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Math 231a - Algebraic Topology
Taught by Peter B. KronheimerNotes by Dongryul Kim
Fall 2016
This course was taught by Peter Kronheimer. We met on Mondays, Wednes-days, and Fridays from 2:00pm to 3:00pm at Science Center 507 and used AllenHatcher’s book Algebraic Topology. There were 19 students enrolled and therewas one final paper. The course assistance for this course was Meng Guo.
Contents
1 August 31, 2016 51.1 The first homology of T 2 . . . . . . . . . . . . . . . . . . . . . . . 5
2 September 2, 2016 72.1 Simplicial complex . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Homology for an abstract simplicial complex over Z/2 . . . . . . 8
3 September 7, 2016 103.1 Simplicial homology with G coefficients . . . . . . . . . . . . . . 103.2 Singular homology . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4 September 9, 2016 134.1 Singular homology . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Functorial aspects of homology . . . . . . . . . . . . . . . . . . . 15
5 September 12, 2016 165.1 Exact sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.2 Exact sequence of chain complexes . . . . . . . . . . . . . . . . . 16
6 September 14, 2016 196.1 Relative homology . . . . . . . . . . . . . . . . . . . . . . . . . . 19
7 September 16, 2016 217.1 The five lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217.2 Eilenberg-Steenrod axioms . . . . . . . . . . . . . . . . . . . . . . 21
1 Last Update: August 27, 2018
8 September 19, 2016 238.1 Homotopy and homology . . . . . . . . . . . . . . . . . . . . . . 23
9 September 21, 2016 259.1 Reduced homology . . . . . . . . . . . . . . . . . . . . . . . . . . 259.2 Homology of spheres . . . . . . . . . . . . . . . . . . . . . . . . . 26
10 September 23, 2016 2810.1 Barycentric subdivision . . . . . . . . . . . . . . . . . . . . . . . 28
11 September 26,2016 3011.1 Excision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3011.2 Mayer-Vietoris sequence . . . . . . . . . . . . . . . . . . . . . . . 31
12 September 28, 2016 3312.1 Retracts and the Brouwer fixed point theorem . . . . . . . . . . . 33
13 September 30, 2016 3613.1 Identification spaces and wedge sums . . . . . . . . . . . . . . . . 3613.2 Degree of a map . . . . . . . . . . . . . . . . . . . . . . . . . . . 3713.3 Attaching cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
14 October 3, 2016 3814.1 CW complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3814.2 Cellular homology . . . . . . . . . . . . . . . . . . . . . . . . . . 39
15 October 5, 2016 4015.1 Equivalence of singular and celluar homology . . . . . . . . . . . 4015.2 The infinite dimensional case . . . . . . . . . . . . . . . . . . . . 41
16 October 7, 2016 4216.1 Computing cellular homology . . . . . . . . . . . . . . . . . . . . 4216.2 Homology of projective space . . . . . . . . . . . . . . . . . . . . 43
17 October 12, 2016 4417.1 Computing more homologies . . . . . . . . . . . . . . . . . . . . . 4417.2 Cochains and cohomology . . . . . . . . . . . . . . . . . . . . . . 45
18 October 14, 2016 4618.1 Singular cohomology . . . . . . . . . . . . . . . . . . . . . . . . . 46
19 October 17, 2016 4819.1 Cohomology in terms of homology . . . . . . . . . . . . . . . . . 48
20 October 19, 2016 5020.1 The Ext groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2
21 October 21, 2016 5221.1 Naturality and the universal coefficients theorem . . . . . . . . . 5221.2 Relative cohomology groups . . . . . . . . . . . . . . . . . . . . . 52
22 October 24, 2016 5422.1 The cup product . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
23 October 26, 2016 5623.1 Cohomology of Euclidean space minus the origin . . . . . . . . . 56
24 October 28, 2016 5824.1 Cohomology of projective space . . . . . . . . . . . . . . . . . . . 58
25 October 31, 2016 6025.1 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6025.2 Simple version of Kunneth theorem . . . . . . . . . . . . . . . . . 61
26 November 2, 2016 6226.1 Orientation of manifolds . . . . . . . . . . . . . . . . . . . . . . . 62
27 November 4, 2016 64
28 November 7, 2016 6528.1 The cap product . . . . . . . . . . . . . . . . . . . . . . . . . . . 6528.2 Unimodular pairings . . . . . . . . . . . . . . . . . . . . . . . . . 65
29 November 9, 2016 6729.1 Proof of Poincare duality for smooth manifolds . . . . . . . . . . 67
30 November 11, 2016 6930.1 Alexander duality . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
31 November 14, 2016 7131.1 Traditional Alexander duality . . . . . . . . . . . . . . . . . . . . 7131.2 Cohomology with compact support . . . . . . . . . . . . . . . . . 72
32 November 16, 2016 7332.1 Homotopy groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
33 November 18, 2016 7533.1 Some properties of homotopy groups . . . . . . . . . . . . . . . . 75
34 November 21, 2016 7634.1 Exactness of the long exact sequence . . . . . . . . . . . . . . . . 76
35 November 28, 2016 7735.1 Smooth approximation . . . . . . . . . . . . . . . . . . . . . . . . 7735.2 Computing homotopy groups . . . . . . . . . . . . . . . . . . . . 77
3
36 November 30, 2016 7936.1 Classifying framed manifolds . . . . . . . . . . . . . . . . . . . . 79
37 December 2, 2016 8137.1 Fiber bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
4
Math 231a Notes 5
1 August 31, 2016
This is a introduction to algebraic topology, and the textbook is going to bethe one by Hatcher. The book really tries to bring the material to life by lotsexamples and the pdf is available from the author’s website. Chapter 1 is aboutfundamental groups and covering spaces, and is dealt in Math 131. We willstart with homology, but we need are going refer to the Chpater 1 material ifneeded.
1.1 The first homology of T 2
What I am going to do today is to talk about homology, beginning with ex-amples. Consider the 2-dimensional torus, which we will going to think of as aglued square.
We are going to talk about the first homology of this 2-torus.In the development of the subject, this was divided into little triangles like
this:
This triangulation has vertices, edges, triangles, which are respectively 0-simplices, 1-simplices, and 2-simplices. The standard k-simplex ∆k (in Rk+1)is the convex hull of the basis vectors e0, . . . , ek. Sometimes you would want totalk about ∆−1 ⊆ R0. This is the convex hull of {} and is thus ∅ by convention.
Let Sk be the set of all k-simplices in our triangulated T 2. A 1-chain (withZ/2 coefficient) is a subset of S1. Likewise, a 0-chain is a subset of S0, and ani-chain is a subset of Si. Let us denote by Ci the set of i-chains. This is called“with Z/2 coefficeints” because Ci is the vector space over Z/2 with basis Si.That is, the element of Ci looks like∑
σ∈Si
λσσ
with λσ ∈ {0, 1} and it then corresponds to the subset {σ : λσ = 1} ⊆ Si. Thisalternative viewpoint allows us to add 1-chains.
Math 231a Notes 6
Definition 1.1. A 1-chain is a 1-cycle if every vertex is an enedpoint of aneven number of 1-simplices in the chain.
For instance, this is a 1-cycle that is interesting:
The reason I call this interesting is because it is not a boundary. This 1-cycle,for instance is a boundary of a 2-chain.
On the other hand, you can easily convince yourself that the first 1-cycle is nota boundary.
Definition 1.2. For a 2-simplex, the boundary is defined as ∂T = σ1 + σ2 +σ3 ∈ C1 where σi are the three sides of T . For a 2-chain T1 + · · · + Tl, itsboundary is defined as
∂(T1 + · · ·+ Tl) =
l∑i=1
∂Ti.
We have thus defined a linear map ∂ : C2 → C1. The boundaries in C1
are the image im ∂ = B1 ⊆ C1. We also note that 1-cycles are the kernel of themap ∂ : C1 → C0 defined similarly as σ 7→ v1 + v2 and
σ1 + σ2 + · · ·+ σl 7→∑
all endpoints.
Confusingly both maps C2 → C1 and C1 → C0 are called ∂, and if you want todistinguish them, you can call them ∂2 : C2 → C1 and ∂1 : C1 → C0.
Definition 1.3. We define the first homology H1 as the quotient vector spaceker ∂1/ im ∂2.
That is, I am intereseted in the 1-cycles, but I am not interested in the 1-cycles that are boundaries. It is not so hard to compute the first homology ofthe torus.
Proposition 1.4. For T 2, H1 is a vector space of dimension 2 over Z/2.
Math 231a Notes 7
2 September 2, 2016
Today I want to define homology in the context of a simplicial complex.
2.1 Simplicial complex
There are two slightly different ways of looking at it. A simplicial complex isa topological space, obtained by gluing simplices together.
v0
v1
v2 v3
This has one 2-simplex {v0, v1, v2}, and four 1-simplices {v2, v3}, {v0, v2},{v0, v1}, {v1, v2}. There are four 0-simplex {v0}, {v1}, {v2}, {v3}.
Now an abstract simplicial complex consisting of a set V of “vertices”and a set S of “simplices” such that
• each simplex s ∈ S is a finite subset of V and s 6= ∅, (If s has k + 1elements, call s a k-simplex and write dim(s) = k. We denote Sk = {s ∈S : dim(s) = k}.)
• if s ∈ S and ∅ 6= t ⊆ s then t ∈ S.
We can make the definition shorter, if we allow ∅ also to be a simplex andjust do not give V .
An abstract simplicial complex is a set S such that every s ∈ S is finiteand s ∈ S and t ⊆ s implies t ∈ S.
Now given an abstract simplicial complex K = (V, S), we can constructthe corresponding topological space. If V is finite, then take the vector spaceRV ∼= Rn with V as basis, if n = #V :
RV =∑v∈V
λvv
for λv ∈ R. For s ∈ S with s = {v0, . . . , vk}, we set |s| to be the convex hull ofv0, . . . , vk in RV . Then set
|K| =⋃s∈S|s| ⊆ Rn.
Now a slight modification is needed if V is infinite. Then RV becomes
RV =
{∑v∈V
λvv
}
Math 231a Notes 8
with λv = 0 for all but finitely many v. We can similarly define |K| ⊂ RV .This embeds |K| into some huge Euclidean space, but we still need to get atopological sapce out of it. We do it so that a map f : |K| → Y is continuousif and only if the restriction to every simplex |s| is. Alternatively, we can setC ⊆ |K| to be closed if and only if C ∩ |s| is closed in |s| for every s ∈ S.
2.2 Homology for an abstract simplicial complex over Z/2Consider an abstract simplicial complex K = (V, S). Write F = Z/2.
For k ≥ 0, let Sk be the set of the k-simplices, and let
Ck = FSk = Pfinite(Sk).
These are the k-chains.We are now going to define the boundary map ∂k : Ck → Ck−1. First define
it on a basis vector s ∈ Sk as
∂ks =∑
t⊆s, dim(t)=s
t.
In other words, if s = {v0, . . . , vk} then
∂ks = {v1, . . . , vk}+ {v0, v2, . . . , vk}+ · · ·+ {v0, . . . , vk−1}
=
k∑i=0
{v0, . . . , vi, . . . , vk}.
Also, for several simplices, we define
∂k(x1 + · · ·+ sl) =
l∑j=1
∂ksj .
Definition 2.1. A k-chain γ ∈ Ck is a k-cycle if ∂kγ = 0. We denote
Zk = ker ∂k = {k-cycles} ⊆ Ck.
Definition 2.2. A k-chaing γ ∈ Ck is a k-boundary if γ = ∂k+1β for someβ ∈ Ck+1. We denote
Bk = im ∂k+1 = {k-boundary} ⊆ Ck.
Lemma 2.3. Boundaries are cycles, i.e., Bk ⊆ Zk, i.e., ∂k∂k+1β = 0 for allβ ∈ Ck+1.
Proof. We can check ∂k∂k+1β = 0 for a basis vector in Ck+1, which is a (k+ 1)-simplex s = {v0, . . . , vk+1}.
We have
∂k+1s =
k+1∑i=0
{v0, . . . , vi, . . . , vk+1}
Math 231a Notes 9
and then
∂k∂k+1s =
k+1∑i=0
∂k{v0, . . . , vi, . . . , vj , . . . , vk+1}
=
k+1∑i=0
k+1∑j=0, j 6=i
{v0, . . . , vi, . . . , vj , . . . , vk+1}
= 2∑
0≤i<j≤k+1
{v0, . . . , vi, . . . , vj , . . . , vk+1} = 0
because 2 = 0 in F = Z/2.
One remark to make is that for k < 0 we define Ck = 0. So ∂0 : C0 → C−1
is the zero map.
Definition 2.4. For an abstract simplicial complex K, we define its k-th ho-mology with Z/2 coefficients as
Hk(K;F) = Zk/Bk = ker ∂k/ im ∂k+1.
Let us go back to our torus with 9 vertices.
The 0-cycles are all the 0-chains, and the 0-boundaries are 0-chains {v1}+ · · ·+{vl} with l even. So in this case,
H0 =F9
B0
∼= F
is 1-dimensional. The first homologyH1 is 2-dimensional. ForH2, the boundaryis B2 = 0 because there are no 3-simplices in this simplicial complex, and Z2 isisomorphic to F because it has two elements: 0 and the sum of all 1-simplices.
Math 231a Notes 10
3 September 7, 2016
In doing Z-coefficients, or more generally G coefficients for an abelian group G,we need an ordering ≤ on V for an abstract simplicial complex K = (V, S).This ordering has to be a total ordering, or at least, restricts to a total orderon each s ∈ S.
3.1 Simplicial homology with G coefficients
We also use a slightly different notation. For a k-simplex s = {v0, . . . , vk}. Wewrite
s = [v0, . . . , vk]
to mean s = {v0, . . . , vk} and v0 < v1 < · · · < vk. We need this because wewant to take care of the orientation of the simplices. For instance, we want∂[v0, v1] = {v1} − {v0}.
The chain groups with Z coefficients will be
Ck = Ck(K;Z) = ZSk
which is the free abelian group with generators Sk. For coefficients in G, wedefine
Ck(K;G) ={∑
finite
ass}
for as ∈ G.Now for every k, we are going to define the boundary map ∂k : Ck → Ck−1
as
∂k(as) = ∂k(a[v0, . . . , vk]) =
k∑i=0
(−1)ia[v0, . . . , vi, . . . , vk].
We don’t need G to be a ring because we aren’t actually multiplying anything.
Lemma 3.1. The map ∂k ◦ ∂k+1 : Ck+1 → Ck−1 is zero.
Proof. Let us use Z coefficients. Let s = [v0, . . . , vk+1]. Then
∂k+1s =
k+1∑i=0
(−1)i[v0, . . . , vi, . . . , vk+1]
Math 231a Notes 11
and then
∂k∂k+1s =
k+1∑i=0
(−1)i∂k[v0, . . . , vi, . . . , vk+1]
=∑i
∑j<i
(−1)i(−1)j [v0, . . . , vj , . . . , vi, . . . , vk+1]
+∑i
∑j>i
(−1)i(−1)j+1[v0, . . . , vi, . . . , vj , . . . , vk+1]
=∑i<j
((−1)i+j + (−1)i+j+1)[v0, . . . , vj , . . . , vi, . . . , vk+1] = 0.
For example if we took [v0, v1, v2] then
∂ = [v1, v2]− [v0, v2] + [v0, v1],
∂∂ = [v2]− [v1]− [v2] + [v0] + [v1]− [v0] = 0.
Define γ ∈ Ck to be a cycle if ∂kγ = 0 and a boundary if γ = ∂k+1β forsome β. The lemma is then equivalent to Bk ⊆ Zk.
Definition 3.2. The kth homology group is
Hk(K;G) = ker ∂k/ im ∂k+1 = Zk/Bk.
We further call two cycles γ and γ′ are homologous if γ−γ′ is a boundary.A cycle γ gives a homology class [γ] = γ +Bk ∈ Zk/Bk = Hk(K;G).
3.2 Singular homology
In the book, Hatcher uses the notion of a ∆-complex instead of a simplicialcomplex. This gives him more flexibility to build spaces from a fewer numberof simplices.
A triangulation of a topological space X is an abstract simplicial complexK and a homeomorphism |K| → X. A lot of simplicial complexes admit atriangulation, but there are topological spaces that are not homeomorphic toany realization of a simplicial complex. One can ask, if K and K ′ are abstractsimplicial complexes that both triangulate X, i.e., there are homeomorphisms|K| → X and |K ′| → X, is it true that Hk(K;G) ∼= Hk(K ′;G)? For instance,are the homologies of the tetrahedron and the octahedron the same? The answeris yes, although it is not obvious.
