-connected base. Assume thatHs(B) = 0 for s < p where p ≥ 1. Let ∗ ∈ B be a chosen basepoint. Denote byF the fiber π−1(∗). Assume Ht(F ) = 0 for t < q, where q ≥ 1. We would liketo use the Serre spectral sequence to understand H∗(E). As always, we willassume that π1(B) acts trivially on H∗(F ).

Recall that the Serre spectral sequence runs

E2s,t =Hs(B;Ht(F ))⇒Hs+t(E).

Our assumptions imply that E20,0 = Z, and E2

0,t = 0 for t < q. Moreover, E2s,0 = 0

for s < p. In particular, E20,q+t = Hq+t(F ) and E2

p+k,0 = Hp+k(B) — the rest ofthe spectral sequence is mysterious.

By sparsity, the first possible differential is dp ∶ Hp(B) → Hp−1(F ), anddp+q ∶ Hp+1(B) → Hp(F ). In the mysterious zone, there are differentials thathit E2

p,q.Again by sparsity, the only differential is ds ∶ Ess,0 → Es0,s−1 for s < p + q − 1.

This is called a transgression. It is the last possible differential which has achance at being nonzero. This means that the cokernel of ds is E∞

0,s−1. Thereis also a map E∞

s,0 → Ess,0. We obtain a mysterious composite

0→ E∞s,0 → Ess,0 ≃Hs(B) dsÐ→ Es0,s−1 ≃Hs−1(F )→ E∞

0,s−1 → 0. (5.4)

Let n < p + q − 1. Recall that FsHn(E) = im(H∗(π−1(sks(B))) → H∗(E)),so F0Hn(E) = E∞

0,n. Here, we are using the fact that F−1H∗(E) = 0. Inparticular, there is a map E∞

0,n →Hn(E). By our hypotheses, there is only oneother potentially nonzero filtration in this range of dimensions, so we have ashort exact sequence:

0→ F0Hn(E) = E∞0,n →Hn(E)→ E∞

n,0 → 0 (5.5)

Splicing the short exact sequences (5.4) and (5.5), we obtain a long exactsequence:

Hp+q−1(F )→ ⋯→Hn(F )→Hn(E)→Hn(B) transgressionÐÐÐÐÐÐÐ→Hn−1(F )→Hn−1(E)→ ⋯

This is called the Serre exact sequence. In this range of dimensions, homologybehaves like homotopy.


63 Edge homomorphisms, transgression

Recall the Serre spectral sequence for a fibration F → E → B has E2-pagegiven by

E2s,t =Hs(B;Ht(F ))⇒Hs+t(E).

If B is path-connected, Ht(F ) = 0 for t < q, Hs(B) = 0 for s < p, and π1(B)acts trivially on H∗(F ), we showed that there is a long exact sequence (theSerre exact sequence)

Hp+q−1(F ) Ð→Hp+q−1(E)→Hp+q−1(B)→Hp+q−2(F )→ ⋯ (5.6)

Let us attempt to describe the arrow marked by .Let (Erp,q, dr) be any spectral sequence such that Erp,q = 0 if p < 0 or q < 0;

such a spectral sequence is called a first quadrant spectral sequence. The Serrespectral sequence is a first quadrant spectral sequence. In a first quadrantspectral sequence, the d2-differential d2 ∶ E2

0,t → E2−2,t+1 is zero, since E2

s,t

vanishes for s < 0. This means that Ht(F ) = H0(B;Ht(F )) = E20,t surjects

onto E30,t. Arguing similarly, this surjects onto E4

0,t. Eventually, we find thatEr0,t ≃ Et+2

0,t for r ≥ t + 2. In particular,

Et+20,t ≃ E∞

0,t ≃ gr0Ht(E) ≃ F0Ht(E),

which sits inside Ht(E). The composite

E20,t =Ht(F )→ E3

0,t → ⋯→ Et+20,t ⊆ F0Ht(E)→Ht(E)

is precisely the map ! Such a map is known as an edge homomorphism.The map F → E is the inclusion of the fiber; it induces a map Ht(F ) →

Ht(E) on homology. We claim that this agrees with . Recall that F0Ht(E) isdefined to be im(Ht(F0E)→Ht(E)). In the construction of the Serre spectralsequence, we declared that F0E is exactly the preimage of the zero skeleton.Since B is simply connected, we find that F0E is exactly the fiber F .

To conclude the proof of the claim, consider the following diagram:

F //

F

F //

E

∗ // B

The naturality of the Serre spectral sequence implies that there is an inducedmap of spectral sequences. Tracing through the symbols, we find that thisobservation proves our claim.

The long exact sequence (5.6) also contains a map Hs(E) → Hs(B). Thegroup FsHs(E) = Hs(E) maps onto grsHs(E) ≃ E∞

s,0. If F is connected, then

63. EDGE HOMOMORPHISMS, TRANSGRESSION 203

Hs(B) = Hs(B;H0(F )) = E2s,0. Again, the d2-differential d2 ∶ E2

s+2,−1 → E2s,0

is trivial (since the source is zero). Since E3 = kerd2, we have an injectionE3s,0 → E2

s,0. Repeating the same argument, we get injections

E∞s,0 = Es+1

s,0 → ⋯→ E2s,0 → E2

s,0 =Hs(B).

Composing with the map Hs(E)→ E∞s,0 gives the desired map Hs(E)→Hs(B)

in the Serre exact sequence. This composite is also known as an edge homo-morphism.

As above, this edge homomorphism is the map induced by E → B. This canbe proved by looking at the induced map of spectral sequences coming fromthe following map of fiber sequences:

F //

∗

E //

B

B // B

The topologically mysterious map is the boundary map ∂ ∶ Hp+q−1(B) →Hp+q−2(F ). Such a map is called a transgression. Again, let (Ers,t, dr) be afirst quadrant spectral sequence. In our case, E2

n,0 = Hn(B), at least F isconnected. As above, we have injections

i ∶ Enn,0 → ⋯→ E3n,0 → E2

n,0 =Hn(B).

Similarly, we have surjections

s ∶ E20,n−1 → E3

0,n−1 → ⋯→ En0,n−1.

There is a differential dn ∶ Enn,0 → En0,n−1. The transgression is defined as thelinear relation (not a function!) E2

n,0 → E20,n−1 given by

x↦ i−1dns−1(x).

However, the reader should check that in our case, the transgression is indeeda well-defined function.

Topologically, what is the origin of the transgression? There is a mapHn(E,F ) π∗Ð→ Hn(B,∗), as well as a boundary map ∂ ∶ Hn(E,F ) → Hn−1(F ).We claim that:

imπ∗ = im(Enn,0 →Hn(B) = E2n,0), ∂ kerπ∗ = ker(Hn−1(F ) = E2

0,n−1 → En0,n−1).

Proof sketch. Let x ∈ Hn(B). Represent it by a cycle c ∈ Zn(B). Lift it to achain in the total space E. In general, this chain will not be a cycle (considerthe Hopf fibration). The differentials record this boundary; let us recall the


geometric construction of the differential. Saying that the class x survives tothe En-page is the same as saying that we can find a lift to a chain σ in E,with dσ ∈ Sn−1(F ). Then dn(x) is represented by the class [dc] ∈ Hn−1(F ).This is precisely the trangression.

Informally, we lift something from Hn(B) to Sn(E); this is well-defined upto something in F . In particular, we get an element in Hn(E,F ). We send it,via ∂, to an element of Hn−1(F ) — and this is precisely the transgression.

An example

We would like to compare the Serre exact sequence (5.6) with the homotopyexact sequence:

∗→ πp+q−1(F )→ πp+q−1(E)→ πp+q−1(B) ∂Ð→ πp+q−2(F )→ ⋯

There are Hurewicz maps πp+q−1(X) → Hp+q−1(X). We claim that there is amap of exact sequences between these two long exact sequences.

Hp+q−1(E) π∗ // Hp+q−1(B)∂// Hp+q−2(F ) // ⋯

πp+q−1(E) π∗//

h

OO

πp+q−1(B)

h

OO

// πp+q−2(F ) //

h

OO

⋯

The leftmost square commutes by naturality of Hurewicz. The commutativityof the righmost square is not immediately obvious. For this, let us draw in theexplicit maps in the above diagram:

Hp+q−1(E,F )

vv ''Hp+q−1(E) π∗ // Hp+q−1(B)

∂// Hp+q−2(F ) // ⋯

πp+q−1(E) π∗//

''

h

OO

πp+q−1(B)

h

OO

// πp+q−2(F ) //

h

OO

⋯

πp+q−1(E,F )

CC

33

≅ s

OO

The map marked s is an isomorphism (and provides the long arrow in the abovediagram, which makes the square commute), since

πn(E,F ) = πn−1(hofib(F → E)) = πn−1(ΩB) = πn(B).

Let us now specialize to the case of the fibration

ΩX → PX →X.

64. SERRE CLASSES 205

Assume that X is connected, and ∗ ∈ X is a chosen basepoint. Let p ≥ 2, andsuppose that Hs(X) = 0 for s < p. Arguing as in §62, we learn that the Serrespectral sequence we know that the homology of ΩX begins in dimension p− 1since PX ≃ ∗, so q = p−1. Likewise, if we knew Hn(ΩX) = 0 for n < p−1, thenthe same argument shows that Hn(X) = 0 for n < p.

A surprise gust: the Hurewicz theorem

The discussion above gives a proof of the Hurewicz theorem; this argument isdue to Serre.

Theorem 63.1 (Hurewicz, Serre’s proof). Let p ≥ 1. Suppose X is a pointedspace with πi(X) = 0 for i < p. Then Hi(X) = 0 for i < p and πp(X)ab →Hp(X)is an isomorphism.

Proof. Let us assume the case p = 1. This is classical: it is Poincaré’s theo-rem. We will only use this result when X is a loop space, in which case thefundamental group is already abelian.

Let us prove this by induction, using the loop space fibration. By assump-tion, πi(ΩX) = 0 for i < p − 1. By our inductive hypothesis, Hi(ΩX) = 0 fori < p − 1, and πp−1(ΩX) ≃Ð→ Hp−1(ΩX). By our discussion above, we learn

that Hi(X) = 0 for i < p. The Hurewicz map πp(X) hÐ→ Hp(X) fits into acommutative diagram:

πp−1(ΩX) ≃ // Hp−1(ΩX)

πp(X)

≃

OO

h// Hp(X)

≃ transgression

OO

It follows from the Serre exact sequence that the transgression is an isomor-phism.