Also, which topological spaces X are triangulable? It is known that smoothmanifolds are, and this was proved by Whitehead. But not all topologicalmanifolds are triangulable. It was known for 30 years that not all 4-manifoldsare triangulable, and it was recently shown by C. Manolescu that this is alsothe case for dimensions at least five.
Math 231a Notes 12
So you do not want to restrict yourselves to simplicial complexes. Singularhomology defines Hk(X;G) for a topological space X not dependent on a choiceof triangulation.
Let ∆k ⊆ Rk+1 be a standard k-simplex. Consider the set
Σk(X) = {σ : ∆k → X}
of all continuous maps. We this the set of singular k-simplices.Now for an abelian group G, we construct the singular k-chains
Ck(X;G) ={ ∑σ∈Σk(X)
aσσ}
where aσ = 0 except for a finite number of σ and this is just a formal sum. IfG = Z, then Ck(K;Z) = ZΣk.
We define the group homomorphisms ∂k : Ck(X;Z) → Ck−1(X;Z). Givenan element σ ∈ Σk(X), i.e., σ : ∆k → X, we want to define ∂kσ. The intuitionis that you would want it to make like the boundary of a simplex.
Given v0, . . . , vn ∈ RN , there is a unique affine-linear map ∆n → RN thatsends ei to vi. Call this map [v0, . . . , vn] : ∆n → RN . Using this notation, givenσ : ∆k → X the faces of σ are
σ ◦ [e0, . . . , ei, . . . , ek] : ∆k−1 → X.
Then we define
∂kσ =∑
(−1)iσ ◦ [e0, . . . , ei, . . . , ek].
Math 231a Notes 13
4 September 9, 2016
It is worth abstracting what we are doing and looking at it algebraically.A chain complex is a collection of abelian groups Ck with k ∈ Z and
morphisms ∂k : Ck → Ck−1 satisfying ∂k∂k+1 = 0. Let us all write this (C∗, ∂).The homology of (C∗, ∂) is defined as
Hk(C∗, ∂) = ker ∂k/ im ∂k+1 = Zk/Bk.
We will call them cycles and boundaries even if it is just algebra.
4.1 Singular homology
Recall that the singular chain complex for a topological space X is defined as
Ck(X;G) =⊕
σ∈Σk(X)
G, Σk = {σ : ∆k → X}
and Ck = 0 for k < 0. The boundary map is given as
∂k(aσ) =∑i
(−1)iaσ ◦ [e0, . . . , ei, . . . , ek].
Lemma 4.1. For k ≥ 1, ∂k ◦ ∂k+1 = 0.
Proof. Let us compute ∂k∂k+1(aτ). It is
∂k∂k+1(aτ) =∑i
∑j
(sign)aτ [e0, . . . , ei, . . . , ej , . . . , ek+1]
with the sign canceling out. So we are done.
Definition 4.2. The Hk(X;G) is the kth homology of the singular chain com-plex of X. We also denote Hk(X) = Hk(X;Z).
Usually the chain complex groups are enormous. A point is one of the rarecases we can actually compute the homology just from the definition.
Example 4.3. Let X be a single point. Then Σ is a set with a single element,and so Ck(X;G) = ⊕σ∈G ∼= G. Then the chain complex will look like
· · · G G G G 0 0 · · · .∂3 ∂2 ∂1 ∂0 ∂−1
The boundary map will be given by
∂k(aσk) =
k∑i=0
(−1)kaσk−1 =
{0 if k odd,
a if k odd.
Math 231a Notes 14
Then the chain complex is
· · · G G G G 0 0 · · · .∂3=0 ∂2=1 ∂1=0 ∂0 ∂−1
Now we can compute the homology. The 0th homology is given as
H0(pt, G) = ker ∂0/ im ∂1 = G/0 = G.
For k ≥ 1 odd,Hk(pt, G) = ker ∂k/ im ∂k+1 = G/G = 0.
For k ≥ 2 even,
Hk(pt, G) = ker ∂k/ im ∂k+1 = 0/0 = 0.
So the conclusion is that
Hk =
{G if k = 0
0 if k 6= 0.
If X =⋃mX
m are the path components of X, then Σk(X) =∐m Σk(Xm).
Then Ck(X) =⊕Ck(Xm), and we also have Zk =
⊕m Zk(Xm) and Bk =⊕
mBk(Xm). Likewise the obvious thing Hk(X) =⊕
mHk(Xm) happens forhomology.
Using this, we can compute the homology for a finite topological space, or aspace whose path components are points.
Let us compute H0(X) in general. Assume that X is path-connectedd first.We see that
Σ0(X) = {maps ∆0 → X} = X
and so C0(X) is the collections of points in X with Z multiplicities. The cyclesare
Z0(X) = ker(∂0 : C0 → C−1 = 0) = C0.
A 0-chain is a boundary if and only if the sum of the coefficients is zero. Thisis because we can connect any two points with a path. So
B0(X) = {∑
axx :∑
ax = 0}.
Then the homology is
H0(X;G) = Z0(X)/B0(X) ∼= G.
So in the general case, H0(X;G) =⊕
path componentsG.
Math 231a Notes 15
4.2 Functorial aspects of homology
One good thing about singular homology is that the construction is functorial.For two topological spaces X and Y and a continuous map f : X → Y , thisinduces a map
f ◦ (−) : Σk(X)→ Σk(Y )
σ 7→ f ◦ σ.
These maps gives us a chain map C∗(X) → C∗(Y ), i.e., a map for each ksatisfying
f# ◦ ∂Xk = ∂Yk ◦ f#.
This gives a map f∗ : Hk(X) → Hk(Y ) because f# : Zk(X) → Zk(Y ) andf# : Bk(X)→ Bk(Y ).
Two evident facts are that (f ◦ g)∗ = f∗ ◦ g∗ and if f = idX then f∗ :Hk(X) → Hk(X) is also the identity map. This is what allows us to say thathomology is functorial.
Math 231a Notes 16
5 September 12, 2016
5.1 Exact sequences
Let Ai be abelian groups for i ∈ Z, and let fi : Ai → Ai−1 be homomorphisms.
· · · A3 A2 A1 A0 · · ·f3 f2 f1
This sequence is called exact if ker fi = im fi+1 (in Ai) for all i. If it is true forone particular i, we call that it is exact at i. For example, the sequence
· · · R2 R2 R2 · · ·fi+1 fi, fi = fi+1 = · · · =
(0 10 0
)is exact. Also, the sequence
Z/2 Z/4 Z/2α β
with α : x 7→ 2x and β : y → y is exact at the middle group.The sequence
0 A B Cα β
being exact at A means that kerα = im 0 = 0, i.e., that α is injective. Likewisethe exactness of
A B C 0β 0
at C means imβ = ker 0 = 0, i.e., β is onto. So if
0 A B 0α
is exact, then α is an isomorphism.A short exact sequence is a sequence
0 A B C 0
That is exact at A,B,C. Then A → B is injective and B → C is surjective.In this case we can replace A by A ∼= α(A) ⊆ B. We can also identify C =im(B → C) = B/A. So up to isomorphism, the short exact sequence can bealso written as
0 A B B/A 0.i j
5.2 Exact sequence of chain complexes
Consider a chain complex (C∗, ∂C). This is not an exact sequence because
∂k∂k+1 = 0 only means that ker ∂k ⊇ im ∂k+1. A chain map f : B∗ → C∗
Math 231a Notes 17
between chain complexes B∗, C∗ means, for every k a map fk : Bk → Ck suchthat ∂Ck ◦ fk = fk−1 ◦ ∂Bk .
· · · Bk Bk−1 · · ·
· · · Ck Ck−1 · · ·
∂Bk
fk fk−1
∂Ck
The formula is simply saying that “both routes are equal”, i.e., the diagramcommutes.
A short exact sequence of chain complexes
0 A∗ B∗ C∗ 0i j
consists of three chain complexes, two chain maps (i, j) with i injective (ikinjective for all k) and j surjective and im i = ker j (i.e., im ik = ker jk for allk). This can be drawn as
0 0 0
· · · Ak+1 Ak Ak−1 · · ·
· · · Bk+1 Bk Bk−1 · · ·
· · · Ck+1 Ck Ck−1 · · ·
0 0 0
∂
i
∂
i i
∂
j
∂
j j
∂ ∂
The rows are chain complexes of abelian groups and the columns are short exactsequences, and the diagram commutes.
Remenber that Hk(A) = ker ∂k/ im ∂k+1. If α ∈ ker ∂k is a cycle, we denoteits homology class as [α] ∈ Hk(A). In the diagram, we can construct a map
Hk(C) Hk−1(A)∂∗
as follows. Have [γ] ∈ Hk(C) represented by γ ∈ Ck with ∂γ = 0. Because jis surjective, ther eexists a β ∈ Bk such that γ = jβ. Because j∂β = ∂jβ =∂γ = 0. So ∂β ∈ ker j = im i. So there exists an α with iα = ∂β. Also α is acycle: i∂α = ∂iα = ∂∂β = 0 implies ∂α = 0 since i is injective. So there existsa homology class [α] ∈ Hk−1(A). We are going to define
∂∗ : [γ] 7→ [α].
Math 231a Notes 18
We need to show that this is independent of choice. First we chose γ.Suppose [γ] = [γ′] so that γ′ = γ + ∂Γ. We made a second choice β withjβ = γ. If jβ′ = jβ = γ then β − β′ ∈ ker j = im i so β − β′ = i(a). Then∂β− ∂β′ = ∂(ia) = i∂a. Then there exists a unqiue α and α′ with iα = ∂β andiα = ∂β′. Then ∂β − ∂β′ = i(α − α′) = i∂a so α − α′ = ∂a. So [α] = [α′]. I’llleave the independence of the choice of γ.
Now note that i induces the map i∗ : Hk(A) → Hk(B) and j induces themap j∗ : Hk(B)→ Hk(C).
Proposition 5.1. The following sequence is exact:
Hk(A) Hk(B) Hk(C)
Hk−1(A) Hk−1(B) Hk−1(c)
......
...
i∗ j∗
∂∗
i∗ j∗
This is called the long exact sequence.
Math 231a Notes 19
6 September 14, 2016
Given a short exact sequence of chain complexes
0 A∗ B∗ C∗ 0,i j
it induces a long exact sequence
Hk+1(A) Hk+1(B) Hk+1(C)
Hk(A) Hk(B) Hk(C)
Hk−1(A) · · · .
i∗ j∗
∂∗
i∗ j∗
∂∗
i∗
We want to show that im ⊆ ker and im ⊇ ker for (i∗, j∗), (j∗, ∂∗), and (∂∗, i∗).For instance, let us prove im(i∗) ⊇ ker(j∗). Given [β] ∈ Hk(B) with β ∈ Bk,
we have ∂β = 0. Suppose that j∗[β] = [jβ] = 0. We must show that [β] ∈ im(i∗).Note that [jβ] = 0 means that jβ = ∂γ for some γ ∈ Ck+1. Because j is onto,there exists an ω ∈ Bk+1 with jω = γ. Then ∂jω = j∂ω = ∂γ. So j(β−∂ω) = 0and then β − ∂ω ∈ im i because ker j = im i, and so β − ∂ω = iα. Now α is acycle because i∂α = ∂β − ∂∂ω = 0 and so ∂α = 0. Because β − ∂ω = i∂, thisshows that [β] ∈ im(ik). The other ones can be proved similarly.
6.1 Relative homology
Let us temporarily get back to topology. Let X be a topological space and A bea topological subspace with the inclusion i : A → X. This induces a injectioni] : C∗(A) ↪→ C∗(X). This is a starting of a short exact sequence
0 C∗(A) C∗(X) C∗(X)/C∗(A) 0i# j#
of chain complexes. We denote
C∗(X,A) = C∗(X)/C∗(A).
A chain in this complex will be a chain γ in X and ignores chains σ : ∆ → Xwhose images are in A. The γ represents a cycle in C∗(X,A) if ∂γ consists ofsimplices in A. The relative homology groups are defined as
H∗(X,A) = H∗(C∗(X,A), ∂).
Because we have a short exact sequence of chain complexes and we havedone all the algebra, we have a long exact sequence.
Hk(A) Hk(X) Hk(X,A)
Hk−1(A) Hk−1(X) · · ·
i∗ j∗
∂∗
Math 231a Notes 20
In this context, the map ∂∗ has a more geometric interpretation. Start witha relative cycle in (X,A), and choose a chain β in X representing the cycle.Take ∂β which lines in A. That is the image of the cycle in ∂∗.
Math 231a Notes 21
7 September 16, 2016
7.1 The five lemma
Lemma 7.1 (Five lemma). Suppose we have a commutative diagram (of abeliangroups, for example)
G1 G2 G3 G4 G5
H1 H2 H3 H4 H5
p1
f1
p2
f2
p3
f3
p4
f4 f5
q1 q2 q3 q4
and suppose that the rows are exact. If the maps f1, f2, f4, f5 are isomorphism,then f3 is also an isomorphism.
Proof. We first prove that f3 is injective. Suppose that f3(g3) = 0. Considerg4 = p3(g3). Then f4(g4) = f4(p3(g3)) = q3(f3(g3)) = 0. Because f4 is anisomorphism, this shows that g4 = 0. So g3 ∈ ker p3 = im p2. Let g3 = p2(g2) forsome g2 ∈ G2. Let us set h2 = f2(g2). Then q2(h2) = 0 and so there exists andh1 with q1(h1) = h2. In fact, because f1 is onto there exists g1 ∈ G1 such thatf1(g1) = h1. Consider p1(g1) = g′2. Then f2(g′2) = f2(p1(g1)) = q1(f1(g1)) =q1(h1) = h2. So p1(g1) = g′2 = g2. Now g3 = p2(g2) = p2(p1(g1)) = 0.
You can show surjectivity by yourself.
If you finish the proof, you will have used that f1 is onto, f2, f4 are isomor-phisms, and f5 is injective.
A pair is a pair (X,A) with A ⊆ X. We regard X also as a pair (X, ∅). Amap of pairs f : (X,A) → (Y,B) is a map f : X → Y with f(A) ⊆ B. Forexample, (X, ∅)→ (X,A) is a map of pairs with the identity map, and also theinclusion map induces a map i : (A, ∅)→ (X, ∅).
A map f : (X,A) → (Y,B) induces a map f∗ : Hk(X,A) → Hk(Y,B)between homologies. Using the map j : (X, ∅)→ (X,A) and i : (A, ∅)→ (X, ∅),the long exact sequence can be interpreted as
· · · Hk(A) Hk(X) Hk(X,A) · · · .i∗ j∗
7.2 Eilenberg-Steenrod axioms
In the context of our class, the Eilenberg-Steenrod axioms are more likeproperties of singular homology.
(ES0) Hk assignes to each pair (X,A) an abelian group Hk(X,A).
(ES1) To each f : (X,A)→ (Y,B) there is a map f∗ : Hk(X,A)→ Hk(Y,B) foreach k.
(ES2) (g ◦ f)∗ = g∗ ◦ f∗.(ES3) (id(X,A))∗ = id on Hk(X,A) for all k.
Math 231a Notes 22
(ES4) H0(point) = Z and Hk(point) = 0 for k 6= 0.
(ES5) For all (X,A) and k, there exist ∂∗ : Hk(X,A)→ Hk−1(A) that form thelong exact sequence.
(ES6) ∂∗ is natural for all maps of pairs, i.e., for all k and all f : (X,A)→ (Y,B)the following commutes:
Hk(X,A) Hk−1(A)
Hk(Y,B) Hk−1(B)
∂∗
f∗ f∗
∂∗
Consider two maps f0, f1 : X → Y . A homotopy from f0 to f1 is a(continuous) map
F : I ×X → Y, I = [0, 1]
such that F (0, x) = f0(x) and F (1, x) = f1(x) for all x ∈ X. We also writeF (t, x) = ft(x).
For maps of pairs, consider f0, f1 : (X,A) → (Y,B). A homotopy from f0
to f1 is a map F : (I ×X, I ×A)→ (Y,B) such that it is f0 at t = 0 and f1 att = 1.
(ES7) If f0 and f1 are homotopy maps of pairs (X,A) → (Y,B) then (f0)∗ =(f1)∗ as maps Hk(X,A)→ Hk(Y,B) for all k.
We would like to prove this fact for singular homology. We need an algebraiccriterion for two chain maps. f0 : A∗ → B∗ and f1 : A∗ → B∗ to give equalmaps (f0)∗ = (f1)∗ : Hk(A)→ Hk(B).