64 Serre classes

Definition 64.1. A class C of abelian groups is a Serre class if:

1. 0 ∈C.

2. if I have a short exact sequence 0 → A → B → C → 0, then A&C ∈ C ifand only if B ∈C.

Some consequences of this definition: a Serre class is closed under isomor-phisms (easy). A Serre class is closed under subobjects and quotients, becausethere is a short exact equence

0→ A B → B/A→ 0.


Consider an exact sequence A → B → C (not necessarily a short exact se-quence). If A,C ∈C, then B ∈C because we have a short exact sequence:

coker i

// 0

Ai //

Bp //

;;

C 0

0 // kerp

==

Some examples are in order.

Example 64.2. 1. C = 0, and C the class of all abelian groups.

2. Let C be the class of all torsion abelian groups. We need to check thatC satisfies the second condition of Definition ??. Consider a short exactsequence

0→ AiÐ→ B

pÐ→ C → 0.

We need to show that B is torsion if A and C are torsion. To see this, letb ∈ B. Then p(b) is killed by some integer n, so there exists a ∈ A suchthat i(a) = nb. SInce A is torsion, it follows that b is torsion, too.

3. Let P be a set of primes. Define:

CP = A ∶ if p /∈ P, then p ∶ A ≃Ð→ A, i.e., A is a Z[1/p]-module

Let Z(P) = Z[1/p ∶ p /∈ P] ⊆Q.

For instance, if P is the set of all primes, then CP is the Serre class of allabelian groups. If P is the set of all primes other than `, then CP is theSerre class consisting of all Z[1/`]-modules. If P = `, then C` =∶ C`

is the Serre class of all Z(`)-modules. If P = ∅, then C∅ is all rationalvector spaces.

4. If C and C′ are Serre classes, then so is C∩C′. For instance, Ctors ∩Cfgis the Serre class Cfinite. Likewise, Cp ∩ Ctors is the Serre class of allp-torsion abelian groups.

Here are some straightforward consequences of the definition:

1. If C is a chain complex, and Cn ∈C, then Hn(C) ∈C.

2. Suppose F∗A is a filtration on an abelian group. If A ∈C, then grnA ∈Cfor all n. If F∗A is finite and grnA ∈C for all n, then A ∈C.

3. Suppose we have a spectral sequence Er. If E2s,t ∈C, then Ers,t ∈C for

r ≥ 2. It follows that if Er is a right half-plane spectral sequence, thenEs+1s,t ↠ Es+2

s,t ↠ ⋯↠ E∞s,t ∈C.

64. SERRE CLASSES 207

Thus, if the spectral sequence comes from a filtered complex (which isbounded below, such that for all n there exists an s such that FsHn(C) =Hn(C), i.e., the homology of the filtration stabilizes), then E∞

s,t = grsHs+t(C).This means that if the E2

s,t ∈C for all s + t = n, then Hn(C) ∈C.

To apply this to the Serre spectral sequence, we need an additional axiom forDefinition 64.1:

2. if A,B ∈C, then so are A⊗B and Tor1(A,B).

All of the examples given above satisfy this additional axiom.

Terminology 64.3. f ∶ A→ B is said to be a C-epimorphism if coker f ∈C, aC-monomorphism if ker f ∈ C, and a C-isomorphism if it is a C-epimorphismand a C-monomorphism.

Proposition 64.4. Let π ∶ E → B be a fibration and B path connected, suchthat the fiber F = π−1(∗) is path connected. Suppose π1(B) acts trivially onH∗(F ).

Let C be a Serre class satisfiying Axiom 2. Let s ≥ 3, and assume thatHn(E) ∈ C where 1 ≤ n < s − 1 and Ht(B) ∈ C for 1 ≤ t < s. Then Ht(F ) ∈ Cfor 1 ≤ t < s − 1.

Proof. We will do the case s = 3, for starters. We’re gonna want to relatethe low-dimension homology of these groups. What can I say? We knowthat H0(E) = Z since it’s connected. I have H1(E) → H1(B), via π. Thisis one of the edge homomorphisms, and thus it surjects (no possibility for adifferential coming in). I now have a map H1(F ) → H1(E). But I have apossible d2 ∶H2(B)→H1(F ), which is a transgression that gives:

H2(B) ∂Ð→H1(F )→H1(E)→H1(B)→ 0

Let me take a step back and say something general. You might be interestedin knowing when something in Hn(F ) maps to zero in Hn(E). I.e., what’sthe kernel of Hn(F ) → Hn(E). The sseq gives an obstruction to being anisomorphism. The only way that something can be killed by Hn(F )→Hn(E)is described by:

ker(Hn(F )→Hn(E)) =⋃ (im of dr hitting Er0,n)

You can also say what the cokernel is: it’s whatever’s left in E∞s,t with s+ t = n.

These obstruct Hn(F )→Hn(E) from being surjective.In the same way, I can do this for the base. If I have a class in Hn(E), that

maps to Hn(B), the question is: what’s the image? Well, the only obstructionis the possibility is that the element in Hn(B) supports a nonzero differential.Thus:

im(Hn(E) π∗Ð→Hn(B)) =⋂ (ker(dr ∶ Err,0 → ⋯))


Again, you can think of the sseq as giving obstructions. And also, the obstruc-tion to that map being a monomorphism that might occur in lower filtrationalong the same total degree line.

Back to our argument. We had the low-dimensional exact sequence:

H2(B) ∂Ð→H1(F )→H1(E)→H1(B)→ 0

Here p = 3, so we have H2(B) ∈ C and H1(E) ∈ C. Thus H1(F ) ∈ C. That’sthe only thing to check when p = 3.

Let’s do one more case of this induction. What does this say? Now I’ll dop = 4. We’re interested in knowing if E2

0,3 ∈ C. There are now two possibledifferentials! I have H2(F ) = E2

0,2 ↠ E30,2. This quotient comes from d2 ∶

E22,1 → E2

0,2. Now, d3 ∶ E33,0 → E3

0,2 which gives a surjection E30,2 ↠ E4

0,2 ≃E∞

0,2 H2(E). Now, our assumptions were that E22,1,E

33,0,H2(E) ∈ C. Thus

E30,2 ∈C and so E2

0,2 =H2(F ) ∈C. Ta-da!

We’re close to doing actual calculations, but I have to talk about the mul-tiplicative structure on the Serre sseq first.

65 Mod C Hurewicz, Whitehead, cohomology spectralsequence

We had Cfg and Ctors, and

CP = A∣` ∶ A ≃Ð→ A, ` /∈ P, Cp =Cp, Cp′ =Cnot p

Another one is Cp′ ∩Ctors, which consists of torsion groups such that p is anisomorphism on A. There is therefore no p-torsion, and it has only prime-to-ptorsion. This is the same thing as saying that A⊗Z(p) = 0.

Theorem 65.1 (Mod C Hurewicz). Let X be simply connected and C a Serreclass such that A,B ∈ C implies that A⊗B,Tor1(A,B) ∈ C (this is axiom 2).Assume also that if A ∈ C, then Hj(K(A,1)) = Hj(BA) ∈ C for all j > 0.(This is valid for all our examples, and is what is called Axiom 3.)

Let n ≥ 1. Then πi(X) ∈ C for any 1 < i < n if and only if Hi(X) ∈ C forany 1 < i < n, and πn(X)→Hn(X) is a mod C isomorphism.

Example 65.2. For 1 < i < n, the group Hi(X) is:

1. torsion;

2. finitely generated;

3. finite;

4. − ⊗Z(p) = 0

if and only if πi(X) for 1 < i < n.

65. MOD C HUREWICZ, WHITEHEAD, COHOMOLOGY SPECTRALSEQUENCE 209

Proof. Look at ΩX → PX →X. Then π1ΩX ∈C. Look at Davis+Kirk.

There’s a Whitehead theorem that comes out of this, that I want to statefor you.

Theorem 65.3 (ModCWhitehead theorem). Let C be a Serre class satisfyingaxioms 1, 2, 3, and:

(2′) A ∈C implies that A⊗B ∈C for any B.

This is satisfied for all our examples except Cfg.Suppose I have f ∶ X → Y where X,Y are simply connected. Suppose

π2(X) → π2(Y ) is onto. Let n ≥ 2. Then πi(X) → πi(Y ) is a C-isomorphismfor 2 ≤ i ≤ n and is a C-epimorphism for i = n, with the same statement forHi.

These kind of theorems help us work locally at a prime, and that’s super.You’ll see this in the next assignment, which is mostly up on the web. You’llalso see this in calculations which we’ll start doing in a day or two.

Change of subject here. Today I’m going to say a lot of things for which Iwon’t give a proof. I want to talk about cohomology sseq.

Cohomology sseq

We’re building up this powerful tool using spectral sequences. We saw howpowerful the cup product was, and that is what cohomology is good for. Incohomology, things get turned upside down:

Definition 65.4. A decreasing filtration of an object A is

A ⊇ ⋯ ⊇ F −1A ⊇ F 0A ⊇ F 1A ⊇ F 2A ⊇ ⋯ ⊇ 0

This is called “bounded above” if F 0A = A. Write grsA = F sA/F s+1A.

Example 65.5. Suppose X is a filtered space. So there’s an increasing filtra-tion ∅ = F−1X ⊆ F0X ⊆ ⋯. Let R be a commutative ring of coefficients. ThenI have S∗(X), where the differential goes up one degree. Define

F sS∗(X) = ker(S∗(X)→ S∗(Fs−1X))

For instance, F 0S∗(X) = S∗(X). Thus this is a bounded above decreasingfiltration . My computer will run out

of juice soon, TEXthis uplater!

My computer will run outof juice soon, TEXthis uplater!Example 65.6. Let X = E πÐ→ B = CW-complex with π1(B) acting trivially

on Ht(F ). Then FsE = π−1(sksB). Thus I get a filtration on S∗(E), and

F sH∗(X) = ker(H∗(X)→H∗(Fs−1X))

Doing everything the same as before, we get a cohomology spectral sequence.Here are some facts.


1. First, you have Es,tr (note that indices got reversed). There’s a differentialdr ∶ Es,tr → Es+r,t−r+1

r , so that the total degree of the differential is 1.

2. You discover thatEs,t2 ≃Hs(B;Ht(F ))

3. and Es,t∞ ≃ grsHs+t(E).