Definition 7.2. Two chain maps f0 and f1 are called chain homotopic ifthere exist maps Kk : Ak → Bk+1 such that
∂Bk+1 ◦Kk +Kk−1 ◦ ∂Ak = f1 − f0
for every k.
If f0 and f1 are chain homotopic, then (f0)∗ = (f1)∗. To see this, considerany [α] ∈ Hk(A) so α ∈ Ak with ∂Ak α = 0. Applying the equation, we see that
f1α− f0α = ∂BKα.
Then [f1α] = [f0α] in Hk(B).
Math 231a Notes 23
8 September 19, 2016
8.1 Homotopy and homology
Proposition 8.1. If f0, f1 : X → Y are homotopic (to a homotopy F : I×X →Y ), then the chain maps (f0)#, (f1)# : C∗(X) → C∗(Y ) are chain homotopic.That is, there is a Kk : Ck(X) → Ck+1(Y ) with (f1)# − (f0)# = ∂k+1Kk +Kk−1∂k as maps Ck(X)→ Ck(Y ). So (f1)∗ = (f0)∗ as maps Hk(X)→ Hk(Y ).
Basically what this proposition is require us to do is to find a way to subdi-vide I ×∆k into ∆k+1. This way, we can use the homotopy to get a k+ 1-cyclethat has f1(γ)− f0(γ) as the boundary. The vertices of I ×∆k are vi = (0, ei)and wi = (1, ei) for 0 ≤ i ≤ k. Then I ×∆k is a unit of k + 1-simplices:
[v0, w0, w1, . . . , wk], . . . , [v0, . . . , vj , wj , . . . , wk], . . . , [v0, . . . , vk, wk].
Although this is the geometric picture to keep in mind, it is not going to usedin the proof.
Proof. We are going to define K = Kk : Ck(X)→ Ck+1(Y ) on a basis elementσ : ∆k → X. From σ, we get a map δ : I×∆k → I×X and F ◦ σ : I×∆k → Y .Consdier for j = 0, . . . , k the map
F ◦ σ[v0, . . . , vj , wj , . . . , wk] : ∆k+1 → Y.
Now define
Kσ =
k∑j=0
(−1)jFσ[v0, . . . , vj , wj , . . . , wk].
We need to verify ∂Kσ +K∂σ = f1σ − f0σ. First
∂Kσ =
k∑j=0
(−1)j( j∑i=0
(−1)iFσ[v0, . . . , vi, . . . , vj , wj , . . . , wk]
+
k∑i=j
(−1)i+1Fσ[v0, . . . , vj , wj , . . . , wj , . . . , wk]
)and the other way round is
K∂σ = K
k∑i=0
(−1)iσ[e0, . . . , ei, . . . , ek]
=
k∑i=0
i−1∑j=0
(−1)(i+ j)Fσ[v0, . . . , vj , wj , . . . , wi, . . . , wk]
+
k∑i=0
k∑j=i+1
(−1)i+j−1Fσ[v0, . . . , vi, . . . , vj , wj , . . . , wk].
Math 231a Notes 24
So when we add them, we get
∂Kσ +K∂σ =
k∑j=0
Fσ[v0, . . . , vj−1, wj , . . . , wk]
−k∑j=0
Fσ[v0, . . . , vj , wj+1, . . . , wk]
= Fσ[w0, w1, . . . , wk]− Fσ[v0, v1, . . . , vk] = f1σ − f0σ.
This finishes the proof.
So homotopic maps give rise to the same maps on homology.
Definition 8.2. A map f : X → Y is a homotopy-equivalence if there existsa map g : Y → X such that g ◦ f is homotopic to 1X and f ◦ g is homotopic to1Y .
Corollary 8.3. If f : X → Y is a homotopy-equivalence, then the correspondingmap f∗ on homology is an isomorphism for all k.
Proof. There exists a g as above. Because g ◦ f is homotopic to 1X , we haveg∗ ◦ f∗ = 1Hk(X). Smilarly f∗ ◦ g∗ = 1Hk(Y ). Thus f∗ is an isomorphism.
Definition 8.4. X is contractible if the map X → point is a homotopyequivalence.
Corollary 8.5. If X is contractible, then H0(X) = Z and Hk(X) = 0 fork > 0.
For instance, RN , the open and closed balls, and cones, are all contractibleand so have the same homology to a point.
Math 231a Notes 25
9 September 21, 2016
If f : X → Y is a homotopy equivalence, then f∗ : H∗(X) → H∗(Y ) is anisomorphism. Likewise, we have the following theorem.
Theorem 9.1. If f : (X,A) → (Y,B) is a map of pairs, and f : X → Y andf |A : A → B are both homotopy equivalences, then f∗ : H∗(X,A) → H∗(Y,B)is an isomorphism.
Proof. Use the five lemma on
Hk(A) Hk(X) Hk(X,A) Hk−1(A) Hk−1(X)
Hk(B) Hk(Y ) Hk(Y,B) Hk−1(B) Hk−1(Y )
and we immediately get the result.
9.1 Reduced homology
We define the reduced homology Hk(X). This is going to satisfy
Hk(X) ∼=
{Hk(X) for k 6= 0,
H0(X) for k = 0
if X is nonempty.There are two definitions of the reduced homology that turns out to be
equivalent. The first one uses augmented chain complexes for X, which aredefined as
Ck(X) =
Ck(X) k ≥ 0,
Z k = −1,
0 k ≤ −2,
with the maps
· · · C1(X) C0(X) Z 0,∂ ε
where ε is given by ε :∑aixi 7→
∑ai.
There is a short exact sequence
0 E∗ C∗(X) C∗(X) 0,
where E∗ is the chain complex with E−1 = Z and Ek = 0 for k 6= −1. Then wehave an exact sequence
· · · 0H0(X) H0(X) H−1(E∗) = Z H−1(X) = 0 · · · ,
Math 231a Notes 26
where H−1(X) = 0 because X is nonempty.So we have an short exact sequence
0 H0(X) H0(X) Z 0.i p
If there exists an q with p ◦ q = id, then we say that the sequence splitsnad H0(X) = H0(X) ⊕ Z. So we need to choose a base point x0 and sendq : a 7→ [ax0].
An alternative definition for the reduced homology is H0(X) = H0(X,x0)for a base point x0. This works because the point x0 gives a map H0(x0) ∼=Z→ H0(X), which is what we needed.
9.2 Homology of spheres
To compute these, we need the excision axiom. Let (X,A) be a pair and letZ ⊂ A satisfy closX(Z) ⊂ intX(A). Then i : (X \ Z,A \ Z) ↪→ (X,A) in anatural way.
Proposition 9.2 (Excision axiom). The map i∗ on homology H∗(X\Z,A\Z)→H∗(X,A) is an isomorphism (under the hypothesis clos(Z) ⊂ int(A)).
Before proving this, let us see how it is used. Let Bn = Bn(0, 1) ⊆ Rn bethe closed ball. Define Sn−1 = {x ∈ Rn : |x| = 1}. For instance, S0 has twopoints.
Theorem 9.3. For n > 0, Hn(Sn) = Z and H0(Sn) and Hk(Sn) = 0 fork 6= n, 0.
Over reduced homology this becomes tidier.
Theorem 9.4. For n ≥ 0,
Hk(Sn) =
{Z k = n
0 k 6= n.
By the way, for reduced homology, C∗(X)/C∗(A) = C∗(X)/C∗(A) so thereis a long exact sequence with H∗(A), H∗(X), H∗(X,A).
Proof. Let us prove the theorem by induction. The case n = 0 is trivial. Assume
Hk(Sn−1 =
{Z k = n− 1
0 otherwise.
We must prove Hk(Sn) ∼= Hk−1(Sn−1). We know that Hk(Bn) = 0 for all kbecause Bn is homotopy equivalent to a point. From the long exact sequenceat (Bn, Sn−1),
Hk(Bn) = 0 Hk(Bn, Sn−1) Hk−1(Sn−1) Hk−1(Bn) = 0.
Math 231a Notes 27
So Hk(Bn, Sn−1) ∼= Hk−1(Sn−1).Now the map (Bn, Sn−1) ↪→ (Bn(0, 1.5), A) is an homotopy equivalence, and
so H∗(Bn, Sn−1) ∼= H∗(B
n(1.5), A). Next we can map Bn around the ball andsee
(Bn(0, 1.5), A) ∼= ({xn+1 ≤ −1/2}, {−1/2 ≤ xn+1 ≤ 0}) ⊂ Sn.
By excision, this has the same homology as (Sn, {xn+1 ≤ 0}). Because {xn+1 ≤0} is homotopy equivalence, we get
Hk−1(Sn−1) ∼= Hk(Bn, Sn−1) ∼= Hk(Sn, {xn+1 ≤ 0})∼= Hk(Sn,point) ∼= Hk(Sn).
This proves the theorem.
We haven’t proved excision, and we are going to do it next class. Thisinvolves some actual epsilons and deltas, as opposed to what we have beendoing so far.
Math 231a Notes 28
10 September 23, 2016
10.1 Barycentric subdivision
The idea is that every cycle γ is homologous to a cycle γ′ whose simplices are“small”.
Figure 1: Subdivision of the 2-simplex
Let ∆n = [e0, e1, . . . , en] and v : {0, . . . , n} → {e0, . . . , en} be a bijection.For 0 ≤ j ≤ n, let
bjv = centroid of [v(j), . . . , v(n)].
So b0v is always the center and bnv is one of the vertices. Now this bijection vpicks out a simplex
bv = [b0v, . . . , bnv ]
in ∆n that can be also thought of as an embedding bv : ∆n → ∆n.Now define a subdivision operator S : Cn(X) → Cn(x) for all n on
generators as
Sσ =∑v
(−1)v(−1)vσ ◦ bv =∑v
(−1)vσ[b0v, . . . , bnv ].
Proposition 10.1. The map S is chain-homotopy to the identity 1 : C∗(X)→C∗(X), i.e., there exists an K such that 1 − S = ∂K + K∂ and K : Cn(X) →Cn+1(X).
This will imply that S is a chain map and [Sγ] = [γ] for all cycles γ.
Proof. Define K as
Kσ =
n∑j=0
∑v
v(j)<···<v(n)
(−1)j(−1)vσ[b0v, . . . , bjv, v(j), . . . , v(n)].
Now we have to verify that scary formula.
Math 231a Notes 29
Let ∂i denote “omit the ith vertex counting from 0”. Then
∂Kσ = ∂
n∑j=0
∑v(j)<···<v(n)
(−1)j(−1)vσ[b0v, . . . , bjv, v(j), . . . , v(n)]
=
n∑j=0
n+1∑i=0
∑v(j)<···<v(n)
(−1)i + j(−1)vσ∂i[b0v, . . . , bjv, v(j), . . . , v(n)]
= A+B + C +D
where
A = terms with i = j = 0, B = terms with i = n+ 1, j = n
C = terms with i = 0, j ≥ 0 D = terms with i ≥ 1 that is not B.
The terms A and B are
A = σ∂0[b0v, v(0), . . . , v(n)] = σ[v(0), . . . , v(n)] = σ[e0, . . . , en] = σ.
B =∑v
−(−1)vσ∂n+1[b0v, . . . , bnv , v(n)] = −
∑v
(−1)vσ[b0v, . . . , bnv ] = −Sσ.
C give K∂σ, and D cancel in pairs.
Chain homotopy is preserved under composition, and so there exist Lm suchthat
1− Sm = ∂Lm + Lm∂,
where Lm = Lm−1 +KSm−1 is defined recursively.
Math 231a Notes 30
11 September 26,2016
For a k-simplex ∆ = [v0, . . . , vk] in RN , denote by S∆ be its barycentric subdivi-sion. Let L = diam ∆ be its longest 1-simplex. Let ` = max{diam δ : δ ∈ S∆}.Then it is an exercise to show
` ≤ k
k + 1L.
So as a corollary, if `m = max{diam δ : δ ∈ Sm∆}, then `m → 0 as m→∞.Let X be a topological space and U = {Uα}α∈A be a cover such that intUα
cover X. Say a singular simplex σ : ∆k → X is U-small if im(σ) ⊆ Uα forsome α ∈ A. A chain γ is U -fine if all its simplices are U -small. Then there isa subset ΣUk (X) ⊆ Σk(X) of small simplices and the corresponding subgroupCUk (X) ⊆ Ck(X) of fine chains. Then there is an inclusion of chain maps
ϕ : (CU∗ , ∂) ↪→ (C∗, ∂).
Proposition 11.1. The inclusion ϕ gives isomorphisms
ϕ∗ : HUk (X)→ Hk(X).
Proof. If [γ] ∈ Hk(X) and γ ∈ Ck(X), then there exists an m such that Smγ isU -fine. Why is this? Given σ : ∆k → X consider the sets σ−1(Uα) for α ∈ Acovering ∆. This is an open cover of a compact subset, and so there exists anε > 0, which is called the Lebesgue number, such that for all balls K ⊆ ∆k ofdiameter ε there exists an α with K ⊆ σ−1(intUα). So there exists an m suchthat Smσ ∈ CUk (X). Ditto, for every chain γ there exists an Smγ ∈ CUk (X).
Now Sm is a chain map and is chain-homotopic to 1. So in Hk(X), [γ] =[Smγ]. The latter is in the image of HU
k (X), and this shows that ϕ∗ is surjective.To show injectivity, we consider [ω] ∈ HU
k with ω ∈ CUk that maps to 0 inHk. Then ω ∈ ∂β for some β ∈ Ck+1(X). Then Smω = Sm∂β = ∂Smβ. Form� 0, Smβ is U -fine. Also
ω − Smω = (∂Lm + Lm∂)ω = ∂Lmω
is the boundary of a U -fine chain. So
ω = Smω + ∂Lmω = ∂(Smβ + Lmω) = ∂ζ,
where ζ is U -fine.
11.1 Excision
Let (X,A) be a pair, and let Z ⊂ A be a subset with clX(Z) ⊆ intX(A). Thisis simply saying that
X = int(X \ Z) ∪ int(A).
Math 231a Notes 31
Consider U = {U1, U2} with U1 = X \ Z and U2 = A. There is a short exactsequence
0 C∗(A) CU∗ (X)CU∗ (X)C∗(A) 0.
This gives a long exact sequence, and we can compare it with the ordinary longexact seqence.
· · · H∗(A) HU∗ (A) HU
∗ (X,A) · · ·
· · · H∗(A) H∗(X) H∗(X,A) · · ·
∼= ϕ∗
Applying the five lemma, we see that the chain map
CU∗ (X)
C∗(A)→ C∗(X)
C∗(A)
gives isomorphisms in homology. The left hand side is
CU∗ (X)
C∗(A)=C∗(A) + C∗(X \ Z)
C∗(A)=
C∗(X \ Z)
C∗(X \ Z) ∩ C∗(A)=C∗(X \ Z)
C∗(A \ Z).
This shows that
Hk(X \ Z,A \ Z) ∼= Hk(X,A).
This is called excision.
11.2 Mayer-Vietoris sequence
Suppose X = U1 ∪U2 and furthermore X = intU1 ∪ intU2. The chain complexC∗(U1) computes H∗(U1), and the chain complex C∗(U1) ⊕ C∗(U2) computesH∗(U1) ⊕ H∗(U2). Inside C∗(X), the interior sum C∗(U1) + C∗(U2) computesHU∗ (X) ∼= H∗(X), where U = {U1, U2}.
Now there is a natural map
C∗(U1)⊕ C∗(U2)→ C∗(U1) + C∗(U2); (γ1, γ2) 7→ γ1 + γ2.
The kernel is pairs with γ2 = −γ1, and thus must be a chain in U1 ∩ U2. So wehave a short exact sequence
0 C∗(U1 ∩ U2) C∗(U1)⊕ C∗(U2) CU∗ (X) 0.
Then we have a long exact sequence
Hk(U1 ∩ U2) Hk(U1)⊕Hk(U2) Hk(X)
HK−1(U1 ∩ U2) Hk−1(U1)⊕Hk−1(U2) Hk−1(X)
Math 231a Notes 32
This is a very useful sequence in computing homology. It is worth notingthat you can make the exactly same exact sequence by looking at the reducedhomology instead.
Math 231a Notes 33
12 September 28, 2016
If X = intU1 ∪ intU2, then there is a short exact sequence
0 C∗(U1 ∩ U2) C∗(U1)⊕ C∗(U2) CU∗ (X) 0.
This gives a map ∂∗ : Hk(X) = HUk (X) → Hk−1(U1 ∩ U2), mapping a k-chain
to the k − 1-chain that “cuts” it into two pieces, one lying in U1 and the otherlying in U2.
Sometimes X = A1 ∪A2 but X 6= intA1 ∪ intA2. In this case, suppose thatA1 ⊆ U1, A2 ⊆ U2, and A1 ∩ A2 ⊂ U1 ∩ U2 are three homotopy equivalences,and X = int(U1) ∪ int(U2). Then by the homotopy equivalences, we have thesame Mayer-Vietoris sequence
Hk(A1 ∩A2) Hk(A1)⊕Hk(A2) Hk(X) Hk−1(A1 ∩A2).
Let us now compute the homology of the sphere. Consider Sn ⊆ Rn+1, withSn = Bn+ ∪Bn−,
Bn+ = {xn+1 ≥ 0}, Bn− = {xn+1 ≤ 0}.
The intersection is Bn+ ∩Bn− = Sn−1 ⊆ Rn. Since
H∗(Bn+)⊕ H∗(Bn−) = 0,
the map Hk(Sn) → Hk−1(Sn−1) is an isomorphism for all n and k. So again,we obtain
Hk(Sn) =
{Z if n = k
0 otherwise..
12.1 Retracts and the Brouwer fixed point theorem
A subset A ⊆ X is a retract of X if there exists a retraction r : X → A suchthat r|A = id. For instance, {(0, 0)} is a retract of {0, 1} × [0, 1].
A deformation retract is a homotopy ft : X → X such that f0 = idX ,f1(X) = A, and f1(X) = A with f1|A = idA.
Let us now consider the case X = Bn and A = Sn−1. Is A a retract of X?The answer in this case is no, i.e., there does not exist a map r : Bn → Sn−1
such that r(x) = x for all x ∈ Sn−1.
Proof. Suppose r exists. Then r : Bn → Sn−1 and i : Sn−1 → Bn satisfy r ◦ i =idSn−1 . Apply reduced homology Hn−1 and get r∗ : Hn−1(Bn)→ Hn−1(Sn−1)and i∗ : Hn−1(Sn−1)→ Hn−1(Bn). Then the composition
Z 0 Zi∗ r∗
must be identity on Z. This is a contradiction.
Math 231a Notes 34
Corollary 12.1 (Brouwer fixed point theorem). If f : Bn → Bn is a continuousmap, then f has a fixed point, i.e., there exists an x with f(x) = x.
Proof. Suppose f(x) 6= x for all x. We construct from f a retraction r : Bn →Sn−1 as follows: there exists a unique t ≥ 1 such that
st(x) = tx+ (1− t)f(x)
has norm 1, and define r(x) = st(x). This is clearly a tract, and therefore weget a contradiction.
In the Mayer-Vietoris sequence the map ∂∗ is natural. If f : X → Y isa continuous map with f(Ui) ⊆ Vi for i = 1, 2, and X = int(U1) ∪ int(U2),Y = int(V1) ∪ int(V2), then the diagram
Hk(X) Hk−1(U1 ∩ U2)
Hk(Y ) Hk−1(V1 ∩ V2)
∂∗
f∗
∂∗
commutes.Consider Sn ⊆ Rn+1 and f : Sn → Sn given by
(x1, . . . , xn+1) 7→ (−x1, x2, . . . , xn+1). (†)
Lemma 12.2. On Hn, f∗ : Z→ Z is given by d 7→ −d.
Proof. Recall from the Mayer-Vietoris sequence
Hn(Sn) Hn−1(Sn−1)
Hn(Sn) Hn−1(Sn−1)
∂∗∼=
f∗ f∗
∂∗∼=
If f∗ = −1 for Sn−1, then it is too for Sn. So it suffices to check for n = 0. Thisis obvious.
A linear map A : Rn → Rn means that ATA = I. Then detA = ±1. Denote
O(n) = all orthogonal transformations,
SO(n) = {A ∈ O(n) : detA = 1}.
Any A0 in O(n) can be joined by a path At to either I or †. This is becauseyou can match each axis one by one.
Corollary 12.3. An orthogonal matrix A, giving a map A : Sn−1 → Sn−1
applies to homology A∗ : Hn−1(Sn−1) → Hn−1(Sn−1) is 1 or −1 according asdetA = 1 or −1.
Math 231a Notes 35
Corollary 12.4. If detA = −1 then A : Sn−1 → Sn−1 is not homotopic to theidentity.
Corollary 12.5. If n is odd, the map Sn−1 → Sn−1 sending x 7→ −x is nothomotopic to 1.
Corollary 12.6. If n is odd, on Sn−1 there does not exist a nowhere vanishing(continuous) vector field.
Math 231a Notes 36
13 September 30, 2016
13.1 Identification spaces and wedge sums
For a closed subset A ⊆ X, the identification space X/A is defined as iden-tifying all points on A to one point a. Let us call X/A = X. Then there is anatural map
p∗ : H∗(X,A)→ H∗(X, a).
We show that this is an isomorphism in good cases.Suppose there exists an U such that A ⊆ intX(U) and A is a strong de-
formation retract of U . That is, there exists a homotopy ft : U → U suchthat f0 = idU and f1 is a traction of U to A and ft|A = idA. This gives us ahomotopy ft : U → U which deformation retracts U to a ∈ U .
We then have a diagram
(X \A,U \A) (X \ {a}, U \ {a})
(X,U) (X,U)
p′
p
By excision, the two vertical maps give isomorphisms on homology, and p′ is anhomeomorphism. Thus p also gives an isomorphism on homology
p∗ : H∗(X,U)→ H∗(X,U).
But the left hand side is isomorphic to H∗(X,A) and the right hand side isisomorphic to H∗(X, {a}) because they are homotopic.
For two topological spaces (X,x) and (Y, y) with base points, we define theirwedge sum as
X ∨ Y =X q Y{x, y}
.
Likewise, for a collection of spaces (Xα, xα), we define∨α
Xα =
∐Xα
{xα : α ∈ A}.
Suppose for each α there exists a neighborhood Uα of xα that strongly defor-mation retracts to xα. Then
H∗
(∨Xα
)= H∗
(∐Xα
/∐{xα}
)∼= H∗
(∐Xα,
∐{xα}
)=⊕
H∗(Xα).
The simplest case the bouquet of spheres X =∨Sαn . There are pathological
cases like attaching two Hawaiian earrings at the accumulation points.
Math 231a Notes 37
There are natural maps πβ :∨Xα → Xβ given by mapping all other spaces
to the base point of Xβ and mapping Xβ to itself by the identity, and thisinduces the projection map
(πβ)∗ : H∗
(∨Xα
)→ H∗(Xβ).
Likewise there is a natural inclusion map iβ : Xβ →∨Xα that induces the map
(iβ)∗ : H∗(Xβ)→ H∗
(∨Xα
).
13.2 Degree of a map
Any continuous map f : Sn → Sn gives a map f∗ : Hn(Sn) → Hn(Sn), whichis actually Z→ Z. This is multiplication by d for some d, and this d = deg f iscalled the degree of f .
The degree of f : Sn → Sn can be computed by picking a q ∈ Sn and lookingat f−1(q) ⊆ Sn and “counting the points in f−1(q) with multiplicities”.
Let ω ∈ Hn(Sn) be a generator. Then
ω ∈ Hn(Sn) = Hn(Sn,point) = Hn(Sn, U) = Hn(Sn, Sn \ q) ∼= Hn(B,B \ q)
by excision. Let ωq be the generator of Hn(B,B \ q) that corresponds to ω.Now suppose f−1(q) = P = {p1, . . . , pN} and let Bi 3 pi be disjoint balls
with f(Bi) ⊆ B. Like ωq, there exists an element ωpi ∈ Hn(Bi, Bi \ pi) that isa generator. Then f induces a map (Bi, Bi \ pi)→ (B,B \ q), and thus we canlet
(fi)∗(ωpi) = diωq.
This di is called the local degree at pi.
Proposition 13.1. In this case, d =∑i di is the degree of f .
Proof. Consider W = Sn \ (⋃Bi), and identify W to one point. Then Sn/W
will be a bouquet of spheres, each of one corresponding to the Bi. In the targetspace, this is the identifying Z = Sn \ B to one point. The map between theidentification spaces will be adding up the projection maps. It follows thatd =
∑i di.
13.3 Attaching cells
Consider a space Y and a map ϕ : Sn−1 → Y where Sn−1 is the boundary ofthe closed n-ball Dn. Now form
Y ∪ϕ en = Y ∪Dn/(x ∈ Dn ∼ ϕ(x) ∈ Y if x ∈ Sn−1).
This is the space obtained form Y by attaching an n-cell.
Math 231a Notes 38
14 October 3, 2016
Recall that for a map Sn−1 = ∂Dn → Y , we can attach an n-cell to obtain
X = Y ∪ϕ en = Y qDn/ ∼ .
We can also attach many n-cells. For α ∈ A where A may be infinite, considermaps ϕα : Sn−1 → Y . Then we can attach n-cells at the same time by taking
X =Y q (
∐αD
nα)
∼where x ∈ ∂Dn
α ∼ ϕα(x) ∈ Y.
Then X is obtained from Y by attaching n-cells. A subset K ⊆ X is open(respectively closed) if and only if its inverse image in Y q (
∐Dnα) is open
(respectively closed).
14.1 CW complexes
Definition 14.1. A CW complex X is defined by its skeleta ∅ = X−1 ⊆ X0 ⊆X1 ⊆ · · · and X =
⋃nX
n, where Xn is obtained from Xn−1 by attaching n-cells, indexed by An. In this case, K ⊆ X is closd if and only if K ∩ Xn isclosed in Xn for all n.
Example 14.2. The sphere Sn is a CW complex because we can attach ann-cell to a point to obtain Sn. This is not the only way to make the sphehre.For example S2 can be obtained by attaching two 1-cells to two 0-cells to get aS1 and then attaching two 2-cells.
Consider the pair (Xn+1, Xn). For each α ∈ An+1, let oα ∈ en+1α be the
center. Then
U = Xn ∪ (enα \ oα)α = complement of {oα : α ∈ An+1}
is open inXn+1 and moreover U deformation retracts toXn. HenceHk(Xn, Xn−1) =Hk(Xn/Xn−1). But Xn/Xn−1 is just a bouquet of spheres
∨α∈An S
nα. Thus
Hk(Xn/Xn−1) =
{⊕α Z if k = n,
0 otherwise.
Now the long exact sequence of the pair (Xn, Xn−1) is given by
Hm+1(Xn−1) Hm+1(Xn) Hm+1(Xn/Xn−1)
Hm(Xn−1) Hm(Xn) Hm(Xn/Xn−1)
Hm−1(Xn−1) · · ·
Math 231a Notes 39
In particular, we have the maps jn : Hn(Xn) → Hn(Xn/Xn−1) coming fromthe pair (Xn, Xn−1) and the maps ∂n : Hn+1(Xn+1/Xn) → Hn(Xn). Thenthe composite map is
(jn ◦ ∂n) : Hn+1(Xn+1/Xn)→ Hn(Xn/Xn−1)
that will look like⊕
α∈An+1Zα →
⊕β∈An Zβ .
14.2 Cellular homology
Now define the cellular chain complex of X as
CCWn (X) = Hn(Xn/Xn−1) ∼=
⊕α∈An
Z
with
∂CWn+1 = jn ◦ ∂n : CCW
n+1(X)→ CCWn (X).
Proposition 14.3. ∂CW ◦ ∂CW = 0 and
ker(∂CWn )
im(∂CWn+1)
=ker(jn−1 ◦ ∂n−1)
im(jn ◦ ∂n)∼= Hn(X).
For example, consider Sn = e0 ∪ en for n ≥ 2. Then CCWk (X) ∼= Z for
k = 0, n and CCWk (X) ∼= 0 for k 6= 0, n. It follows that Hk(Sn) ∼= Z for k = 0, n
and 0 otherwise.We are going to suppose today that X is finite dimensional, i.e., XN =
XN+1 = · · · for some N . Note that Hm(Xk/Xk−1) 6= 0 only if m = k. Thenfrom the long exact sequence we see that Hm(Xk−1) → Hm(Xk) is injective ifm+ 1 6= k and surjective if m 6= k. Thus
Hn(Xn+1) ∼= Hn(Xn+2) ∼= · · · ∼= Hn(X)
if X is finite dimensional. Likewise Hn(Xn−1) ∼= Hn(Xn−2) ∼= · · · ∼= 0.We have an exact sequence
Hn+1(Xn+1/Xn) Hn(Xn) Hn(Xn+1) 0∂n+1
so Hn(Xn+1/Xn) ∼= Hn(Xn)/ im(∂n+1). Likewise
0 = Hn(Xn−1) Hn(Xn) Hn(Xn/Xn−1)j
so j is injective.
Math 231a Notes 40
15 October 5, 2016
Recall that a CW complex is defined by a increasing chain of skeleta
X0 ↪→ X1 ↪→ · · · ↪→ Xk ↪→ Xk+1 ↪→ · · · .
The dimension of X is dimX = N where N is the minimal number satisfyingXN = XN+1 = · · · = X. There is an exact sequence
Hn+1(∨Sk+1) Hn(Xk) Hn(Xk+1) Hn(
∨Sk+1)
and so Hn(Xk)→ Hn(Xk+1) is an isomorphism unless n = k+1 or n+1 = k+1.So for a fixed n, the sequence Hn(X0)→ Hn(X1)→ · · · will look like
0 · · · 0 Hn(Xn) Hn(Xn+1) Hn(Xn+2) · · · .∼=
So all the interesting stuff happens there in the middle.
15.1 Equivalence of singular and celluar homology
From the pair (Xn+1, Xn), we get maps
Xn Xn+1 Xn+1/XnI J
and these give an exact sequences
Hn+1(Xn+1) Hn+1(Xn+1/Xn) Hn(Xn) Hn(Xn+1),
Hn(Xn) Hn(Xn/Xn−1) Hn−1(Xn−1) Hn−1(Xn),
Hn−1(Xn−1) Hn−1(Xn−1/Xn−2) Hn−2(Xn−2) Hn−2(Xn−1).
jn+1 ∂n+1 in+1
jn ∂n in
jn−1 ∂n−1 in−1
Then
Hn(X) = Hn(Xn+1) = Hn(Xn)/ im ∂n+1∼= jn(Hn(Xn))/jn(im(∂n+1))
= im jn/ im(jn ◦ ∂n+1) = ker ∂n/ im(jn ◦ ∂n+1)
= ker(jn+1 ◦ ∂n)/ im(jn ◦ ∂n+1).
The good thing about this is that jn−1 ◦ ∂n and jn ◦ ∂n+1 are the same upto index shifting. So we can define
CCWn (X) = Hn(Xn/Xn−1), ∂CW
n = jn−1 ◦ ∂n : CCWn (X)→ CCW
n−1(X).
This is a chain complex because ∂CWn ◦ ∂CW
n = jn−1 ◦ ∂n ◦ jn ◦ ∂n+1 = 0. Thenwhat we did can be written as:
Math 231a Notes 41
Proposition 15.1.
Hn(X) =ker(∂CW
n )
im(∂CWn+1)
.
This equation is still true for n = 0, if we use unreduced throughout, exceptthat we interpret
CCW0 (X) = H0(X0, ∅) = H0(X0).
15.2 The infinite dimensional case
What if X is infinite dimensional? It might be the case that Hn(Xn+1) =Hn(Xn+2) = · · · 6= Hn(X). In fact, we are okay. Let me give a more generalstatement. Let Y be an increasing union of Y0 ⊆ Y1 ⊆ · · · and Y =
⋃Yn.
Suppose K ⊆ Y is closed if and only if K ∩Yn is closed in Yn for all n, and alsothat every compact K ⊆ Y is contained in Yn for some n. How can Hn(Y ) beexpressed in terms of Hn(Yk)?
We construct the limit
lim−→k
Hn(Yk) = G.
The elements of G are equivalence classes in∐kHn(Yk), where g ∈ Hn(Yk)
and g′ ∈ Hn(Y ′k) satisfy g ∼ g′ if they become equal in Hn(Yk′′) for somek′′ > max{k, k′}.
Proposition 15.2. In this situation, Hn(Y ) ∼= lim−→kHn(Yk).
Proof. Since there are maps Yk ↪→ Y , there are maps Hn(Yk) → Hn(Y ). Sothere is a map
Φ : lim−→Hn(Yk)→ Hn(Y ).
Now we need to show that Φ is an isomorphism.First we show that Φ is onto. Think of a [γ] ∈ Hn(Y ) represented by
γ =
N∑i=1
aiσi
for σi : ∆n → Y . Then⋃i imσi ⊆ K is compact. So there exists an k such that
K ⊆ Yk. Regard γ as giving γk ∈ Cn(Yk). Then Φ maps to the correspondingequivalence to [γ].