4.

66 A few examples, double complexes, Dress sseq

Way back in 905 I remember computing the cohomology ring of CPn usingPoincaré duality. Let’s do it fresh using the fiber sequence

S1 → S2n+1 →CPn

where S1 acts on S2n+1. Here we know the cohomology of the fiber and thetotal space, but not the cohomology of the base. Let’s look at the cohomologysseq for this. Then

Es,t2 =Hs(CPn;Ht(S1)) ≃Hs(CPn)⊗Ht(S1)RightarrowHs+t(S2n+1)

The isomorphism Hs(CPn;Ht(S1)) ≃ Hs(CPn) ⊗ Ht(S1) follows from theUCT.

We know at least that CPn is simply connected by the lexseq of homotopygroups. I don’t have to worry about local coefficients. Let’s work with thecase S5. We know that CPn is simply connected, so the one-dimensionalcohomology is 0. The only way to kill E0,1

2 is by sending it via d2 to E2,02 . Is

this map surjective? Yes, it’s an isomorphism.Now I’m going to give names to the generators of these things; see the below

diagram. E2,12 is in total degree 3 and so we have to get rid of it. I will compute

d2 on this via Leibniz:

d2(xy) = (d2x)y − xd2y = (d2x)y = y2

which gives (iterating the same computation):

0 1 2 3 4

0

0 Zx

Z 0

0

Zy

Zxy

0

0

Zy2

Zxy2

d2d2 d2d2

This continues until the end where you reach Zxy?? which is a permanent cyclesince it lasts until the E∞-page.

66. A FEW EXAMPLES, DOUBLE COMPLEXES, DRESS SSEQ 211

Another example: let Cm be the cyclic group of order m sitting inside S1.How can we analyse S2n+1/Cm =∶ L? This is the lens space. We have a mapS2n+1/Cm → S2n+1/S1 = CPn. This is a fiber bundle whose fiber is S1/Cm.The spectral sequence now runs:

E2s,t =Hs(CPn)⊗Ht(S1/Cm)⇒Hs+t(L)

We know the whole E2 term now:

0 1 2 3 4

0

0 Z

Z 0

0

Z

Z

0

0

Z

Zmm

In cohomology, we have something dual:

0 1 2 3 4

0

0 u

1 0

0

y

uy

0

0

y2

uy2

mm mm

What’s the ring structure? We get that H∗(L) = Z[y, v]/(my, yn+1, yv, v2)where ∣v∣ = 2n + 1 and ∣y∣ = 2. By the way, when m = 1, this is RP2n+1. Thisis a computation of the cohomology of odd real projective spaces. Rememberthat odd projective spaces are orientable and you’re seeing that here becauseyou’re picking up a free abelian group in the top dimension.

Double complexes

As,t is a bigraded abelian group with dh ∶ As,t → As−1,t and dv ∶ As,t → As,t−1

such that dvdh = dhdv. Assume that (smt) ∶ s+ t = n,As,t ≠ 0 is finite for anyn. Then

(tA)n = ⊕s+t=n

As,t

Under this assumption, there’s only finitely many nonzero terms. I like thispersonally because otherwise I’d have to decide between the direct sum andthe direct product, so we’re avoiding that here. It’s supposed to be a chaincomplex. Here’s the differential:

d(as,t) = dhas,t + (−1)sdvas,t

Then d2 = 0, as you can check.


Question 66.1. What is H∗(tA∗)?

Define a filtration as follows:

Fp(tA)n = ⊕s+t=n,s≤p

As,t ⊆ (tA)n

This kinda obviously gives a filtered complex. Let’s compute the low pages ofthe sseq. What is grs(tA)? Well

grs(tA)s+t = (Fs/Fs−1)s+t = As,t

This associated graded object has its own differential grs(tA)s+t = As,tdvÐ→

As,t−1 = grs(tA)s+t−1. Let E0s,t = grs(tA)s+t = As+t, so that d0 = dv. Then

E1 = H(E0s,t, d

0) = H(As,t;dv) =∶ Hvs,t(A). So computing E1 is ez. Well,

what’s d1 then?To compute d1 I take a vertical cycle that and the differential decreases

the ... by 1, so that d1 is induced by dh. This means that I can write E2s,t =

Hhs,t(Hv(A)).

Question 66.2. You can also do ′E2s,t =Hv

s,t(Hh(A)), right?

Rather than do that, you can define the transposed double complex ATt,s =

As,t, and dTh(as,t) = (−1)sdv(as,t) and dTv (as,t) = (−1)tdhas,t. When I set thesigns up like that, then

tAT ≃ tA

as complexes and not just as groups (because of those signs). Thus, you get aspectral sequence

TE2s,t =Hv

s,t(Hh(A))

converging to the same thing. I’ll reserve telling you about Dress’ constructionuntil Monday because I want to give a double complex example. It’s not ...it’s just a very clear piece of homological algebra.

Example 66.3 (UCT). For this, suppose I have a (not necessarily commuta-tive) ring R. Let C∗ be a chain complex, bounded below of right R-modules,and let M be a left R-module. Then I get a new chain complex of abeliangroups via C∗ ⊗RM . What is H(C∗ ⊗RM)? I’m thinking of M as some kindof coefficient. Let’s assume that each Cn is projective, or at least flat, for alln.

Shall we do this?Let M ← P0 ← P1 ← ⋯ be a projective resolution of M as a left R-module.

Then H∗(P∗)≃Ð→M . Form C∗ ⊗R P∗: you know how to do this! I’ll define As,t

to be Cs⊗RPt. It’s got two differentials, and it’s a double complex. Let’s workout the two sseqs.

Firstly, let’s take it like it stands and take homology wrt P first. I’morganizing it so that C is along the base and P is along the fiber. What is the

67. DRESS SPECTRAL SEQUENCE, LERAY-HIRSCH 213

vertical homology Hv(A∗,∗)? If the C are projective then tensoring with themis exact, so that Hv(As,∗) = Cs ⊗RH∗(P∗), so that E1

s,t =Hvs,t(A∗,∗) = Cs ⊗M

if t = 0 and 0 otherwise. The spectral sequence is concentrated in one row.Thus,

E2s,t =

⎧⎪⎪⎨⎪⎪⎩

Hs(C∗ ⊗RM) if t = 0

0 else

This is canonically the same thing as E∞s,0 ≃Hs(tA).

Let me go just one step further here. The game is to look at the otherspectral sequence, where I do horizontal homology first. Then Hh(A∗,∗) =Ht(C∗)⊗ Ps again because the P∗ are projective. Thus,

E2s,t =Hv(Hh(A∗,∗)) = TorRs (Ht(C),M)⇒Hs+t(C∗ ⊗RM)

That’s the universal coefficients spectral sequence.What happens if R is a PID? Only two columns are nonzero, and E2

0,n =Hn(C)⊗RM and E2

1,n−1 = Tor1(Hn−1(C),M). This exactly gives the universalcoefficient exact sequence.

Later we’ll use this stuff to talk about cohomology of classifying spaces andGrassmannians and Thom isomorphisms and so on.

67 Dress spectral sequence, Leray-Hirsch

I think I have to be doing something tomorrow, so no office hours then. Thenew pset is up, and there’ll be one more problem up. There are two more thingsabout spectral sequences, and specifically the multiplicative structure, that Ihave to tell you about. The construction of the Serre sseq isn’t the one that wegave. He did stuff with simplicial homology, but as you painfully figured out,∆s × ∆t isn’t another simplex. Serre’s solution was to not use simplices, butto use cubes. He defined a new kind of homology using the n-cube. It’s morecomplicated and unpleasant, but he worked it out.

Dress’ sseq

Dress made the following variation on this idea, which I think is rather beau-tiful. We have a trivial fiber bundle ∆t → ∆s × ∆t → ∆s. Let’s do with thiswhat we did with homology in the first place. Dress started with some mapπ ∶ E → B (not necessarily a fibration), and he thought about the set of mapsfrom ∆s × ∆t → ∆s to π ∶ E → B. This set is denoted Sins,t(π). This for-gets down to Ss(B). Altogether, this Sin∗,∗(π) is a functor ∆op ×∆op → Set,forming a “bisimplicial set”.

The next thing we did was to take the free R-module, to get a bisimplicialR-module RSin∗,∗(π). We then passed to chain complexes by forming thealternating sum. We can do this in two directions here! (The s is horizontaland t is vertical.) This gives us a double complex. We now get a spectral


sequence! I hope it doesn’t come as a surprise that you can compute thehorizontal – you can compute the vertical differential first, and then taking thehorizontal differential gives the homology of B with coefficients in something.Oh actually, the totalization tRSin∗,∗(π) ≃ RSin∗(E) = S∗(E). We’ll have

E2s,t =Hs(B; crazy generalized coefficients)⇒Hs+t(E)

These coefficients may not even be local since I didn’t put any assumptionson π! This is like the “Leray” sseq, set up without sheaf theory. If π is afibration, then those crazy generalized coefficients is the local system given bythe homology of the fibers. This gives the Serre sseq.

This has the virtue of being completely natural. Another virtue is that Ican form Hom(−,R), and this gives rise to a multiplicative double complex.Remember that the cochains on a space form a DGA, and that’s where the cupproduct comes from. The same story puts a bigraded multiplication on thisdouble complex, and that’s true on the nose. That gives rise a multiplicativecohomology sseq.

This is very nice, but the only drawback is that the paper is in German.That was item one in my agenda.

Leray-Hirsch

This tells you condition under which you can compute the cohomology of atotal space. Anyway. We’ll see.

Let’s suppose I have a fibration π ∶ E → B. For simplicity suppose that Bis path connected, so that gives meaning to the fiber F which we’ll also assumeto be path-connected. All cohomology is with coefficients in a ring R. I havea sseq

Es,t2 =Hs(B;HtF )⇒Hs+t(E)

If you want assume that π1(B) acts trivially so that that cohomology in localcoefficients is just cohomology with coefficients in H∗F . I have an algebramap π∗ ∶ H∗(B) → H∗(E), making H∗(E) into a module over H∗(B). Wehave E∗,t

2 = H∗(B;Ht(F )), and this is a H∗(B)-module. That’s part of themultiplicative structure, since E∗,0

2 = H∗B. This row acts on every other rowby that module structure.