You can show that Φ is injective similarly.
Math 231a Notes 42
16 October 7, 2016
16.1 Computing cellular homology
Consider a CW complex X. We proved that the homology of X is the same asthe homology coming from the chain complex
∂CWn : Hn(Xn/Xn−1)→ Hn−1(Xn−1/Xn−2)
given by ∂CWn = jn−1 ◦ ∂n. But we need to actually describe these maps in
terms of X. We know that Hn(Xn/Xn−1) ∼=⊕
α∈An Zα and so
∂CWn :
⊕α
∈ An →⊕
β∈An−1
Zβ .
Take a particular α0 ∈ An and β0 ∈ An−1. Then there inclusion and projectionmaps
Z⊕
α∈An Zα⊕
β∈An−1Zβ Z.
qα0pβ0
The composition will be an entry in the matrix, and we are interested in com-puting this map, which is represeted by mαβ ∈ Z.
The composition can be described as
Z = Hn(Dnα0, Sn−1α0
) Hn(Xn, Xn−1) Hn−1(Xn−1)
Hn−1(Xn−1/Xn−2) Hn−1(Sn−1β ) ∼= Z.
∂n
jn−1 pβ0
But there is a commutative diagram
Hn(Dnα0, Sn−1α0
) Hn−1(Sn−1α0
)
Hn(Xn, Xn−1) Hn−1(Xn−1).
∂
qα0 (ϕα0)∗
∂n
On topological spaces, these maps corresponds to
Sn−1α0
Xn−1 Xn−1/Xn−2 Dn−1β0
/Sn−2β0
= Sn−1β0
.ϕα0 Jn−1 pβ0
So ×mαβ is the map
Φα0β0 = (pβ0 ◦Hn−1 ◦ ϕα0) : Sn−1 → Sn−1
as it acts on Hn−1(Sn−1). That is, mαβ is the degree of the map Φαβ : Sn−1 →Sn−1.
Math 231a Notes 43
16.2 Homology of projective space
Using this, let us compute the homology of
RPn = {` : ` is a 1-dimensional subspace of Rn+1} = (Rn+1 \ {0})/R∗.CPn = {` : ` is a 1-dimensional subspace of Cn+1} = (Cn+1 \ {0})/C∗.
Another way of thinking of RPn is
RPn = Sn/(v ∼ −v) ={v ∈ Rn+1 : ‖v‖ = 1, vn+1 ≥ 0}
(v1, . . . , vn, 0) ∼ (−v1, . . . ,−vn, 0)
So there is an onto map Q : Dn+ → RPn with Q|Sn−1 : Sn−1 → RPn−1. So
RPn = RPn−1 ∪ψn−1en, where ψn−1 : Sn−1 → RPn−1 is given by u 7→ {u,−u}.
So
RPn = RP0 ∪ e1 ∪ e2 ∪ · · · ∪ en = e0 ∪ e1 ∪ · · · ∪ en.
Now we see that
CCWk (RPn) =
{Z, if k = 0, 1, . . . , n
0 if k < 0 or k > n.
Then the celluar complex looks like
Z Z · · · Z.∂CWn ∂CW
n ∂CW1
We now need to compute the boundary maps.
Proposition 16.1. In CCW∗ (RPn), the map ∂CW
m is 0 if m is odd and 2 if m iseven.
Proof. We want to look at the map Sm−1 → RPm−1/RPm−2 = Sm−1. Thedegree of this map cna be computed by adding the local degrees. Because themap is 2-to-1, it is either 1 + 1 = 2 or 1− 1 = 0. If m is odd then Sm−1 is notorientable and it turns out the two have different signs. If m is even then thetwo has same orientation, so it is 2.
For example, let us compute RP4 and RP5.
Z Z Z Z Z
Hn(RP4) : 0 Z/2 0 Z/2 Z.
2 0 2 0
Z Z Z Z Z Z
Hn(RP5) : Z 0 Z/2 0 Z/2 Z.
0 2 0 2 0
Math 231a Notes 44
17 October 12, 2016
17.1 Computing more homologies
We have computed the homology of RPn. We can give a similar argument forTn = Rn/Zn. There is a natural CW complex structure on Tn given by
DkI = {x ∈ Rn : 0 ≤ xj ≤ 1, xi = 0 for i ∈ I}
with the natural inclusion DkI → Tn. The chain complex will be CCW
k (Tn) =
Z(nk), and also ∂CWk = 0 for all k. So we have
Hk(Tn) ∼= Z(nk).
We have an infinite sequence RPn ⊆ RPn+1 ⊆ · · · . We call the union RP∞,where we may consider it as 1-dimensional subspaces of
R∞ = {x = (x1, x2, . . .) : xi ∈ R, xi = 0 for almost all i}.Then the k-th homology of RP∞ as
Hk(RP∞) =
Z k = 0,
Z/2 k odd, k ≥ 1
0 k even, k 6= 0.
When we talked about RPn, we also talked about CPn. In this case have
CPn = CPn−1 ∪ e2n
because
CPn = {z : ‖z‖}/z ∼ λz = {z : ‖z‖ = 1, zn+1 ∈ R≥0}/ ∼where (z1, . . . , zn, 0) ∼ λ(z1, . . . , zn, 0) for |λ| = 1. So the chain complex lookslike
· · · Z 0 Z 0 Z.So
Hk(CPn) =
{Z k is even with 0 ≤ k ≤ 2n
0 otherwise.
Likewise, we can define CP∞ and its homology groups will be
Hk(CP∞) =
{Z k ≥ 0 is even
0 otherwise.
In passing, it is worth saying that if K is an abstract simplicial complex,then there is the abstract simplicial homology Hk(K) and the singular homologyHk(X) for X = |K|.
Theorem 17.1. Hsimpl.k (K) ∼= Hsing.
k (X).
This is just because simplicial complexes are CW complexes. You need tocheck that the boundary maps on the simplicial chains and CW chains areactually the same, but you can unravel what the boundary maps actually are.
Math 231a Notes 45
17.2 Cochains and cohomology
Now I want to talk about cohomology. A homology is something we pick upwhen we have a chain complex, and a cohomology is something we get from acochain complex.
Definition 17.2. A cochain complex is a collection of abelian groups Cn forn ∈ Z, a collection of homomorphisms δn : Cn → Cn+1 satisfying δn+1 ◦ δn = 0.
For example, the De Rahm complex is a cochain complex. If (Cn, ∂) is a chaincomplex, then we get an uninteresting cochain complex by setting Cn = C−n.
Math 231a Notes 46
18 October 14, 2016
A cochain complex (C∗, δ) is a collection of abelian groups Cn and maps δn :Cn → Cn+1 such that δn+1 ◦ δn = 0. The cohomology of is defined as
Hn(C∗) =ker δn
im δn−1
where ker δn are called the cocycles and im δn−1 are called the coboundaries. Acochain map ψ : A∗ → B∗ is a collection of maps ψn : An → Bn satisfyingψn ◦ δAn−1 = δBn ψn. Such ψ gives a map ψ∗ : Hn(A∗)→ Hn(B∗).
If A and B are abelian groups, then
Hom(A,B) = {homomorphisms A→ B}
is an abelian group by the addition (h+h′)(a) = h(a)+h′(a). A homomorphismf : A → A′ gives a map f∗ : Hom(A′, B) → Hom(A,B) by composition g 7→g ◦ f . So this is a “contravariant functor”, sending A to Hom(A,B) and f :A→ A′ to f∗ : Hom(A′, B)→ Hom(A,B). It satisfies (f1 ◦ f2)∗ = f∗2 ◦ f∗1 and(1A)∗ = 1Hom(A,B).
Fix an abelian group G and let (C∗, ∂) be a chain complex. Now define acochain complex
Cn = Hom(Cn, G), δ = ∂∗ : Cn → Cn+1.
What is ker δ? a ∈ Hom(Cn, G) means δa = 0 means a ◦ ∂ = 0 means aannihilates boundaries, i.e., a|Bn = 0.
18.1 Singular cohomology
Let X be a topological space, and let Cn(X) be the nth singular chain groupwith Z coefficients. These form a singular chain complex (C∗(X), ∂).
For a fixed G, define a cochain complex
(C∗(X;G), δ)
using the general Hom(−, G) idea, i.e., Cn(X;G) = Hom(Cn(X), G).What is a ∈ Cn(X;G) : Cn(X) → G? This is uniquely determined by
a(σ) ∈ G for all σ ∈ Σn(X). Then we would have
a(n1σ1 + · · ·+ nkσk) =∑
nia(σi).
As we have said, a is a cocycle if and only if a(γ) = 0 whenever γ is a boundary.A map f : X → Y gives a chain map f# : C∗(X) → C∗(Y ), and f∗ :
Hn(X)→ Hn(Y ). Then f# gives a cochain map
f# : C∗(Y ;G)→ C∗(X;G) and f∗ : Hn(Y ;G)→ Hn(X;G).
Math 231a Notes 47
We call the cohomology Hn(X;G) coming from (C∗(X;G), δ) singular coho-mology. To actually compute singular cohomology, use the CW chain complex(because it’s write-downable) and construct the cochain complex–which we hopecomputes the singular cohomology. (In fact, it does.)
Example 18.1. Consider S4 = e0 ∪ e4. The chain complex is
CCW∗ (S4) : Z 0 0 0 Z.
and so the cochain complex is
C∗CW(S4;G) : G 0 0 0 G.
So the cohomology is
H∗CW(S4;G) =
{G, k = 0, 4
0, otherwise.
Example 18.2. Let us compute for RP4. We have
CCW∗ : Z Z Z Z Z
C∗CW : G G G G G.
0 ×2 0 ×2
0 ×2 0 ×2
So in the case of G = Z,
H∗Cw(RP4) : Z 0 Z/2 0 Z/2.
Later we will see from the universal coefficients theorem that:
• “If you now H∗(X), then you know H∗(X).”
• If f : X → Y gives isomorphisms f∗ : Hn(X)→ Hn(Y ), then it also givesisomorphisms f∗ : Hn(Y ;G)→ Hn(X;G).
For instance, if f : X → Y is a homotopy-equivalence then f∗ is an isomorphismon cohomology.
For pairs, we define
C∗(X,A;G) = Hom(C∗(X,A);G)
and define Hn(X,A;G). The excision axiom and the universal coefficients the-orem tells us that if (X,A) is a pair with cl(Z) ⊆ int(A) then
Hn(X,A;G)→ Hn(X \ Z,A \ Z;G)
is an isomorphism.
Math 231a Notes 48
19 October 17, 2016
We are going to prove the universal coefficients theorem in stages. Begin witha chain complex (C∗, ∂). Assume that Cn is a free abelian group for each n.This is the case for the singular chain complex. Write Hn = Hn(C∗, ∂). Wethen have a cochain complex (C∗(G), δ) with Cn(G) = Hom(Cn, G). ThenHn(G) = Hn(C∗(G), δ).
In general Hn(G) 6= Hom(Hn, G) as we have seen in the case of the realprojective space, but there is a map Φ : Hn(G)→ Hom(Hn, G), given by
[a] 7→ ([γ] 7→ a(γ)).
Is this well-defined? We have ∂γ = 0 and δa = 0, i.e., a ◦ ∂ = 0. If γ′ = γ + ∂β,then
a(γ′) = a(γ + ∂β) = a(γ) + (a ◦ ∂)β = a(γ).
If a′ = a+ δb = a+ b ◦ ∂, then
a′(γ) = a(Γ) + (b ◦ ∂)γ = a(γ) + 0.
19.1 Cohomology in terms of homology
Proposition 19.1. (Still assuming that Cn is free abelian for all n,) Φ issurjective. Also Φ is injective if Hn−1 is free abelian.
Proof. Let us show surjectivity first. Let ψ : Hn → G. Then we get a ψ : Zn →G such that ψ|Bn = 0. We now want to expend ψ : Zn → G to all of Cn, to getϕ : Cn → G with ϕ|Zn = ψ.
We have a short exact sequence
0 Zn Cn Bn−1 0.
This short exact sequence splits, because Bn−1 is a free abelian group, becauseBn−1 ⊆ Cn−1. Then Cn = Zn ⊕ Bn−1 and so we can extend ψ to ϕ byϕ(γ, β) = ψ(γ).
Then ϕ : Cn → G with ϕ|Bn = ψ|Bn = 0 is a cocycle, so we have [ϕ] ∈Hn(G). An Φ([ϕ]) = ψ because for all [γ] ∈ Hn,
Φ([ϕ])[γ] = ϕ(γ) = ψ(γ) = ψ[γ].
This just means that Φ([ϕ]) = ψ.Now assume that Hn−1 is free abelian. We will show that Φ is injective.
Consider [a] ∈ Hn(G) with Φ([a]) = 0, i.e., a(γ) = 0 for all [γ] ∈ Hn, i.e.,a|Zn = 0. We want so show that [a] = 0, i.e., a = δb, i.e., a = b ◦ ∂ for someb : Cn−1 → G.
0 Zn Cn Bn−1 0
Math 231a Notes 49
Now a|Zn = 0, so a gives b′ : Bn−1 → G, which will be given by a(γ) = (b′◦∂)(γ).We need to extend b′ : Bn−1 → G to b : Cn−1 → G. But the short exact sequence
Bn−1 Zn−1 Hn−1
splits and so we can extend from Bn−1 to Zn−1, and then extend Zn−1 toCn−1.
Math 231a Notes 50
20 October 19, 2016
Suppose that every chain group in a chain complex (Cn, δ) is free. Define Hn
comming from Hom(C∗, G) = C∗(G). We have shown that
Hn(G) Hom(Hn, G)Φ
is surjective, and injective too if Hn−1 is free. If Hn−1 is not free, then whatis ker Φ? To show injectivity, recall that we had to extend a map B → G toZ → G and then to C → G.
We have an exact sequence
Hom(Zn−1, G) Hom(Bn−1, G) Hn(G) Hom(Hn, G) 0.i∗ b 7→b◦∂ Φ
So we get
ker Φ =Hom(Bn−1, G)
i∗Hom(Zn−1, G).
20.1 The Ext groups
The general situation is, when we have an exact sequence
0 B Z H 0,i
with B adn Z free abelian groups, examining
T =Hom(B,G)
i∗Hom(Z,G).
Example 20.1. Suppose Z = Z and B = 3Z. Let G = Z. In this case,H = Z/3, and
T =Hom(3Z,Z)
i∗Hom(Z,Z)=
Z3Z∼= Z/3.
Proposition 20.2. The group T depends only on H and G, and not on Z andB.
Consider H and a homorphism f : H → H. Choose generators for H andwe get B ↪→ Z � H.
B Z H
B Z H
i
h
π
g f
i π
Fine g : Z → Z so that the square commutes (this is not unique), and this gis going to map ker π into kerπ. This gives a map h : B → B, and this gives amap
h∗ : Hom(B,G)→ Hom(B,G).
Math 231a Notes 51
Moreover, h∗(i∗Hom(Z,G)) ⊆ i∗Hom(Z,G). So we get a map
h∗ :Hom(B,G)
i∗Hom(Z,G)= T → Hom(B,G)
i∗Hom(Z, G)= T .
We have made a choice of g here, but you can prove as an exercise that this h∗
does not depend on the choice of g.Now take H = H with f = id, but Z and Z still different. Then from the
diagram
B Z H T
B Z H T
B Z H T
i
h′
π
g′ 1
i
h
π
g 1
i π
we see that the composite T → T → T is the identity on T and likewiseT → T → T is the identity on T . Therefore T ∼= T . So we can write
T = T (H,G)
and further f : H ′ → H gives a map f∗ : T (H,G) → T (H ′, G). This T isusually called Ext, and givng B → Z → H, we define
Ext(H,G) =Hom(B,G)
i∗Hom(Z,G).
Example 20.3. For example,
Ext(Z/n,Z) ∼= Z/n, Ext(Z,Z) = 0, Ext(Z, G) = 0.
Because Ext(H1⊕H2, G) ∼= Ext(H1, G)⊕Ext(H2, G), if H is finitely gener-ated, then
Ext(H,Z) = Ext(Zr ⊕ (torsion),Z) ∼= (torsion).
Corollary 20.4. Going to our situation with cohomology, there exists a shortexact sequence
0 Ext(Hn−1, G) Hn(G) Hom(Hn, G) 0.Φ
In this particular case, you can prove that the sequence splits. So in fact,
Hn(G) ∼= Hom(Hn, G)⊕ Ext(Hn−1, G).