Everything in the bottom row is a permanent cycle, i.e., survives to theE∞-page. In other words

H∗(B) = E∗,02 ↠ E∗,0

3 ↠ ⋯↠ E∗,0∞

Each one of these surjections is an algebra map.What the multiplicative structure is telling us is that E∗,0

r is a gradedalgebra acting on E∗,t

r . Thus, E∗,t∞ is a module for H∗(B).

Really I should be saying that it’s a module for H∗(B;H0(F )). Can Iguarantee that the π1(B)-action on F is trivial. We know that F → ∗ inducesan iso on H0 (that’s part of being path-connected). So if you have a fibration

67. DRESS SPECTRAL SEQUENCE, LERAY-HIRSCH 215

whose fiber is a point, there’s no possibility for an action. This fibration looksthe same as far as H0 of the fiber is concerned. Thus the π1(B)-action is trivialon H0(F ), so saying that it’s a H∗(B)-module is fine.

Where were we? We have module structures all over the place. In particu-lar, we know that H∗(E) is a module over H∗(B) as we saw, and also E∗,t

∞ isa H∗(B)-module. These better be compatible!

Define an increasing filtration on H∗(E) via FtHn(E) = Fn−tHn(E). Forinstance, F0H

n(E) = FnHn(E). What is that? In our picture, we have theassociated quotients along the diagonal on Es,t∞ given by s + t = n. In the end,since we know that Fn+1Hn(E) = 0, it follows that

F0Hn(E) = FnHn(E) = En,0∞ = im(π∗ ∶Hn(B)→Hn(E))

With respect to this filtration, we have

grtH∗(E) = E∗,t

∞

I learnt this idea from Dan Quillen. It’s a great idea. This increasing filtrationF∗H

∗(E) is a filtration by H∗(B)-modules, and grtH∗(E) = E∗,t

∞ is true asH∗B-modules. It’s exhaustive and bounded below.

This is a great perspective. Let’s use it for something. Let me give you theLeray-Hirsch theorem.

Theorem 67.1 (Leray-Hirsch). Let π ∶ E → B.

1. Suppose B and F are path-connected.

2. Suppose that Ht(F ) is free1 of finite rank as a R-module.

3. Also suppose that H∗(E) ↠ H∗(F ). That’s a big assumption; it’s dualis saying that the homology of the fiber injects into the homology of E.This is called “totally non-homologous to zero” – this is a great phrase, Idon’t know who invented it.

Pick an R-linear surjection σ ∶ H∗(F ) → H∗(E); this defines a map σ ∶H∗(B)⊗R H∗(F ) → H∗(E) via σ(x⊗ y) = π∗(x) ∪ σ(y). This is the H∗(B)-linear extension. Then σ is an isomorphism.

Remark 67.2. It’s not natural since it depends on the choice of σ. It tellsyou that H∗(E) is free as a H∗(B)-module. That’s a good thing.

Proof. I’m going to use our Serre sseq

Es,t2 =Hs(B;HtF )⇒Hs+t(E)

Our map H∗(E)→H∗(F ) is an edge homomorphism in the sseq, which meansthat it factors as H∗(E) → E0,∗

2 = H0(B;H∗(F )) ⊆ H∗(F ). Since H∗(E) →H∗(F ), we have H0(B;H∗(F )) ≃ H∗(F ). Thus the π1(B)-action on F istrivial.

1Everything is coefficients in R


Question 67.3. What’s this arrow H∗(E)→ E0,∗2 ? We have a map H∗(E)→

H∗(E)/F 1 = E0,∗∞ . This includes into E0,∗

2 .

Now you know that the E2-term is Hs(B;Ht(F )). By our assumption onH∗(F ), this is Hs(B) ⊗R Ht(F ), as algebras. What do the differentials looklike? I can’t have differentials coming off of the fiber, because if I did thenthe restriction map to the fiber wouldn’t be surjective, i.e., that dr ∣E0,∞

r= 0.

The differentials on the base are of course zero. This proves that dr is zero onevery page by the algebra structure! This means that E∞ = E2, i.e., E∗,t

∞ =H∗(B)⊗Ht(F ).

Now I can appeal to the filtration stuff that I was talking about, so thatE∗,t∞ = grtH

∗(E). Let’s filter H∗(B) ⊗H∗(F ) by the degree in H∗(F ), i.e.,Fq =⊕t≤qH

∗(B)⊗Ht(F ). The map σ ∶H∗(B)⊗H∗(F )→H∗(E) is filtrationpreserving, and it’s an isomorphism on the associated graded. This is theidentification H∗(B) ⊗ Ht(F ) = E∗,t

∞ = grtH∗(E). Since the filtrations are

exhaustive and bounded below, we conclude that σ itself is an isomorphism.

68 Integration, Gysin, Euler, Thom

Today there’s a talk bythe onethe onlyJEAN-PIERRE SERREOK let’s begin.

Umkehr

Let π ∶ E → B be a fibration and suppose B is path-connected. Suppose thefiber has no cohomology above some dimension d. The Serre sseq has nothingabove row d.

Let’s look at Hn(E). This happens along total degree n. We have thisneat increasing filtration that I was talking about on Monday whose associatedquotients are the rows in this thing. So I can divide out by it (i.e I divide outby Fd−1H

n(E)). Then I get

Hn(E)↠Hn(E)/Fd−1Hn(E) = En−d,d∞ ↣ En−d,d2 =Hn−d(B;Hd(F ))

That’s because on the E2 page, at that spot, there’s nothing hitting it, butthere might be a differential hitting it. There it is; here’s another edge homo-morphism.

Remark 68.1. This is a wrong-way map, also known as an “umkehr” map.It’s also called a pushforward map, or the Gysin map.

We know from the incomprehensible discussion that I was giving on Mondaythat this was a filtration of modules over H∗(B), so that this map Hn(E) →Hn−d(B;Hd(F )) is a H∗(B)-module map.

68. INTEGRATION, GYSIN, EULER, THOM 217

Example 68.2. F is a compact connected d-manifold with a givenR-orientation.Thus Hd(F ) ≃ R, given by x ↦ ⟨x, [F ]⟩. There might some local cohomologythere, but I do get a map Hn(E;R) → Hn−d(B;R). This is such a map, andit has a name: it’s written π! or π∗. I’ll write π∗.

Of course, if π1(B) fixes [F ] ∈Hd(F ;R), thenR-cohomology isR-cohomology.Thus our map is now Hn(E;R) → Hn−d(B;R). Sometimes it’s also called apushforward map. Note that we also get a projection formula

π∗(π∗(b) ∪ e) = b ∪ π∗(e)

where π∗ is the pushforward, e ∈ Hn(E) and b ∈ Hs(B). Others call thisFrobenius reciprocity.

Gysin

Suppose H∗(F ) =H∗(Sn−1). In practice, F ≅ Sn−1, or even F ≃ Sn−1. In thatcase, π ∶ E → B is called a spherical fibration Then the spectral sequence iseven simpler ! It has only two nonzero rows!

Let’s pick an orientation for Sn−1, to get an isomorphism Hn−1(Sn−1). Wellthe spectral sequence degenerates, and you get a long exact sequence

⋯→Hs(B) π∗Ð→Hs(E) π∗Ð→Hs−n+1(B;R) dnÐ→Hs+1(B) π∗Ð→Hs+1(E)→ ⋯

That’s called theGysin sequence2. Because everything is a module overH∗(B),this is a lexseq of H∗(B)-modules.

Let me be a little more explicit. Suppose we have an orientation. We nowhave a differentialH0(B)→Hn(B). We have the constant function 1 ∈H0(B),and this maps to something in B. This is called the Euler class, and is denotede.

Since dn is a module homomorphism, we have dn(x) = dn(1 ⋅x) = dn(1) ⋅x =e ⋅ x where x is in the cohomology of B. Thus our lexseq is of the form

⋯→Hs(B) π∗Ð→Hs(E) π∗Ð→Hs−n+1(B;R) e⋅−Ð→Hs+1(B) π∗Ð→Hs+1(E)→ ⋯

Some facts about the Euler class

Suppose E → B has a section σ ∶ B → E (so that πσ = 1B). So, if it came froma vector bundle, I’m asking that there’s a nowhere vanishing cross-section ofthat vector bundle. Let’s apply cohomology, so that you get σ∗π∗ = 1H∗(B).Thus π∗ is monomorphic. In terms of the Gysin sequence, this means thatHs−n(B) e⋅−Ð→Hs(B) is zero. But this implies that

e = 0

2pronounced Gee-sin


Thus, if you don’t have a nonzero Euler class then you cannot have a section!If your Euler class is zero sometimes you can conclude that your bundle has asection, but that’s a different story.

The Euler class of the tangent bundle of a manifold when paired with thefundamental class is the Euler characteristic. More precisely, if M is orientedconnected compact n-manifold, then

⟨e(τM), [M]⟩ = χ(M)

That’s why it’s called the Euler class. (He didn’t know about spectral sequencesor cohomology.)

Time for Thom

This was done by Rene Thom. Let ξ be a n-plane bundle over X. I can lookat H∗(E(ξ),E(ξ) − section). If I pick a metric, this is H∗(D(ξ), S(ξ)), whereD(ξ) is the disk bundle3 and S(ξ) is the sphere bundle. If there’s no point-setannoyance, this is H∗(D(ξ)/S(ξ)).

If X is a compact Hausdorff space, then ... The open disk bundle D0(ξ) ≃E(ξ). This quotient D(ξ)/S(ξ) = E(ξ)+ since you get the one-point compact-ification by embedding into a compact Hausdorff space (D(ξ) here) and thenquotienting by the complement (which is S(ξ) here). This is called the Thomspace of ξ. There are two notations: some people write Th(ξ), and some people(Atiyah started this) write Xξ.

Example 68.3 (Dumb). Suppose ξ is the zero vector bundle. Then yourfibration is π ∶ X → X. What’s the Thom space? The disk bundle is X, andthe boundary of a disk is empty, so Th(0) =X0 =X ⊔ ∗.

The Thom space is a pointed space (corresponding to ∞ or the point whichS(ξ) is collapsed to).

I’d like to study its cohomology, because it’s interesting. There’s no otherjustification. Maybe I’ll think of it as the relative cohomology.

So, guess what? We’ve developed sseqs and done cohomology. Anythingelse we’d like to do to groups and functors and things?

Let’s make the spectral sequence relative!I have a path connected B, and I’ll study:

F0

// F

E0

// E

B B

3D(ξ) = v ∈ E(ξ) ∶ ∣∣v∣∣ ≤ 1.

68. INTEGRATION, GYSIN, EULER, THOM 219

Then if you sit patiently and work through things, we get

Es,t2 =Hs(B;Ht(F,F0))⇒s Hs+t(E,E0)

Note that ⇒s means that s determines the filtration.Let’s do this with the Thom space. We have D(ξ) ≃Ð→ X. That isn’t very

interesting. In our case, we have an incredibly simple spectral sequence, whereeverything on the E2-page is concentrated in row n. Thus the E2 page is thecohomology of

Hs+n(Th(ξ)) =Hs+n(D(ξ), S(ξ)) ≃Hs(B;R)

where R =Hn(Dn, Sn−1). This is a canonical isomorphism of H∗(B)-modules.Suppose your vector bundle ξ is oriented, so that R = R. Now, if s = 0, then

I have 1 ∈H0(B). This gives u ∈Hn(Th(ξ)), which is called the Thom class.The cohomology of B is a free module of rank one over H∗(B), so that

H∗(Th(ξ)) is also a H∗(B)-module that is free of rank 1, generated by u.Let me finish by saying one more thing. This is why the Thom space is

interesting. Notice one more thing: there’s a lexseq of a pair

⋯→ Hs(Th(ξ))→Hs(D(ξ))→Hs(S(ξ))→ Hs+1(Th(ξ))→ ⋯

We have synonyms for these things:

⋯→Hs−n(X)→Hs(X)→Hs(S(ξ))→Hs−n+1(X)→ ⋯

And aha, this is is exactly the same form as the Gysin sequence. Except, ohmy god, what have I done here?

Yeah, right! In the Gysin sequence, the map Hs−n(X) → Hs(X) wasmultiplication by the Euler class. The Thom class u maps to some e′ ∈Hn(X)via Hn(D(ξ), S(ξ))→Hn(D(ξ)) ≃Hn(X). And the map Hs−n(X)→Hs(X)is multiplication by e′. Guess what? This is the Gysin sequence.

You’ll explore more in homework.I’ll talk about characteristic classes on Friday.

Chapter 6

Characteristic classes

69 Grothendieck’s construction of Chern classes

Generalities on characteristic classes

We would like to apply algebraic techniques to study G-bundles on a space.Let A be an abelian group, and n ≥ 0 an integer.

Definition 69.1. A characteristic class for principal G-bundles (with valuesin Hn(−;A)) is a natural transformation of functors Top→Ab:

BunG(X) cÐ→Hn(X;A)

Concretely: if P → Y is a principal G-bundle over a space X, and f ∶ X → Yis a continuous map of spaces, then

c(f∗P ) = f∗c(P ).

The motivation behind this definition is that BunG(X) is still rather mys-terious, but we have techniques (developed in the last section) to compute thecohomology groups Hn(X;A). It follows by construction that if two bundlesover X have two different characteristic classes, then they cannot be isomor-phic. Often, we can use characteristic classes to distinguish a given bundlefrom the trivial bundle.

Example 69.2. The Euler class takes an oriented real n-plane vector bundle(with a chosen orientation) and produces an n-dimensional cohomology classe ∶ Vectorn (X) = BunSO(n)(X) → Hn(X;Z). This is a characteristic class. Tosee this, we need to argue that if ξ ↓ X is a principal G-bundle, we can pullthe Euler class back via f ∶ X → Y . The bundle f∗ξ ↓ Y has a orientation if ξdoes, so it makes sense to even talk about the Euler class of f∗ξ. Since all ofour constructions were natural, it follows that e(f∗ξ) = f∗e(ξ).

Similarly, the mod 2 Euler class is e2 ∶ Vectn(X) = BunO(n)(X)→Hn(X;Z/2Z)is another Euler class. Since everything has an orientation with respect toZ/2Z, the mod 2 Euler class is well-defined.

221

222 CHAPTER 6. CHARACTERISTIC CLASSES

By our discussion in §58, we know that BunG(X) = [X,BG]. Moreover, aswe stated in Theorem 51.8, we know that Hn(X;A) = [X,K(A,n)] (at least ifX is a CW-complex). One moral reason for cohomology to be easier to computeis that the spaces K(A,n) are infinite loop spaces (i.e., they can be deloopedinfinitely many times). It follows from the Yoneda lemma that characteristicclasses are simply maps BG→K(A,n), i.e., elements of Hn(BG;A).

Example 69.3. The Euler class e lives in Hn(BSO(n);Z); in fact, it is e(ξ),the Euler class of the universal oriented n-plane bundle over BSO(n). Asimilar statement holds for e2 ∈Hn(BO(n);Z/2Z). For instance, if n = 2, thenSO(2) = S1. It follows that

BSO(2) = BS1 =CP∞.

We know that H∗(CP∞;Z) = Z[e] — it’s the polynomial algebra on the “uni-versal” Euler class! Similarly, O(1) = Z/2Z, so

BO(1) = BZ/2 =RP∞.

We know that H∗(RP∞;F2) = F2[e2] — as above, it is the polynomial algebraover Z/2Z on the “universal” mod 2 Euler class.

Chern classes

These are one of the most fundamental example of characteristic classes.

Theorem 69.4 (Chern classes). There is a unique family of characteristicclasses for complex vector bundles that assigns to a complex n-plane bundle ξover X the nth Chern class c(n)k (ξ) ∈H2k(X;Z), such that:

1. c(n)0 (ξ) = 1.

2. If ξ is a line bundle, then c(1)1 (ξ) = −e(ξ).

3. The Whitney sum formula holds: if ξ is a p-plane bundle and η is aq-plane bundle (and if ξ ⊕ η denotes the fiberwise direct sum), then

c(p+q)k (ξ ⊕ η) = ∑

i+j=kc(p)i (ξ) ∪ c(q)j (η) ∈H2k(X;Z).

Moreover, if ξn is the universal n-plane bundle, then

H∗(BU(n);Z) ≃ Z[c(n)1 ,⋯, c(n)n ],

where c(n)k = c(n)k (ξn).

This result says that all characteristic classes for complex vector bundles aregiven by polynomials in the Chern classes because the cohomology of BU(n)gives all the characteristic classes. It also says that there are no universal alge-braic relations among the Chern classes: you can specify them independently.

69. GROTHENDIECK’S CONSTRUCTION OF CHERN CLASSES 223

Remark 69.5. The (p + q)-plane bundle ξp × ξq = pr∗1ξp ⊕ pr∗2ξq over BU(p) ×BU(q) is classified by a map BU(p)×BU(q) µÐ→ BU(p+ q). The Whitney sumformula computes the effect of µ on cohomology:

µ∗(c(n)k ) = ∑i+j=k

c(p)i × c(q)j ∈H2k(BU(p) ×BU(q)),

where, you’ll recall,x × y ∶= pr∗1x ∪ pr∗2y.

The Chern classes are “stable”, in the following sense. Let ε be the trivialone-dimensional complex vector bundle, and let ξ be an n-dimensional vectorbundle. What is c(n+q)k (ξ⊕εq)? For this, the Whitney sum formula is valuable.

The trivial bundle is characterized by the pullback:

X ×Cn = nε //

Cn

X // ∗

By naturality, we find that if k > 0, then c(n)k (nε) = 0. The Whitney sum

formula therefore implies that

c(n+q)k (ξ ⊕ εq) = c(n)k (ξ).

This phenomenon is called stability: the Chern class only depends on the“stable equivalence class” of the vector bundle (really, they are only defined on“K-theory”, for those in the know). For this reason, we will drop the superscripton c(n)k (ξ), and simply write ck(ξ).

Grothendieck’s construction

Let ξ be an n-plane bundle. We can consider the vector bundle π ∶ P(ξ)→ X,the projectivization of ξ: an element of the fiber of P(ξ) over x ∈ X is a lineinside ξx, so the fibers are therefore all isomorphic to CPn−1.

Let us compute the cohomology of P(ξ). For this, the Serre spectral se-quence will come in handy:

Es,t2 =Hs(X;Ht(CPn−1))⇒Hs+t(P(ξ)).

Remark 69.6. Why is the local coefficient system constant? The spaceX neednot be simply connected, but BU(n) is simply connected since U(n) is simplyconnected. Consider the projectivization of the universal bundle ξn ↓ BU(n);pulling back via f ∶ X → BU(n) gives the bundle π ∶ P(ξ) → X. The map onfibers H∗(P(ξn)f(x)) → H∗(P(ξn)x) is an isomorphism which is equivariantwith respect to the action of the fundamental group of π1(X) via the mapπ1(X)→ π1(BU(n)) = 0.


Because H∗(CPn−1) is torsion-free and finitely generated in each dimen-sion, we know that

Es,t2 ≃Hs(X)⊗Ht(CPn−1).

The spectral sequence collapses at E2, i.e., that E2 ≃ E∞, i.e., there are nodifferentials. We know that the E2-page is generated as an algebra by elementsin the cohomology of the fiber and elements in the cohomology of the base.Thus, it suffices to check that elements in the cohomology of the fiber surviveto E∞. We know that

E0,2t2 = Z⟨xt⟩, and E0,2t+1

2 = 0,

where x = e(λ) is the Euler class of the canonical line bundle λ ↓CPn−1.In order for the Euler class to survive the spectral sequence, it suffices

to come up with a two dimensional cohomology class in P(ξ) that restrictsto the Euler class over CPn−1. We know that λ itself is the restriction ofthe tautologous line bundle over CP∞. There is a tautologous line bundleλξ ↓ P(ξ), given by the tautologous line bundle on each fiber. Explicitly:

E(λξ) = (`, y) ∈ P(ξ) ×X E(ξ)∣y ∈ ` ⊆ ξx.

Thus, x is the restriction e(λξ)∣fiber of the Euler class to the fiber. It followsthat the class x survives to the E∞-page.

Using the Leray-Hirsch theorem (Theorem 67.1), we conclude that

H∗(P(ξ)) =H∗(X)⟨1, e(λξ), e(λξ)2,⋯, e(λξ)n−1⟩.

For simplicity, let us write e = e(λξ). Unforunately, we don’t know what en

is, although we do know that it is a linear combination of the ek for k < n. Inother words, we have a relation

en + c1en−1 +⋯ + cn−1e + cn = 0,

where the ck are elements of H2k(X). These are the Chern classes of ξ. Byconstruction, they are unique!

To prove Theorem 69.4(2), note that when n = 1 the above equation reads

e + c1 = 0,

as desired.