Take G = Z, and suppose that Hn, Hn−1 are finitely generated. ThenExt(Hn−1,Z) ∼= Tor(Hn−1). If Hn = Zs ⊕ Tor(Hn) then
Hn(Z) ∼= (free part of Hn)⊕ (torsion part of Hn−1).
Math 231a Notes 52
21 October 21, 2016
21.1 Naturality and the universal coefficients theorem
A chain map f : X → Y induces a map f∗ : Hk(X)→ Hk(Y ) between homologyand f∗ : Hk(Y ;G)→ Hk(X;G) between cohomology. Then we have a diagram:
0 Ext(Hn−1(X), G) Hn(X;G) Hom(Hn(X);G) 0
0 Ext(Hn−1(Y ), G) Hn(Y ;G) Hom(Hn(Y );G) 0
∼= f∗ f∗ (f∗)∗ ∼=
Naturally states that this diagram commutes. This can be checked by mappingelements. In particular, by the five lemma, if f∗ : Hk(X) → Hk(Y ) is anisomorphism for k = n − 1 and k = n, then f∗ : Hn(Y ;G) → Hn(X;G) is anisomorphism too.
Let X be a CW complex and Xk the skeleta for k ∈ N.
Proposition 21.1. The inclusion i : Xn+1 ↪→ X gives an isomorphism i∗ :Hn(X;G)→ Hn(Xn+1;G) (ditto Hk → Hk for k < n).
Corollary 21.2. You can compute Hk(X) from the CW cochain complex Hom(CCW∗ (X)mG).
21.2 Relative cohomology groups
Let us omit from now on (sometimes). The relative cohomology is definedas the cohomology of a cochain complex
Cn(X,A) = Hom(Cn(X,A), G) = Hom(Cn(X)
Cn(A), G)
= AnnCn(A)(in Cn(X) = Hom(Cn(X), G)).
Because Cn(X) = Cn(A)⊕K, we have a short exact sequence
0 Cn(X,A) Cn(X) Cn(A) 0.
Corollary 21.3. There exists a long exact sequence in cohomology
Hn(X) Hn(A) Hn+1(X,A)
Hn+1(X) Hn+1(A) · · · .
i∗ δ∗
j∗
i∗ δ∗
Example 21.4. Consider the space X which is a disjoint union of points,X =
⋃α∈A point. Then H0(X) =
⊕α∈A Z, and also H0(X) =
∏α∈A Z.
Math 231a Notes 53
Why do we bother about cohomology? If X is a topological space, then thespace
H∗(X) =
∞⊕0
Hi(X)
is a ring (with coefficients Z or R a commutative ring). There is the cupproduct
Hi ×Hj → Hi+j ; (a, b) 7→ a ` b
given by
(a ` b)[v0, . . . , vi+j ] = a[v0, . . . , vi]b[vi, . . . , vi+j ].
Math 231a Notes 54
22 October 24, 2016
22.1 The cup product
As we have defined last time, there is a natural map `: Cl(X) × Cm(X) →Cl+m(X), given by
(a ` b)(σ) = a(σ[e0, . . . , el])b(σ[el, . . . , el+m]).
If a ∈ C0(X) is the one that is given by a(σ) = 1 for all σ, which is the generatorof H0(X) if X is path-connected, then
(a ` b)(τ) = a(τ [e0])b(τ [e0, . . . , em]) = b(τ).
So a ` b = b.
Example 22.1. Consider the 2-dimensional torus, with the standard simplicialcomplex structure. Define the cochain a as a(σ) = 1 if σ crosses La and a(σ) = 0
La
Lb
otherwise. Likewise define b. Then
(a ` b)[u, x, w] = a[u, x]b[x,w]
and it is nonzero just on one 2-simplex.
Lemma 22.2. For a ∈ Cl(X) and b ∈ Cm(X),
δ(a ` b) = δa ` b+ (−1)la ` δb.
Proof. We simply evaluate it on a simplex:
δ(a ` b)τ = (a ` b)∂τ = (a ` b)n+1∑i=0
(−1)iτ [e0, . . . , ei, . . . , en+1]
=
i∑i=0
aτ [e0, . . . , ei, . . . , el+1] · bτ [el+1, . . . , en+1]
+
n+1∑i=l+1
(−1)iaτ [e0, . . . , el]bτ [el, . . . , ei, . . . , en+1]
= (δa)τ [e0, . . . , el+1] · bτ [el+1, . . . , en+1] + (−1)laτ [e0, . . . , el] · (δb)+ (−1)laτ [e0, . . . , el] · (δb)τ [el, . . . , en+1]
=((δa ` b) + (−1)la ` δb
)τ.
Math 231a Notes 55
This lemma tells us that (cocycle) ` (cocycle) = (cocycle) and also tells usthat (cocycle) ` (coboundary) = (coboundary), i.e., a ` (δc) = (−1)lδ(a ` c).So this is well-defined as a map H l(X)×Hm(X)→ H l+m(X).
What about H∗(X,A)? Suppose we have A,B ⊆ X and int(A) ∪ int(B) =A∪B. Then we can define a “`” as H l(X,A)×Hm(X,B)→ H l+m(X,A∪B).Start with chains a ∈ Cl(X,A), i.e., a ∈ Cl(X) and a annihilates Cl(A), i.e.,a(σ) = 0 for σ an l-simplex in A. So
(a ` b)(τ) = a[e0, . . . , el]b[el, . . . , en] = 0
if τ(∆n) ⊆ A or if τ(∆n) ⊆ B. This means that (a ` b) is in both in theannihilator of Cn(A) and the annihilator of Cn(B). So (a ` b) annihilatesCn(A) + Cn(B) ⊆ Cn(X). Temporarily denote Cn(A + B) = Cn(A) + Cn(B),and C∗(X,A+B) be the annihilator of C∗(A+B) in C∗(X). So we have got
`: Cl(X,A)× Cm(X,B)→ Cl+m(X,A+B).
Now we have a cochain map C∗(X,A ∪ B) → C∗(X,A + B). This givesisomorphisms in cohomology groups, because (i) C∗(A + B) → C∗(A ∪ B), bybarycentric subdivision, gives isomorphisms in homology, (ii) by the five lemma,gives the same for C∗(X,A+B), (iii) by the universal coefficient theorem, givesisomorphisms on cohomology. Using this isomorphism, we get
`: H l(X,A)×Hm(X,B)→ H l+m(X,A ∪B).
Math 231a Notes 56
23 October 26, 2016
I haven’t mentioned last time, but the cup product is associative:
(a1 ` a2 ` · · · ` an)σ = a1(σ[e0, . . . , el1 ]) · a2(σ[el1 , . . . , el1+l2 ]) · · · · · · .
23.1 Cohomology of Euclidean space minus the origin
The homology and cohomology of the complex projective space CPn is
Hk(CPn) =
{Z k = 0, 2, . . . , 2n
0 otherwise,Hk(CPn) =
{Z k = 0, 2, . . . , 2n
0 otherwise,
by the universal coefficients theorem. Let gi ∈ H∗(CPn) be the generator indimension 2i for 0 ≤ i ≤ k.
Proposition 23.1. h = g1 ∈ H2(CPn) generates the ring, i.e., gi = hi = h `· · · ` h.
Proposition 23.2. For RPn with Z/2 coefficients,
H∗(RPn;Z/2) ∼=(Z/2)[h]
〈hn+1〉
for h ∈ H1(RPn;Z/2).
We have
Hn(Rn,Rn \ 0) = Z, Hn(Rn,Rn \ 0) = Hom(Z,Z) = Z.
Let us look at the case n = 1. The map σ : ∆1 → R that passes through0 represents the generator [σ] ∈ H1(R,R \ 0). Then there is a cocycle a ∈C1(R,R \ 0) with a(σ) = 1. This [a] generates H1(R,R \ 0) = Z.
In Rn, let p1 : Rn → R be the projection map to the first coordinate. LetA1 = {x1 6= 0} ⊆ Rn. So p1 : (Rn, A1) → (R,R \ 0). Pull back a by theprojection and define a1 = p∗1(a). Define pi and ai similarly for i = 2, . . . , n.
Then
b = a1 ` a2 ` · · · ` an ∈ Cn(Rn, A1 + · · ·+An)
and so [b] ∈ Hn(R, A1 + · · ·+An) ∼= Hn(R, A1 ∪ · · · ∪An).
Lemma 23.3. [b] is the generator for Hn(Rn, A1 + · · ·+An) ∼= Z.
Proof. To prove this, it suffices to find a [γ] ∈ Hn(R, A1+· · ·+An) with b(γ) = 1.Note that we want γ to have boundary in C∗(A1) + · · · + C∗(An), because itneeds to be a cycle. So we try the n-dimensional cube [−1, 1], divided up tosimplices in an appropriate way. Explicitly, this can be written as
γ =∑π∈Sn
(sgnπ)(some simplex)
and if you evaluate at b it becomes 1 for one and 0 for all the other ones.
Math 231a Notes 57
Let us now look at RPn, which is the equivalence classes of nonzero vectorsin Rn+1, denoted by [x0, . . . , xn]. Next time we are going to do a similar thingfor RPn.
Math 231a Notes 58
24 October 28, 2016
24.1 Cohomology of projective space
Let us compute the cohomology ring of RPn with Z/2 coefficients. The CWcomplex has cells in dimension 0, 1, . . . , n and the chain complex is given by
R R R · · · R,×0 ×2 ×0
where R is a ring. For Z/2, 2 = 0 and so Hk(RPn,Z/2) = Z/2 for 0 ≤ k ≤ n.Let us write Hk(−) for Hk(−;Z/2) for a while. Where h is the generator forH1(RPn), we would want to show that Hi(RPn) is generated by gi = hi.
For eeach i, the set {xi = 0} ⊆ RPn is a hyperplane and is isomorphic toRPn−1. So we get a decomposition
RPn = ({x0 = 0}) ∪ U,
where U ∼= Rn. For each i, let Ai = {xi 6= 0}. Then A0 = U , and Ai ∼= Rn foreach i. Also U ∩A1, . . . , U ∩An is just like last time.
We have
H1(RPn) = H1(RPn) = H1(RPn,point) = H1(RPn, Ai).
So there exists a class [ai] ∈ H1(RPn, Ai) such that [ai] 7→ h ∈ H1(RPn) fori = 1, . . . , n. Now take the cup product
[a1] ` · · · ` [an] ∈ Hn(RPn, A1 + · · ·+An) ∼= Hn(RPn, A1 ∪ · · · ∪An)
∼= Hn(RPn,RPn−1) ∼= Z/2.
Because [ai] maps to h in H1(RPn), [a1] ` · · · ` [an] corresponds to h ` · · · ` hin Hn(RPn). So hn 6= 0 if [a1] ` · · · ` [an] is nonzero.
From last time, we have the “old” ai, which now we denote by ai ∈ H1(Rn, {xi 6=0}). From j : (U,U ∩ Ai) ↪→ (RPn, Ai), we get a map j∗ : H1(RPn, Ai) →H1(Rn,Rn ∩Ai), and this is an isomorphism! So
j∗ : [a1] ` · · · ` [an] 7→ [a1] ` · · · ` [an] 6= 0
from last time. This shows that hk is the generator of Hk(RPn). In particular,
H∗(RP∞;Z/2) ∼= (Z/2)[h].
Let us now look at CPn with Z coefficients. We have
H2k(CPn) =
{Z k = 0, . . . , n
0 otherwise.
Let g1 = h be a generator for H2(CPn).
Math 231a Notes 59
Proposition 24.1. The element hk generates H2k(CPn).
Proof. Let Bi = {zi 6= 0} ⊆ Cn. Then
H2(Cn, Bi) ∼= H2(C,C \ {0}) ∼= Z
and so there is a generator [bi]. We want to show that [b1] ∪ · · · ∪ [bn] is agenerator of H2n(Cn,Cn \ {0}) ∼= Z. This is essentially we did for the RPncase.
Math 231a Notes 60
25 October 31, 2016
Proposition 25.1. If [a] ∈ Hi(X;R) and [b] ∈ Hj(X;R) and R is a com-mutative ring, then [a] ` [b] = (−1)ij [b] ` [a]. In other words, H∗(X;R) =⊕Hi(X;R) is a graded commutative ring.
Proof. Let n = i+ j. Define ψ : C∗(X)→ C∗(X) by, on an n-simplex σ,
ψσ = ε(n)σ[en, . . . , e0]; ε(n) = (−1)n(n+1)/2.
We claim that ψ is a chain map and that ψ is chain homotopic to the identity.Assuming this, we have a cochain map ψ∗ : C∗(X;R)→ C∗(X;R) and
ψ∗(a ` b)σ = (a ` b)(ψσ) = ε(n)(a ` b)σ[en, . . . , en]
= ε(n)a(σ[en, . . . , en−i]) · b(σ[ej , . . . , 0])
= ε(n)ε(i)ε(j)(ψ∗a)σ[ej , . . . , en] · (ψ∗b)σ[e0, . . . , ej ]
= (−1)ij(ψ∗(b) ` ψ∗(a))σ.
So in cohomology, [(a ` b)] = (−1)ij([b] ` [a]).
If 2 is invertible in R, e.g. Z/m for m odd or Q, and [a] ∈ Hi with i odd,then [a] ` [a] = −[a] ` [a] and so [a] ` [a] = 0. In contrast, in H∗(RPn,Z/2)and h ∈ H1, all h, h2, . . . , hn are nonzero.
25.1 Tensor products
Definition 25.2. Let V and W be F -vector spaces. We define V ⊗F W as thequotient vectors spaces X/Y , where X is the vector space with basis all symbolsv ⊗ w for v ∈ V and w ∈W , and Y is the vector space generated by
λv ⊗ w = v ⊗ λw = λ(v ⊗ w),(v + v′)⊗ w = v ⊗ w + v′ ⊗ w,v ⊗ (w + w′) = v ⊗ w + v ⊗ w′.
Ditto if R is a commutative ring and V and W are R-modules, V ⊗R W isdefined as generators and relations as above. If R = Z and λ = 1 + · · ·+ 1,
(λv)⊗ w = (v + · · ·+ v)⊗ w = v ⊗ w + · · ·+ v ⊗ w = λ(v ⊗ w)
and so the first condition is not needed.There is always a bilinear map b : H × G → H ⊗ G given by (h, g) 7→
h ⊗ g. This has the universal property in the sense that given a bilinear mapψ : H ×G→ K there exists a ψ : H ⊗G→ K such that ψ = ψ ◦ b.
Example 25.3. Z/2⊗ Z/3 = 0.
Math 231a Notes 61
25.2 Simple version of Kunneth theorem
We want to look at the cohomology of X × Y for topological spaces X and Y .We’ll only look at the case where H∗(Y ;R) is a free R-module of finite rank,where R is a commutative ring. Also we will assume that X is a finite simplicialcomplex.
There are maps p1 : X ×Y → X and p2 : X ×Y → Y . So there is a bilinearmap of R-modules
H∗(X)×H∗(Y )→ H∗(X × Y ); (a1, a2) 7→ p∗1(a1) ` p∗2(a2).
We can regard this as a map κ : H∗(X)⊗R H∗(Y )→ H∗(X × Y ).
Theorem 25.4 (Kunneth theorem). If H∗(Y ) is free and finitely generated asan R-module, then κ is an isomorphism.
This means that
κ :⊕i+j=n
Hi(X)⊗R Hj(Y )→ Hn(X × Y )
is an isomorphism.Let me just sketch the proof. Fix an Y and let Ln(X) be the left hand
side and Rn(X) be the right hand side. Clearly κ : Ln(X) → Rn(X) is anisomorphism if X = ∅ and X = point. The main thing is that if X = U ∪ Vis an open covering, and κ is an isomorphism for U and V and U ∩ V , thenκ is also an isomorphism for X too. This will come from some Mayer-Vietorissequence and the five lemma. The right hand side is just the cohomology andso it is reasonable. For the left hand side, there is a problem is tensoringbecause tensoring does not necessarily preserve exactness. This is where finitelygenerated and free comes in.
Math 231a Notes 62
26 November 2, 2016
Definition 26.1. A manifold of dimension n is a connected Hausdorff spacesuch that for each x ∈M there is a neighborhood U 3 x that is homeomorphicto Rn.
Theorem 26.2. If M is compact and orientable, then Hn(M) ∼= Z.
Theorem 26.3 (Poicare duality). If M is compact and orientable, then forevery k, Hk(M) ∼= Hn−k(M).
26.1 Orientation of manifolds
Note that for every x ∈M ,
Hn(M,M \ {x}) ∼= Hn(R,Rn \ {x}) ∼= Hn−1(Rn \ {x}) ∼= Hn−1(Sn−1) ∼= Z.