70 H∗(BU(n)), splitting principle

Theorem 69.4 claimed that the Chern classes, which we constructed in theprevious section, generate the cohomology of BU as a polynomial algebra.Our goal in this section is to prove this result.

70. H∗(BU(n)), SPLITTING PRINCIPLE 225

The cohomology of BU(n)Recall that BU(n) supports the universal principal U(n)-bundle EU(n) →BU(n). Given any left action of U(n) on some space, we can form the asso-ciated fiber bundle. For instance, the associated bundle of the U(n)-action onCn yields the universal line bundle ξn.

Likewise, the associated bundle of the action of U(n) on S2n−1 ⊆ Cn is theunit sphere bundle S(ξn), the unit sphere bundle. By construction, the fiberof the map EU(n) ×U(n) S

2n−1 → BU(n) is S2n−1. Since

S2n−1 = U(n)/(1 ×U(n − 1)),

we can write

EU(n)×U(n)S2n−1 ≃ EU(n)×U(n)(U(n)/U(n−1)) ≃ EU(n)/U(n−1) = BU(n−1).

In other words, BU(n − 1) is the unit sphere bundle of the tautologous linebundle over BU(n). This begets a fiber bundle:

S2n−1 → BU(n − 1)→ BU(n),

which provides an inductive tool (via the Serre spectral sequence) for computingthe homology of BU(n). In §68, we observed that the Serre spectral sequencefor a spherical fibration was completely described bythe Gysin sequence.

Recall that if B is oriented and S2n−1 → EπÐ→ B is a spherical bundle over

B, then the Gysin sequence was a long exact sequence

⋯→Hq−1(E) π∗Ð→Hq−2n(B) e⋅Ð→Hq(B) π∗Ð→Hq(E) π∗Ð→ ⋯

Let us assume that the cohomology ring of E is polynomial and concentratedin even dimensions. For the base case of the induction, both these assumptionsare satisfied (since BU(0) = ∗ and BU(1) =CP∞).

These assumptions imply that if q is even, then the map π∗ is zero. Inparticular, multiplication by e∣Heven(B) (which we will also denote by e) isinjective, i.e., e is a nonzero divisor. Similarly, if q is odd, then e ⋅Hq−2n(B) =Hq(B). But if q = 1, then Hq−2n(B) = 0; by induction on q, we find thatHodd(B) = 0. Therefore, if q is even, then Hq−2n+1(B) = 0. This implies thatthere is a short exact sequence

0→H∗(B) e⋅Ð→H∗(B)→H∗(E)→ 0. (6.1)

In particular, the cohomology of E is the cohomology of B quotiented by theideal generated by the nonzero divisor e.

For instance, when n = 1, then B =CP∞ and E ≃ ∗. We have the canonicalgenerator e ∈ H2(CP∞); these deductions tell us the well-known fact thatH∗(CP∞) ≃ Z[e].

Consider the surjection H∗(B) π∗Ð→H∗(E). Since H∗(E) is polynomial, wecan lift the generators of H∗(E) to elements of H∗(B). This begets a splitting


s ∶ H∗(E) → H∗(B). The existence of the Euler class e ∈ H∗(B) thereforegives a map H∗(E)[e] sÐ→H∗(B). We claim that this map is an isomorphism.

This is a standard algebraic argument. Filter both sides by powers of e,i.e., take the e-adic filtration on H∗(E)[e] and H∗(B). Clearly, the associatedgraded of H∗(E)[e] just consists of an infinite direct sum of the cohomologyof E. The associated graded of H∗(B) is the same, thanks to the short exactsequence (6.1). Thus the induced map on the associated graded gr∗(s) is anisomorphism. In this particular case (but not in general), we can conclude thats is an isomorphism: in any single dimension, the filtration is finite. Thus, usingthe five lemma over and over again, we see that the map s an isomorphism oneach filtered piece. This implies that s itself is an isomorphism, as desired.

This argument proves that

H∗(BU(n − 1)) = Z[c1,⋯, cn−1].

In particular, there is a map π∗ ∶ H∗(BU(n)) → H∗(BU(n − 1)) which anisomorphism in dimensions at most 2n. Thus, the generators ci have uniquelifts to H∗(BU(n)). We therefore get:

Theorem 70.1. There exist classes ci ∈H2i(BU(n)) for 1 ≤ i ≤ n such that:

• the canonical map H∗(BU(n)) π∗Ð→H∗(BU(n − 1)) sends

ci ↦⎧⎪⎪⎨⎪⎪⎩

ci i < n0 i = n, and

• cn ∶= (−1)ne ∈H2n(BU(n)).

Moreover,H∗(BU(n)) ≃ Z[c1,⋯, cn] .

The splitting principle

Theorem 70.2. Let ξ ↓ X be an n-plane bundle. Then there exists a spaceFl(ξ) πÐ→X such that:

1. π∗ξ = λ1 ⊕⋯λn, where the λi are line bundles on Y , and

2. the map π∗ ∶H∗(X)→H∗(Fl(ξ)) is monic.

Proof. We have already (somewhat) studied this space. Recall that there is avector bundle π ∶ P(ξ)→X such that

H∗(P(ξ)) =H∗(X)⟨1, e,⋯, en−1⟩.

Moreover, in §69, we proved that there is a complex line bundle overP(ξ) whichis a subbundle of π∗ξ. In other words, π∗ξ splits as a sum of a line bundle andsome other bundle (by Corollary 52.11). Iterating this construction proves theexistence of Fl(ξ).

71. THE WHITNEY SUM FORMULA 227

This proof does not give much insight into the structure of Fl(ξ). Rememberthat the frame bundle Fr(ξ) of ξ: an element of Fr(ξ) is a linear, inner-productpreserving map Cn → E(ξ). This satisfies various properties; for instance:

E(ξ) = Fr(ξ) ×U(n) Cn.

Moreover,P(ξ) = Fr(ξ) ×U(n) U(n)/(1 ×U(n − 1)).

The flag bundle Fl(ξ) is defined to be

Fl(ξ) = Fr(ξ) ×U(n) U(n)/(U(1) ×⋯ ×U(1)).

The product U(1) × ⋯ × U(1) is usually denoted Tn, since it is the maximaltorus in U(n). For the universal bundle ξn ↓ BU(n), the frame bundle isexactly EU(n); therefore, Fl(ξn) is just the bundle given by BTn → BU(n).By construction, the fiber of this bundle is U(n)/Tn. In particular, thereis a monomorphism H∗(BU(n)) H∗(BTn). The cohomology of BTn isextremely simple — it is the cohomology of a product of CP∞’s, so

H∗(BTn) ≃ Z[t1,⋯, tn],

where ∣tk ∣ = 2. The ti are the Euler classes of π∗i λi, under the projection mapπi ∶ BTn →CP∞.

71 The Whitney sum formula

As we saw in the previous section, there is an injectionH∗(BU(n))H∗(BTn).What is the image of this map?

The symmetric group sits inside of U(n), so it acts by conjugation onU(n). This action stabilizes this subgroup Tn. By naturality, Σn acts onthe classifying space BTn. Since Σn acts by conjugation on U(n), it acts onBU(n) in a way that is homotopic to the identity (Lemma 58.1). However, eachelement σ ∈ Σn simply permutes the factors in BTn = (CP∞)n; we concludethat H∗(BU(n);R) actually sits inside the invariants H∗(BTn;R)Σn .

Recall the following theorem from algebra:

Theorem 71.1. Let Σn act on the polynomial algebra R[t1,⋯, tn] by permutingthe generators. Then

R[t1,⋯, tn]Σn = R[σ(n)1 ,⋯, σ(n)

n ],

where the σi are the elementary symmetric polynomials, defined via

n

∏i=1

(x − ti) =n

∑j=0

σ(n)i xn−i.


For instance,σ(n)1 = −∑ ti, σ

(n)n = (−1)n∏ ti.

If we impose a grading on R[t1,⋯, tn] such that ∣ti∣ = 2, then ∣σ(n)i ∣ = 2i. It

follows from our discussion in §70 that the ring H∗(BTn)Σn has the same sizeas H∗(BU(n)).

Consider an injection of finitely generated abelian groups M N , withquotient Q. Suppose that, after tensoring with any field, the map M → N anisomorphism. If Q⊗ k = 0, then Q = 0. Indeed, if Q⊗Q = 0 then Q is torsion.Similarly, if Q ⊗ Fp = 0, then Q has no p-component. In particular, M ≃ N .Applying this to the map H∗(BU(n)→H∗(BTn)Σn , we find that

H∗(BU(n);R) ≃Ð→H∗(BTn;R)Σn = R[σ(n)1 ,⋯, σ(n)

n ].

What happens as n varies? There is a map R[t1,⋯, tn] → R[t1,⋯, tn−1] givenby sending tn ↦ 0 and ti ↦ ti for i ≠ n. Of course, we cannot say that this mapis equivariant with respect to the action of Σn. However, it is equivariant withrespect to the action of Σn−1 on R[t1,⋯, tn] via the inclusion of Σn−1 Σnas the stabilizer of n ∈ 1,⋯, n. Therefore, the Σn-invariants sit inside theΣn−1-invariants, giving a map

R[t1,⋯, tn]Σn → R[t1,⋯, tn]Σn−1 → R[t1,⋯, tn−1]Σn−1 .

We also find that for i < n, we have σ(n)i ↦ σ

(n−1)i and σ(n)

n ↦ 0.

Where do the Chern classes go?

To answer this question, we will need to understand the multiplicativity ofthe Chern class. We begin with a discussion about the Euler class. Supposeξp ↓X,ηq ↓ Y are oriented real vector bundles; then, we can consider the bundleξ × η ↓X × Y , which is another oriented real vector bundle. The orientation isgiven by picking oriented bases for ξ and η. We claim that

e(ξ × η) = e(ξ) × e(η) ∈Hp+q(X × Y ).

Since D(ξ × η) is homeomorphic to D(ξ)×D(η), and S(ξ × η) =D(ξ)×S(η)∪S(ξ) ×D(η), we learn from the relative Künneth formula that

H∗(D(ξ × η), S(ξ × η))←H∗(D(ξ), S(ξ))⊗H∗(D(η), S(η)).

It follows thatuξ×η = uξ × uη ∈Hp+q(Th(ξ) ×Th(η));

this proves the desired result since the Euler class is the image of the Thomclass under the map Hn(Th(ξ))→Hn(D(ξ)) ≃Hn(B).

Consider the diagonal map ∆ ∶X →X ×X. The cross product in cohomol-ogy then pulls back to the cup product, and the direct product of fiber bundlespulls back to the Whitney sum. It follows that

e(ξ ⊕ η) = e(ξ) ∪ e(η).

71. THE WHITNEY SUM FORMULA 229

If ξn ↓X is an n-dimensional complex vector bundle, then we defined1

cn(ξ) = (−1)ne(ξR).

We need to describe the image of cn(ξn) under the map H2n(BU(n)) →H2n(BTn)Σn .

Let f ∶ BTn → BU(n) denote the map induced by the inclusion of themaximal torus. Then, by construction, we have a splitting

f∗ξn = λ1 ⊕⋯⊕ λn.

Thus,

(−1)ne(ξ)↦ (−1)ne(λ1 ⊕⋯⊕ λn) = (−1)ne(λ1) ∪⋯ ∪ e(λn).

The discussion above implies that f∗ sends the right hand side to (−1)nt1⋯tn =σ(n)n . In other words, the top Chern class maps to σ(n)

n under the map f∗.Our discussion in the previous sections gives a commuting diagram:

H∗(BU(n)) //

H∗(BTn)Σn

H∗(BU(n − 1)) // H∗(BTn−1)Σn−1

Arguing inductively, we find that going from the top left corner to the bottomleft corner to the bottom right corner sends

ci ↦ ci ↦ σ(n−1)i for i < n.

Likewise, going from the top left corner to the top right corner to the bottomright corner sends

ci ↦ σ(n)i ↦ σ

(n−1)i for i < n.

We conclude that the map f∗ sends c(i)i ↦ σ(i)i .

Proving the Whitney sum formula

By our discussion above, the Whitney sum formula of Theorem 69.4 reducesto proving the following identity:

σ(p+q)k = ∑

i+j=kσ(p)i ⋅ σ(q)

j (6.2)

inside Z[t1,⋯, tp, tp+1,⋯, tp+q]. Here, σ(p)i is thought of as a polynomial in

t1,⋯, tp, while σ(q)i is thought of as a polynomial in tp+1,⋯, tp+q. To derive

1There’s a slight technical snag here: a complex bundle doesn’t have an orientation.However, its underlying oriented real vector bundle does.


Equation (6.2), simply compare coefficients in the following:

p+q∑k=0

σ(p+q)k xp+q−k =

p+q∏i=1

(x − ti)

=p

∏i=1

(x − ti) ⋅p+q∏j=p+1

(x − tj)

= (p

∑i=0

σ(p)i xp−i)

⎛⎝

q

∑j=0

σ(p)j xq−j

⎞⎠

=p+q∑k=0

⎛⎝ ∑i+j=k

σ(p)i σ

(q)j

⎞⎠xp+q−k.

72 Stiefel-Whitney classes, immersions, cobordisms

There is a result analogous to Theorem 69.4 for all vector bundles (not neces-sarily oriented):

Theorem 72.1. There exist a unique family of characteristic classes wi ∶Vectn(X)→Hn(X;F2) such that for 0 ≤ i and i > n, we have wi = 0, and:

1. w0 = 1;

2. w1(λ) = e(λ); and

3. the Whitney sum formula holds:

wk(ξ ⊕ η) = ∑i+j=k

wi(ξ) ∪wj(η)

Moreover:H∗(BO(n);F2) = F2[w1,⋯,wn],

where wn = e2.

Remark 72.2. We can express the Whitney sum formula simply by definingthe total Steifel-Whitney class

1 +w1 +w2 +⋯ =∶ w.

Then the Whitney sum formula is just

w(ξ ⊕ η) = w(ξ) ⋅w(η).

Likewise, the Whitney sum formula can be stated by defining the total Chernclass.

72. STIEFEL-WHITNEY CLASSES, IMMERSIONS, COBORDISMS 231

Remark 72.3. Again, the Steifel-Whitney classes are stable:

w(ξ ⊕ kε) = w(ξ).

Again, Grothendieck’s definition works since the splitting principle holds.There is an injectionH∗(BO(n))H∗(B(Z/2Z)n). To computeH∗(BO(n)),our argument for computing H∗(BU(n)) does not immediately go through, al-though there is a fiber sequence

Sn−1 → EO(n) ×O(n) O(n)/O(n − 1)→ BO(n);

the problem is that n − 1 can be even or odd. We still have a Gysin sequence,though:

⋯→Hq−n(BO(n)) e⋅Ð→Hq(BO(n)) π∗Ð→Hq(BO(n−1))→Hq−n+1(BO(n))→ ⋯

In order to apply our argument for computing H∗(BU(n)) to this case, weonly need to know that e is a nonzero divisor. The splitting principle gave amonomorphism H∗(BO(n)) H∗((RP∞)n). The fact that e is a nonzerodivisor follows from the observation that under this map,

e2 = wn ↦ e2(λ1 ⊕⋯⊕ λn) = t1⋯tn,

using the same argument as in §71; however, t1⋯tn is a nonzero divisor, sinceH∗((RP∞)n) is an integral domain.

Immersions of manifolds

The theory developed above has some interesting applications to differentialgeometry.

Definition 72.4. Let Mn be a smooth closed manifold. An immersion is asmooth map from Mn to Rn+k, denoted f ∶Mn Rn+k, such that (τMn)x (τRn+k)f(x) for x ∈M .

Informally: crossings are allows, but not cusps.

Example 72.5. There is an immersion RP2 R3, known as Boy’s surface.

Question 72.6. When can a manifold admit an immersion into an Euclideanspace?

Assume we had an immersion i ∶Mn Rn+k. Then we have an embeddingf ∶ τM → i∗τRn+k into a trivial bundle over M , so τM has a k-dimensionalcomplement, called ξ such that

τM ⊕ ξ = (n + k)ε.

Apply the total Steifel-Whitney class, we have

w(τ)w(ξ) = 1,


since there’s no higher Steifel-Whitney class of a trivial bundle. In particular,

w(ξ) = w(τ)−1.

Example 72.7. Let M =RPn Rn+k. Then, we know that

τRPn ⊕ ε ≃ (n + 1)λ∗ ≃ (n + 1)λ,

where λ ↓RPn is the canonical line bundle. By Remark 72.3, we have

w(τRPn) = w(τRPn ⊕ η) = w((n + 1)λ) = w(λ)n+1.

It remains to compute w(λ). Only the first Steifel-Whitney class is nonzero.Writing H∗(RPn) = F2[x]/xn+1, we therefore have w(λ) = x. In particular,

w(τRPn) = (1 + x)n+1 =n

∑i=0

(n + 1

i)xi.

It follows thatwi(τRPn) = (n + 1

i)xi.

The total Steifel-Whitney class of the complement of the tangent bundle is:

w(ξ) = (1 + x)−n−1.

The most interesting case is when n is a power of 2, i.e., n = 2s for some integers. In this case, since taking powers of 2 is linear in characteristic 2, we have

w(ξ) = (1 + x)−1−2s = (1 + x)−1(1 + x)−2s = (1 + x)−1(1 + x2s)−1.

As all terms of degree greater than 2s are zero, we conclude that So

w(ξ) = 1 + x + x2 +⋯ + x2s−1 + 2xs = 1 + x + x2 +⋯ + x2s−1.

As x2s−1 ≠ 0, this means that k = dim ξ ≥ 2s − 1. We conclude:

Theorem 72.8. There is no immersion RP2s R2⋅2s−2.

The following result applied to RP2s shows that the above result is sharp:

Theorem 72.9 (Whitney). Any smooth compact closed manifoldMn R2n−1.

However, Whitney’s result is not sharp for a general smooth compact closedmanifold. Rather, we have:

Theorem 72.10 (Brown–Peterson, Cohen). A closed compact smooth n-manifoldMn R2n−α(n), where α(n) is the number of 1s in the dyadic expansion of n.

This result is sharp, since if n = ∑2di for the dyadic expansion, then M =∏iRP2di /R2n−α(n)−1.

72. STIEFEL-WHITNEY CLASSES, IMMERSIONS, COBORDISMS 233

Cobordism, characteristic numbers

If we have a smooth closed compact n-manifold, then it embeds in Rn+k forsome k ≫ 0. The normal bundle then satisfies

τM ⊕ νM = (n + k)ε.

A piece of differential topology tells us that if k is large, then νM ⊕ Nε isindependent of the bundle for some N .

This example, combined with Remark ??, shows that w(νM) is independentof k. We are therefore motivated to think of Stiefel-Whitney classes as comingfrom H∗(BO;F2) = F2[w1,w2,⋯], where BO = limÐ→BO(n). Similarly, Chernclasses should be thought of as coming from H∗(BU ;Z) = Z[c1, c2,⋯]. Thisexa

Definition 72.11. The characteristic number of a smooth closed compactn-manifold M is defined to be ⟨w(νM), [M]⟩.

Note that the fundamental class [M] exists, since our coefficients are in F2,where everything is orientable.

This definition is very useful when thinking about cobordisms.

Definition 72.12. Two (smooth closed compact) n-manifoldsM,N are (co)bordantif there is an (n + 1)-dimensional manifold Wn+1 with boundary such that

∂W ≃M ⊔N.

For instance, when n = 0, the manifold ∗⊔∗ is not cobordant to ∗, but it iscobordant to the empty set. However, ∗⊔∗⊔∗ is cobordant to ∗. Any manifoldis cobordant to itself, since ∂(M × I) = M ⊔M . In fact, cobordism forms anequivalence relation on manifolds.

Example 72.13. A classic example of a cobordism is the “pair of pants”; thisis the following cobordism between S1 and S1 ⊔ S1: add imageadd image

Let us define

ΩOn = cobordism classes of n-manifolds.

This forms a group: the addition is given by disjoint union. Note that everyelement is its own inverse. Moreover, ⊕nΩOn = ΩO∗ forms a graded ring, wherethe product is given by the Cartesian product of manifolds. Our discussionfollowing Definition 72.12 shows that ΩO0 = F2.

Exercise 72.14. Every 1-manifold is nullbordant, i.e., cobordant to the point.

Thom made the following observation. Suppose an n-manifoldM is embed-ded into Euclidean space, and that M is nullbordant via some (n+1)-manifoldW , so that νW ∣M = νM . In particular,

⟨w(νM), [M]⟩ = ⟨w(νW )∣M , [M]⟩.