Definition 26.4. A generator of Hn(M,M \{x}) is called a local orientationat x.
Note that this notion agrees with the usual notion. An elementu ∈ Hn(Rn,Rn\{0}) ∼= Z can be represented by σ : ∆n → Rn with 0 /∈ σ(∂∆n). This is becauseσ induces a map ∂σ : Sn−1 ∼= ∂∆n → Rn \ {0} ∼= Sn−1.
Definition 26.5. The orientation double cover is defined as
M = {µx ∈ Hn(M,M \ {x}) generator for x ∈M}.
The topology is given by, for a ball B ⊆ U ∼= Rn and µB ∈ Hn(M,M \B), thebasis
SµB = {µx : x ∈ B and Hn(M,M \B)→ Hn(M,M \ {x}) maps µB to µx}.
Note that M →M a 2-fold covering map.
Definition 26.6. A orientation of M is a continuous map µ such that π◦µ =id. Equivalently, an orientation is a function x 7→ µx ∈ Hn(M,M \ {x}). If Mhas an orientation, M is called orientable.
Proposition 26.7. Let M be a connected manifold. M is oriented if and onlyif M have two components.
Proof. If M has two components, then M = M q M because it is a 2-foldcovering. Then it is easy to define µ. If µ is an orientation, then so is −µ. Sothe images must be the two components of M .
Note that a 2-forld cover N →M is given by a homomorphism π(M)→ Z/2.So for M →M , we can interpret this as
φ : π1(M)→ Z/2; γ 7→
{0 “M is oriented along γ”
1 else.
Math 231a Notes 63
If π1(M) = 1 then every 2-fold cover has 2 components and so M is ori-entable. Also, the n-torus M = Tn ∼= Rn/Zn is orientable. More generally,every Lie group is orientable because you can push-forward one local orienta-tion to every point. On the other hand, the Mobius band is non-orientable.This is because M →M induces a surjective π1(M)→ Z/2 is surjective.
Theorem 26.8. Assume M is a manifold with dimension n, with an orientationµ. Then for all compact K ⊆M , there exists a unique µK ∈ Hn(M,M \K) sothat for every x ∈ K, the map ρx : Hn(M,M \ K) → Hn(M,M \ {x}) sendsµK 7→ µx.
Corollary 26.9. Let M be a compact manifold. Take K = M . It follows thatthere exists a µM ∈ Hn(M) gives a local orientation at each point.
Definition 26.10. The element µM ∈ Hn(M) is called the fundamental classand is sometimes denoted [M ].
Corollary 26.11. If M is compact and oriented, then Hn(M) 6= 0.
For example, RP2 is not orientable because H2(RP2) = 0. Also the Kleinbottle K = RP2#RP2 is not orientable because H2(K). Finally, RP2 × RP2 isnot orientable because
H4(RP2 × RP2) ∼= H2(RP2)×H2(RP2) ∼= 0
by the Kunneth formula.
Math 231a Notes 64
27 November 4, 2016
I missed class. We proved Theorem 26.8.
Math 231a Notes 65
28 November 7, 2016
Theorem 28.1 (Poincare duality). If M is a compact oriented n-manifold,Hk(M) ∼= Hn−k(M).
We proved that if M is orientable, µx ∈ Hn(M,M \ {x}) is an orientationand K ⊆ M is compact, then there exists a µK ∈ Hn(M,M \ K) such thatµK 7→ µx for all x ∈ K. So if M is compact, we can take K = M and get a“fundamental class” µ = µM ∈ Hn(M) such that µM 7→ µx for all x ∈M .
28.1 The cap product
We define the cap product a: Ck × Cl → Ck−l as the following, for k ≥ l:
(σ, ϕ) 7→ ϕ(σ[e0, . . . , el])σ[el, . . . , ek].
Let us compute (∂σ) a ϕ. We have
(∂σ) a ϕ =
( n∑i=0
(−1)iσ[e0, . . . , ei, . . . , en]
)a ϕ
=
l+1∑i=0
(−1)iϕ(σ[e0, . . . , ei, . . . , el+1])σ[el+1, . . . , en]
+
n∑i=l
(−1)iϕ(σ[e0, . . . , el])σ[el, . . . , ei, . . . , en]
= σ a (δϕ) + (−1)l∂(σ a ϕ).
We can also write this as
∂(σ a ϕ) = (−1)l((∂σ) a ϕ− σ a (δϕ)).
From this, we again see that cycles times cocycles are cycles, boundaries timescocycles are boundaries, and cycles times coboundaries are boundaries.
Corollary 28.2. There is a well-defined cap product a: Hk(X) × H l(X) →Hk−l(X).
28.2 Unimodular pairings
Theorem 28.3 (Poincare duality). If M is a compact oriented n-dimensionalmanifold, and µM ∈ Hn(M) is a fundamental class, then the map
D : H l(M)→ Hn−l(M); ϕ 7→ µM a ϕ
is an isomorphism for all l.
Math 231a Notes 66
Note that clearly a is bilinear. More interesting is the relation between aand `. Recall that there is a map E : Hk(X)×Hk(X) → Z. Let us use 〈•, •〉for this evaluation map. Then
〈b, σ a a〉 = 〈a ` b, σ〉 = 〈a, σ[e0, . . . , el]〉〈b, σ[el, . . . , en]〉.
Definition 28.4. If V1, V2 are free abelian groups of finite rank, a bilinar mapQ : V1 × V2 → Z is a unimodular paring if the corresponding map V1 →Hom(V2,Z) is an isomorphism.
The univeral coefficients theorem states that
E :Hk(X)
Torsion→ Hom(Hk(X),Z)
is an isomorphism if H∗(X) are finitely generated. In other words, there is aunimodular paring between Hk(X)/Tor and Hk(X)/Tor. If Poincare duality istrue, then
Hk/Tor×Hn−k/Tor→ Z; (b, a) 7→ 〈b,D(a)〉 = 〈a ` b, µM 〉
is unimodluar.
Math 231a Notes 67
29 November 9, 2016
Recall that a: Ck(X)× Cl(X)→ Ck−l(X) is defined by
(σ, a) 7→ a(σ[e0, . . . , el])σ[el, . . . , ek].
This induces two kinds of maps:
Ck(X)
Ck(A)× Cl(X)→ Ck−l(X)
Ck−l(A),
Ck(X)
Ck(A)×Ann(Cl(A))→ Ck−l(X).
Then there are maps
Hk(X,A)×H l(X)→ Hk−l(X,A), Hk(X,A)×H l(X,A)→ Hk−l(X).
Let H,K ⊆ G be abelian groups. Then there are the following exact se-quences
0 H ∩K H ⊕K H +K 0
0 GH∩K
GH ⊕
GK
GH+K 0
0 Ann(H +K) Ann(H)⊕Ann(K) Ann(H ∩K) 0.
Let A,B ⊆ X be open subset and let H = C∗(A), K = C∗(B), and G = C∗(X).Then these short exact sequences give Mayer-Vietoris sequences. The first one isthe ordinary Mayer-Vietoris sequence. The second one gives an “upside-down”Mayer-Vietoris sequence:
Hk(X,A ∩B) Hk(X,A)⊕Hk(X,B) Hk(X,A ∪B) Hk−1(X,A ∩B) · · · .
The third one gives the upside-down Mayer-Vietoris sequence for cohomology:
Hk(X,A ∪B) Hk(X,A)⊕Hk(X,B) Hk(X,A ∩B) Hk+1(X,A ∪B) · · · .
29.1 Proof of Poincare duality for smooth manifolds
Let M be a smooth, compact, oriented, n-dimensional manifold. A c-pair isa pair (U+, U) where U+ ⊆ M is open, U ⊆ M is closed, and (U+, U) ∼=(B◦1 , B1/2). A simple cover of M means a collection of c-pairs (U+
i , Ui) fori = 1, . . . , N such that
(a) M =⋃Ni=1 Ui,
(b) for all I = {i1, . . . , ik}, then U+I = U+
i1∩ · · · ∩U+
ikand UI = Ui1 ∩ · · · ∩Uik
either are both empty sets or form a c-pair.
Sets we are interested in are the sets U+i , Ui, their intersections U+
I , UI , andtheir unions
V + = U+I1∪ · · · ∪ U+
Ir, V = UI1 ∪ · · · ∪ UIr .
Math 231a Notes 68
In this case, Hn(V +, V + \ V ) ∼= Hn(M,M \ V ), which is true by excision,contains a fundamental class µV . We have cap product maps
Hn(V +, V + \ V )×Hk(V +, V + \ V )→ Hn−k(V +),
Hn(V +, V + \ V )×Hk(V +)→ Hn−k(V +, V + \ V ).
Using µV , we get
D : Hk(V +, V + \ V )→ Hn−k(V +), D′ : Hk(V +)→ Hn−k(V +, V + \ V ).
Proposition 29.1. D and D′ are isomorphisms.
Proof. We prove by induction on r. For r = 1, we can check that thing workout fine because the pair is isomorphic to (B◦1 , B
◦1 \B1/2).
Now suppose that this is true for r. Let us now prove for r + 1. LetZ = U ∪V and Z = U+∪V + where V is the union of r balls. Then U ∩V = Wand U+ ∩ V + = W+ are unions of r balls. So by the induction hypothesis,DU , DV , DW and D′U , D
′V , D
′W are all isomorphisms. We want to show that DZ
and D′Z are isomorphisms. Let us temporarily write Hk(X|Z) = Hk(X,X \Z).We have two Mayer-Vietoris sequences. The first one is
Hk(M |U ∩ V ) Hk(M |U)⊕Hk(M |U) Hk(M |U ∪ V ) · · · .
Applying excision, we get
Hk(U+ ∩ V +|U ∩ V ) Hk(U+|U)⊕Hk(V +|V ) Hk(U+ ∪ V +|U ∪ V ) · · · .
There is also the ordinary Mayer-Vietoris sequence, and together we can write
Hk(W+|W ) Hk(U+|U)⊕Hk(V +|V ) Hk(Z+|Z) · · ·
Hn−k(W+) Hn−k(U+)⊕Hn−k(V +) Hn−k(Z+) · · ·
DW DU⊕DV DZ
After verifying that this is a commutative diagram, we conclude from the fivelemma that DZ is an isomorphism. We can also do this for D′.
In the case V = V + = M , we get the isomorphisms D : Hk(M) →Hn−k(M).
Math 231a Notes 69
30 November 11, 2016
Let (U+i , Ui) be a simple cover of M , an compact orientable manifold. Let
(V +, V ) =⋃
(U+I , UI). One can inductively show using induction, the Mayer-
Vietoris sequence, and five lemma that H∗(V ) → H∗(V+) and H∗(V +) →
H∗(V ) are isomorphisms.Last time we showed that if V is simple, then
Hk(V +, V + \ V )→ Hn−k(V +), Hk(V +)→ Hn−k(V +, V + \ V )
are isomorphisms. Using what we have remarked right before, we get isomor-phisms
Hk(M,M \ V )→ Hn−k(V ), Hk(V )→ Hn−k(M,M \ V )
by excision.
30.1 Alexander duality
We now want to compare H l(K) to Hn−l(M,M \K) for K ⊆M compact. Weknow this when K is a union of finitely many balls. But K might be unpleasant,and to deal with this, we look at the direct limit. Fix a compact set K, andlook at all the neighborhoods of K. For K ⊆ V2 ⊆ V1, consider the maps
H l(V1)→ H l(V2), Hk(M,M \ V1)→ Hk(M,M \ V2).
Then we might start looking at the direct limits of these systems.Formally, let K ⊆ Rn be a compact set. Consider (V +, V ) a simple pair,
with V a neighborhood of K. We have
H l(V ) ∼= H l(V +) ∼= Hn−l(Rn,Rn \ V ) ∼= Hn−l−1(R \ V ).
Is it true that H l(K) ∼= Hn−l−1(Rn \ K)? The idea is to consider A ={all nbds of U of K}. We have
lim−→U⊇K
Hn−l−1(Rn \ U) = Hn−l−1(Rn \K)
becasue the union of Rn \ U is Rn \K and every compact set in Rn \K lies inRn \ U for some U . Furthermore, we have
lim−→simple V⊃K
Hn−l−1(Rn \K) = lim−→U⊃K
Hn−l−1(Rn \ V )
because for each U there exists a simple (V +, V ) with V ⊆ U .On the other side of the duality isomorphism, we also similarly have
lim−→U⊇K
H l(U) = limsimple V +⊃K
H l(V +).
Math 231a Notes 70
However, it may not be true that
lim−→U⊇K
H l(U) ∼= H l(K).
Certainly, this is an isomorphism if there exists a neighborhood U0 of K thatdeformations retracts to K.
Definition 30.1. A compact set K ⊆ Rn is called taut if
lim−→U⊃K
H l(U) ∼= H l(K)
for every l.
So for taut K, we have an isomorphism
H l(K) ∼= lim−→U
H l(U) ∼= lim−→V +
H l(V +) ∼= lim−→V
Hn−l−1(R \ V ) ∼= Hn−l−1(R \K).
If K is compact and locally contractible, then K is taut. Using this, if K isthe image of any Sn−1 → R. Thus
Z ∼= Hn−1(Sn−1) ∼= H0(R \ Sn−1).
Therefore Rn \ ψ(Sn−1) has two path components, and thus two connectedcomponents.
Math 231a Notes 71
31 November 14, 2016
Last time, we showed that if K ⊆ Rn is compact, then then K being taut impliesthat
Hk(K) ∼= lim−→U
H l(U) ∼= Hn−l−1(Rn \K).
If K is homeomorphic to Sn−1, then we see that Rn \ K has two connectedcomponents. If n = 3 and K ∼= S1, e.g., K is a knot, then
Z ∼= H1(K) ∼= H1(R3 \K).
Even if K is a “wild” knot, this identity holds.
31.1 Traditional Alexander duality
The traditional version of Alexander duality is given by
H l(K) ∼= Hn−l−1(Sn \K)
by adding a point at infinity. Again, this holds when K is taut and nonempty.They are almost equivalent except for ` = 0, where H l(K) = H l(K) ⊕ Z andHn−1(Rn \K) ∼= Hn−1(Sn \K)⊕ Z.
Lemma 31.1. Hn−1(R \K) = Hn−1(Sn \K) ⊕ Z and the Hms are the samefor m 6= n− 1.
Proof. Let x ∈ K and choose R > 0 so that K ⊆ BnR = BnR(x). Then Hn−1(Rn\K) = Hn−1(BnR \K). Note that Sn−1
r = ∂BnR is a retract of BnR \K, given byprojecting to the boundary by the ray going out from x.
Sn−1R BnR \K
i
r
We have a long exact sequence of pairs
Hn−1(Sn−1R ) Hn−1(BnR \K) Hn−1(BnR \K,S
n−1R ) 0.
i∗
r∗
This splits, and so we get
Hn−1(Rn \K) = Hn−1(Bn \K) = Hn−1(Bn \K,Sn−1)⊕ Z
= Hn−1
(Bn \KSn−1
)⊕ Z = Hn−1(Sn \K)⊕ Z.
Sometimes, when you have a notion of homology, and K is compact, youdefine H l(K) as simply
H l(K) = Hn−k−1(Sn \K)
after embedding K in Sn. This look like nuts, but actually happens in algebraictopology.
Math 231a Notes 72
31.2 Cohomology with compact support
Let us now return to the “other” duality:
H l(M,M \ V ) ∼= Hn−l(V+) ∼= Hn−l(V )
for compact sets V arising in simple pairs. Let A be all compact ses K ⊆ Mdirected by inclusions K1 < K2 ⇔ K1 ⊆ K2. (This is interestint only if M isnon-compact.) These V ’s from simple pairs are cofinal in A. Thus
lim−→V
Hn−l(V ) = limK∈A
Hn−l(K) = Hn−l(M).
On the other hand of the duality, we have
lim−→V
H l(M,M \ V ) = lim−→K∈A
H l(M,M \K) 6= H l(M)
in general. For instance, when M = Rn, we have
lim−→K
H l(Rn,Rn \K) = lim−→N→∞
H l(Rn,Rn \BnN ) =
{Z l = n
0 otherwise.
Definition 31.2. The cohomology with compact support is defined as
H lc(X) = lim−→
K compact
H l(X,X \K).
Why are these called cohomology with compact support? The elementsof H l
c(X) are equivalence classes of elements [a] ∈ H l(X,X \ K), where a ∈Cl(X,X \K) = Ann(Cl(X \K)). If you know de Rham cohomology, then theseare cohomology rising from differential forms with compact support.