On the other hand, the boundary map Hn+1(W,M) ∂Ð→ Hn(M) sends the rel-ative fundamental class [W,M] to [M]. Thus

⟨w(νM), [M]⟩ = ⟨w(νM), ∂[W,M]⟩ = ⟨δw(νM), [W,M]⟩.

However, we have an exact sequence

Hn(W ) i∗Ð→Hn(M) δÐ→Hn+1(W,M).

Since w(νM) is in the image of i∗, it follows that δw(νM) = 0. In particular,the characteristic number of a nullbordant manifold is zero. Thus, we find that“Stiefel-Whitney numbers tell all”:

Proposition 72.15. Characteristic numbers are cobordism invariants. Inother words, characteristic numbers give a map

ΩOn → Hom(Hn(BO),F2) ≃Hn(BO).

More is true:

Theorem 72.16 (Thom, 1954). The map of graded rings ΩO∗ → H∗(BO)defined by the characteristic number is an inclusion. Concretely, if w(Mn) =w(Nn) for all w ∈Hn(BO), then Mn and Nn are cobordant.

The way that Thom proved this was by expressing ΩO∗ is the graded homo-topy ring of some space, which he showed is the product of mod 2 Eilenberg-MacLane spaces. Along the way, he also showed that:

ΩO∗ = F2[xi ∶ i ≠ 2s − 1] = F2[x2, x4, x5, x6, x8,⋯]

This recovers the result of Exercise 72.14 (and so much more!).

73 Oriented bundles, Pontryagin classes, Signaturetheorem

We have a pullback diagram

BSO(n) //

double cover

S∞

BO(n) w1

// BZ/2Z

The bottom map is exactly the element w1 ∈ H1(BO(n);F2). It follows thata vector bundle ξ ↓ X represented by a map f ∶ X → BO(n) is orientable iffw1(ξ) = f∗(w1) = 0, since this is equivalent to the existence of a factorization:

BSO(n) //

S∞

X

;;

ξ// BO(n) w1

// BZ/2Z

73. ORIENTED BUNDLES, PONTRYAGIN CLASSES, SIGNATURETHEOREM 235

The fiber sequence BSO(n) → BO(n) → RP∞ comes from a fiber sequenceSO(n) → O(n) → Z/2Z of groups. For n ≥ 3, we can kill π1(SO(n)) = Z/2Z,to get a double cover Spin(n) → SO(n). The group Spin(n) is called the spingroup. We have a cofiber sequence

BSpin(n)→ BSO(n) w2Ð→K(Z/2Z,2).

If w2(ξ) = 0, we get a further lift in the above diagram, begetting a spinstructure on ξ.

Bott computed that π2(Spin(n)) = 0. However, π3(Spin(n)) = Z; killingthis gives the string group String(n). Unlike Spin(n), SO(n), and O(n), thisis not a finite-dimensional Lie group (since we have an infinite dimensionalsummand K(Z,2)). However, it can be realized as a topological group. Theresulting maps

String(n)→ Spin(n)→ SO(n)→ O(n)

are just the maps in the Whitehead tower for O(n). Taking classifying spaces,we get

BString(n)

BSpin(n)

p1/2 // K(Z,4)

BSO(n) //

K(Z/2Z,2)

X

::

CC

GG

ξ// BO(n) w1

// BZ/2Z

Computing the (mod 2) cohomology of BSO(n) is easy. We have a dou-ble cover BSO(n) → BO(n) with fiber S0. Consequently, there is a Gysinsequence:

0→Hq(BO(n)) w1Ð→Hq+1(BO(n)) π∗Ð→Hq+1(BSO(n))→ 0

since w1 is a nonzero divisor. The standard argument shows that

H∗(BSO(n)) = F2[w2,⋯,wn].

However, it is not easy to compute H∗(BSpin(n)) and H∗(BString(n)); theseare extremely complicated (and only become more complicated for higher con-nective covers of BO(n)). However, we will remark that they are concentratedin even degrees.

To define integral characteristic classes for oriented bundles, we will needto study Chern classes a little more. Let ξ be a complex n-plane bundle, and


let ξ denote the conjugate bundle. What is the total Chern class c(ξ)? Recallthat the Chern classes ck(ξ) occur as coefficients in the identity

∑ ci(ξ)e(λξ)n−i = 0,

where λξ ↓ P(ξ). Note that P(ξ) = P(ξ). By construction, λξ = λξ. Inparticular, we find that

e(λξ) = −e(λξ).It follows that

0 =n

∑i=0

ci(ξ)e(λξ)n−i =n

∑i=0

ci(ξ)(−1)n−ie(λξ)n−i = (−1)ne(λξ)n +⋯

This is not monic, and hence doesn’t define the Chern classes of ξ. We do,however, get a monic polynomial by multiplying this identity by (−1)n:

n

∑i=0

(−1)ici(ξ)e(λξ)n−i = 0.

It follows thatci(ξ) = (−1)ici(ξ).

If ξ is a real vector bundle, then

ci(ξ ⊗C) = ci(ξ ⊗C) = (−1)ici(ξ ⊗C).

If i is odd, then 2ci(ξ ⊗C) = 0. If R is a Z[1/2]-algebra, we therefore define:

Definition 73.1. Let ξ be a real n-plane vector bundle. Then the kth Pon-tryagin class of ξ is defined to be

pk(ξ) = (−1)kc2k(ξ ⊗ C) ∈H4k(X;R).

Notice that this is 0 if 2k > n, since ξ ⊗ C is of complex dimension n. TheWhitney sum formula now says that:

(−1)kpk(ξ ⊕ η) = ∑i+j=k

(−1)ipi(ξ)(−1)jpj(η) = (−1)k ∑i+j=k

pi(ξ)pj(η).

If ξ is an oriented real 2k-plane bundle, one can calculate that

pk(ξ) = e(ξ)2 ∈H4k(X;R).

We can therefore write down the cohomology of BSO(n) with coefficients in aZ[1/2]-algebra:

∗ = 2 4 6 8 10 12H∗(BSO(2)) e2 (e2

2)H∗(BSO(3)) p1

H∗(BSO(4)) p1, e4 (e24)

H∗(BSO(5)) p1 p2

H∗(BSO(6)) p1 e6 p2 (e26)

H∗(BSO(7)) p1 p2 p3

73. ORIENTED BUNDLES, PONTRYAGIN CLASSES, SIGNATURETHEOREM 237

Here, pk ↦ e22k. In the limiting case (i.e., for BSO = BSO(∞)), we get a

polynomial algebra on the pi.

Applications

We will not prove any of the statements in this section; it only serves as anoutlook. The first application is the following analogue of Theorem 72.16:

Theorem 73.2 (Wall). Let Mn,Nn be oriented manifolds. If all Stiefel-Whitney numbers and Pontryagin numbers coincide, then M is oriented cobor-dant to N , i.e., there is an (n + 1)-manifold Wn+1 such that

∂Wn+1 =M ⊔ −N.

The most exciting application of Pontryagin classes is to Hirzebruch’s “sig-nature theorem”. Let M4k be an oriented 4k-manifold. Then, the formula

x⊗ y ↦ ⟨x ∪ y, [M]⟩

defines a pairing

H2k(M)/torsion⊗H2k(M)/tors→ Z.

Poincaré duality implies that this is a perfect pairing, i.e., there is a nonsingularsymmetric bilinear form on H2k(M)/torsion ⊗R. Every symmetric bilinearform on a real vector space can be diagonalized, so that the associated matrixis diagonal, and the only nonzero entries are ±1. The number of 1s minus thenumber of −1s is called the signature of the bilinear form. When the bilinearform comes from a 4k-manifold as above, this is called the signature of themanifold.

Lemma 73.3 (Thom). The signature is an oriented bordism invariant.

This is an easy thing to prove using Lefschetz duality, which is a deeptheorem. Hirzebruch’s signature theorem says:

Theorem 73.4 (Hirzebruch signature theorem). There exists an explicit ra-tional polynomial Lk(p1,⋯, pk) of degree 4k such that

⟨L(p1(τM),⋯, p1(τM)), [M]⟩ = signature(M).

The reason the signature theorem is so interesting is that the polynomialL(p1(τM),⋯, p1(τM)) is defined only in terms of the tangent bundle of themanifold, while the signature is defined only in terms of the topology of themanifold. This result was vastly generalized by Atiyah and Singer to theAtiyah-Singer index theorem.

Example 73.5. One can show that

L1(p1) = p1/3.

The Hirzebruch signature theorem implies that ⟨p1(τ), [M4]⟩ is divisible by 3.


Example 73.6. From Hirzebruch’s characterization of the L-polynomial, wehave

L2(p1, p2) = (7p2 − p21)/45.

This imposes very interesting divisibility constraints on the characteristic classesof a tangent bundle of an 8-manifold. This particular polynomial was used byMilnor to produce “exotic spheres”, i.e., manifolds which are homeomorphic toS7 but not diffeomorphic to it.

Bibliography

[Beh06] M. Behrens. Algebraic Topology II. https://ocw.mit.edu/courses/mathematics/18-906-algebraic-topology-ii-spring-2006/,2006.

[Bre93] G. Bredon. Topology and Geometry. Springer, 1993.

[Bro57] E. Brown. Finite Computability of Postnikov Complexes. Ann. ofMath, 65(1):1–20, 1957.

[GJ99] P. Goerss and J. Jardine. Simplicial Homotopy Theory, volume 174of Progress in mathematics (Boston, Mass.). Springer, 1999.

[Hat15] A. Hatcher. Algebraic Topology. Cambridge University Press, 2015.

[Hus94] Husemöller, D. Fibre Bundles. Graduate Texts in Mathematics.Springer, 1994.

[May99] P. May. A Concise Course in Algebraic Topology. Chicago Lecturesin Mathematics. University of Chicago Press, 1999.

[MS74] J. Milnor and J. Stasheff. Characteristic Classes. Princeton UniversityPress and University of Tokyo Press, 1974.

[Str09] Neil Strickland. The category of CGWH spaces. http://math.mit.edu/~hrm/18.906/strickland-category-cgwh.pdf, 2009.

239

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https://ocw.mit.edu/courses/mathematics/18-906-algebraic-topology-ii-spring-2006/

http://math.mit.edu/~hrm/18.906/strickland-category-cgwh.pdf

http://math.mit.edu/~hrm/18.906/strickland-category-cgwh.pdf

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