Anyways, getting back to the duality thing, we get
Theorem 31.3 (Poincare duality). If M is n-dimensional and oriented, then
D : H lc(M)→ Hn−l(M)
is an isomorphism.
For example, for M = Rn, as we have seen,
l.h.s. =
{Z l = n
0 l 6= n,r.h.s. =
{Z n− l = 0
0 n− l 6= 0.
Another example is when M is a compact manifold with boundary. If M isa smooth manifold, then M has a collar, i.e., M ⊃ C ∼= ∂M × [0, 1). LetMε = (M \ C) ∪ (∂M × [ε, 1)) and M◦ = M \ ∂M =
⋃εMε. Then
H lc(M
◦) = lim−→ε→0
H l(M◦,M◦ Mε) = H l(M,∂M).
If M is oriented, then H lc(M
◦) = Hn−l(M◦) = Hn−l(M)4. Therefore we get
Theorem 31.4 (Poincare duality for manifolds with boundary). If M is anoriented manifold with boundary, then
H l(M,∂M) = Hn−l(M) and H l(M) = Hn−l(M,∂M).
Math 231a Notes 73
32 November 16, 2016
Let us now talk about homotopy. Consider triples of spaces (A,B,C) and definemorphisms of triples f : (A,B,C) → (A′, B′, C ′) as a map f : A → A′ withf(B) ⊆ B′ and etc. A homotopy is a map
F : (A,B,C)× I → (A′, B′, C ′),
i.e., ft : (A,B,C)→ (A′, B′, C ′).
32.1 Homotopy groups
Let us write (X,x0) for (X, {x0}). Denote
[(X,A), (Y,B)] = set of homotopy classes of maps.
Often, when the base point is implicit, we are going to write
[X,Y ] = based homotopy classes = [(X,x0), (Y, y0)].
For instance, we define the fundamental group as
π1(X,x0) = [(S1, s), (X,x0)] = [(I, {0, 1}), (X,x0)].
We know want to define higher dimensional analogues of this. We define thenth homotopy group
πn(X,x0) = [(Sn, s0), (X,x0)] = [(In, ∂In), (X,x0)].
The two definitions are equivalent because In/∂In ∼= Sn.For a based pair (X,A, x0) with x0 ∈ A, we now want to define its homotopy
group. The first idea is to consider the homotopy class of maps
(In, ∂In, τ)→ (X,A, x0)
where τ ∈ ∂In is any point. But instead, we can write ∂nI = In−1 ∪ Jn−1 where
In−1 = {(t1, . . . , tn) ∈ In : tn = 0}, Jn−1 = {tn = 1} ∪ (∂In−1 × I).
Now let us define
πn(X,A, x0) = [(In, ∂In, Jn−1), (X,A, x0)].
Write, for f a map, [f ] its homotopy class. Note that there is a map
πn(X,A, x0)→ πn−1(A, x0); [f ] 7→ [f |∂In−1 ].
Let’s look at some special cases. For n = 1, I0 = {0} and J0 = {1}. Soπ1(X,A, x0) is the homotopy classes f : I → X with f(0) ∈ A and f(1) = x0.
Math 231a Notes 74
If X = R2 then A ⊆ X has n path-components, then there are n homotopyclasses.
It is not obvious what π0(X,A, x0) should be, because we don’t know whatJ−1 is. But we do know what π0(X,x0) is. It is
π0(X,x0) = [(I0, ∂I0), (X,x0)] = [(pt, ∅), (X,x0)] = [pt, X]
which is the path components of X.In general πn(X,x0) is a group if n ≥ 1 and πn(X,A, x0) is also a group if
n ≥ 2. Given fi : (In, ∂In, Jn−1)→ (X,A, x0) for i = 1, 2, we define [f1]+[f2] =[f1 + f2], where
(f1 + f2)(t1, . . . , tn) =
{f1(2t1, t2, . . . , tn) t1 ≤ 1/2
f2(2t1 − 1, t2, . . . , tn) t1 ≥ 1/2.
This map is continuous because the values are all x0 where the two functionsf1 and f2 meet. Note that when n = 1, this doesn’t work.
f1 + f2 = f1 f2
Figure 2: Addition law for πn(X,A, x0)
In the non-relative case, this definition works as πn(X,x0) = πn(X,x0, x0)for n ≥ 2. For n = 1, you can define by concatenating paths.
Math 231a Notes 75
33 November 18, 2016
33.1 Some properties of homotopy groups
Last time we have given a group structure on πn(X,A, x0) for n ≥ 2, orA = {x0}and n = 1. The identity element z : In → X is given by z(In) = {x0} and theinverse can be given by
(−f) : (In, ∂In, Jn−1)→ (X,A, x0); (t1, . . . , tn) = f(−t1, t2, . . . , tn).
Then [f ] + [z] = [f ], [f ] + [−f ] = [z], and associativity can be easily checked.Also note that πn(X,A, x0) is abelian if n ≥ 3. Given f : (In, ∂In, J
n−1)→(X,A, x0), define
fr(t1, t2, . . . , tn) = f(1− t1, 1− t2, t3, . . . , tn).
Note that this map preserves ∂In and Jn−1. Also if n ≥ 3, then f and fr
are homeomorphic because In ∼= D2 × In−2 and then you can rotate. Because(f1 + f2)r = fr2 + fr1 , we get [f1] + [f2] = [f1 + f2] = [f2] + [f1]. Note that πn isa functor from triples (X,A, x0) to groups (n ≥ 2) or to pointed sets (n = 1).This because h : (X,A, x0) → (Y,B, x0) gives a h∗ : πn → πn by taking thecomposition [f ] 7→ [h ◦ f ]. You can check that the resulting map is a grouphomomorphism, or sends the preferred point to the preferred point.
Generally, the inclusion A ↪→ X gives i : (A, x0) → (X,x0), i.e., i :(A, x0, x0)→ (X,x0, x0). There is also the map j : (X,x0, x0)→ (X,A, x0). Sowe have so far
πn(A, x0) πn(X,x0) πn(X,A, x0).i∗ j∗
While ago, we pointed out that there is a natural way to get πn(X,A, x0) →πn−1(A, x0). So we can carry on.
πn(A, x0) πn(X,x0) πn(X,A, x0)
πn−1(A, x0) πn−1(X,x0) πn−1(X,A, x0)
...
π0(A, x0) π0(X,x0)
i∗ j∗
i∗ j∗
i∗
On this sequence, “exactness” makes sense because each set has a preferredelement ei ∈ Si, and so we can understand ker as the inverse image of thepreferred element.
Definition 33.1. Let (X,A) be a pair. We say that (X,A) satisfies the ho-motopy extension property if given any homotopy G : I×A→ Y and givenany f0 : X → Y with f0|A = g0, there exists an extension F : I ×X → Y to allof I ×X with F |I×A = G and F |0×X = f0.
Lemma 33.2. (X,A) has the homotopy extension property if and only if Z =(I ×A) ∪ (0×X) ⊆ I ×X is a retract of I ×X.
Math 231a Notes 76
34 November 21, 2016
We were looking at the homotopy extension property. For a pair (X,A), thishas the homotopy extension property if and only if there is a retract I×X → Zwhere Z = I×A∪{0}×X. For example, (X,A) = (In, ∂In) has the homotopyextension property.
More generally, if X is obtained form A by attaching a cell, then (X,A)has the homotopy extension property. This can be proved using the retractionI× In → 0× In∪ I×∂In. Likewise, attaching many n-cells simultaneously alsogives a pair with homotopy extension property.
34.1 Exactness of the long exact sequence
Recall that we have a sequence
πn(A) πn(X) πn(X,A)
πn−1(A) πn−1(X) · · · .
∂ i∗ j∗
∂ i∗ j∗
Let us first show that ∂◦j∗ = 0. This is actually clear because the restrictionof any thing in πn(X,A) that comes from πn(X), to A is the constant map.Now suppose that [g] ∈ ker ∂. Then g : (In, ∂In, Jn−1) → (X,A, x0) andg|In−1×0 : (In−1, ∂In−1) → (A, x0) is homotopic to the constant map. Usingthe homotopy extension property on (In, ∂In), we can extend this homotopy toget a homotopy from g to a map that maps the boundary to x0. This showsthat ker ∂ ⊆ im j∗. The remaining parts can be proven similarly.
One important thing to note is that excision fails with homotopy groups.Here is an example. Let X = S1 ∨ S1 and let A = S1 with the base point x0
being the point where the two circles are glued. Then
π1(X,A) = (Z ∗ Z)/Z
whereas if we let Z = A \ (−ε, ε) then π1(X \ Z,A \ Z) ∼= π1(S1, x0) = Z.Because excision fails, we can’t really compute anything. So the tools we
have are:
• simplicial approximation
• smooth approximation
If we have simplicial complexes K = (V,W ), L = (W,T ) and a simplicial mapϕ : K → L, this gives a map |ϕ| : |K| → |L|. Given finite simplicial complexesand the continuous maps f : |K| → |L|, you can show that there exists asimplicial map ϕ : K [n] → L such that |ϕ| and f are homotopic as maps from|K [n]| = |K| to |L|.
Math 231a Notes 77
35 November 28, 2016
Today we are going to talk about smooth approximation. If P is a smoothmanifold, then Whitney’s theorem says that P embeds in RN smoothly, forsome N .
35.1 Smooth approximation
Proposition 35.1. Suppose—for simplicity—P is a compact manifold, and sois Q. Then any continuous map f : P → Q is homotopic to a smooth one.Furthermore, if f0, f1 are smooth and homotopic by a continuous homotopyF : I × P → Q, then there is also a smooth homotopy F .
So if we denote the set of homotopy classes of maps P → Q by [P,Q], thenin a nutshell, [P,Q] = [P,Q]smooth. In the proof, we are going to use the tubularneighborhood theorem, that states that if Q ⊆ RN is an embedding of a compactsmooth manifold, then there exists a ε0 > 0 such that if 0 < ε < ε0 then
Uε = {x ∈ RN : d(x,Q) ≤ ε}
deformation retracts to Q.
Proof of proposition 35.1. We note that f can be approximated to a polynomialf such that for all x ∈ P ⊆ RN1 , we have |f(x)− f(x)| ≤ ε. So f : P → Uε andf ' f maps to Uε. Then we can use the deformation retraction r : Uε → Q sothat r ◦ f : P → Q gives r ◦ f ' f .
We can use the same trick for the second part, on F : I × P → Q.
There are variants of this proposition. For instance, we can look at basedmaps (P, p0)→ (Q, q0) and require that f is smooth on P \{p0}. More generally,we can show that
[(P,A), (Q,B)] = [(P,A), (Q,B)]smooth on P\A.
Definition 35.2. For a smooth map f : P → Q, a point q ∈ Q is a regularvalue if for all p ∈ f−1(q), dfp : TpP → TqQ is onto.
35.2 Computing homotopy groups
Theorem 35.3 (Sard). Regular values are dense in Q. If f is proper, thenregular values are open and dense in Q.
For instance, if dimP < dimQ, then q regular means f−1(q) = ∅. ThenSard says that f is not onto if dimP < dimQ.
Corollary 35.4. If m < n, then [Sm, Sn] = 0 (i.e., every f is null-homotopic).Ditto for based maps: πm(Sn) = 0.
Math 231a Notes 78
Proof. We have [Sm, Sn] = [Sm, Sn]smooth. Also, a smooth map f : Sm → Sn
is not onto. So there exists a q ∈ Sn that is not in the image of f . So with anabuse of notation, we can write f : Sm → Sn \ q ∼= Rn and Rn is contractible.Thus f is homotopic to a constant map.
What if m ≥ n? What can we say about πm(Sn)? We can think Sm =Rm ∪ ∞ where ∞ is the base-point. Then we are looking at f : Rm → Sn
(assume smooth), with f(x)→∞ as x→∞.Here is how we are going to approach this problem. Regular values are open
and dense in Rn, and so without loss of generality we may assume that 0 isa regular value (else translate the function). Then M = f−1(0) is a smoothmanifold in Rm and its dimension is dimM = m − n. (It is also possible thatM = ∅.) So far, given f we tweaked a little bit and got a compact submanifoldM ⊆ Rm with dim = m− n. But we want extra data on M .
Definition 35.5. For x ∈ M , denote by TxM the tangent space to M andvxM the normal vectors to M at x. Then vxM = (TxM)⊥ in Rm, or vxM =Rm/TxM . A normal framing of M ⊆ Rm is a collection of n vector fieldsalong M with M 3 x 7→ vi(x) ∈ vxM such that (v1, . . . , vn) are a basis for vxMfor all x.
Our manifold M comes with a normal framing v = (v1, . . . , vn) as follows.Let e1, . . . , en ∈ T0Rn be the standard basis in Rn. The derivative dfx mapsvxM → T0Rn isomorphically. Then let vi(x) = (dfx)−1ei.
So the idea is that classifying based maps Sm → Sn (for m ≥ n) up tohomotopy is equivalent to classifying compact normally framed manifolds (M, v)with M ⊆ Rm up to something.
Consider a smooth homotopy F : I × Sm → Sn where f0, f1 for t = 0, 1are based. We then get a normally framed submanifold N ⊆ I × Rm. Thiscoincides at t = 0 and t = 1 with (M0, v0) and (M1, v1). This is called framedcobordism. Indeed, this “up to something” is “up to framed cobordism.”
I am going to outline next time how the other direction works. That is,why a normally framed submanifold up to framed cobordism corresponds tohomotopy classes.
Math 231a Notes 79
36 November 30, 2016
I want to now start talking about the other direction. Suppose we have amanifold M ⊆ Rm with dimM = m−n with a normal framing v = (v1, . . . , vn).From this can we get an element of πm(Sn)?
For a small enough ε > 0, consider the map
τ : M ×Bn → Rm; (x, (y1, . . . , yn)) 7→ x+ ε
n∑i=1
yivi.
The image is a tubular neighborhood U for ε ≤ ε1. Then we get a map
Rm ⊇ U M ×Bn Bn.τ−1
Using this, define a map
f : Rm ∪ {∞} → Bn/Sn−1 ∼= Sn ∼= Rn ∪ {∞};
f(z) =
{f1(z) if z ∈ interior(U)
basepoint if z /∈ interior(U).
You can see that these are inverse operations. So we need to classify framedmanifolds.
36.1 Classifying framed manifolds
Let us look at the case when n = 1. That is, let us look at normally framedm− 1 manifolds.
Consider the case m = 2. Because 1-manifolds are always circles, any ele-ment will be a disjoint union of circles, framed either to the inside or the outside.But any framed circle is cobordant to the empty set. This you can see thoughthe hemisphere that is pointing either outwards or inwards. This shows that
π2(S1) = 0.
In a very similar way, you can prove that
πm(S1) = 0
for m ≥ 2. The proof isn’t very much different.Let us go to the other extreme and look at framed 0-manifolds in Rm. A
compact 0-manifold is a finite collection of points, and a framing is a choice ofan orthonormal (without loss of generality) basis at each point. The points willfall into two categories according to the orientation of the framing. Moreover,if two points are to cancel out in the cobordism, they need to have differentorientations. Thus the number d = (#R−#L) ∈ Z is invariant under cobordism.So we obtain
πm(Sm) = Z.
Math 231a Notes 80
Now let us look at πm(Sm−1). This is looking at 1-dimensional manifolds.It turns out that we may just assume that these are circles that look like S1 ⊆R2 × 0 ⊆ R2 × Rm−2. Let v1 has before (the outward or inward) and letv2, . . . , vm−1 be the standard basis in Rm−2. Let us call this the “0-framing” anddenote it by v0. Then (S1, v0) becomes cobordant to ∅ and gives 0 ∈ πm(Sm−1).
But this gives a way to construct other framings. Given A : S1 → GL(m−1,R, we can define a new framing (vA1 , . . . , v
Am−1) with
vAj (x) =∑
Aij(x)v0i (x).
This is simply rotating the vectors around along the circle. So given any [A] ∈π1(GL(m−1,R)) = π1(SO(m−1)) we get a framed manifold (M, v). We haven’tverified enough stuff to conclude this, but indeed,
π3(S2) ∼= Z ∼= π1(SO(2)), π4(S3) ∼= Z/2 ∼= π1(SO(3)).
Math 231a Notes 81
37 December 2, 2016
37.1 Fiber bundles
Definition 37.1. A fiber bundle is a map p : E → B such that there is a spaceF—called the fiber—such that for all b ∈ B, the map p−1(b) is homeomorphicto F , and for all b ∈ B there is a neighborhood U 3 b with p−1(U) ∼= U × F sothat this commutes:
π−1(U) U × F
U U
ψ
p prU
Examples occur when a group acts freely on a space, e.g., a compact Liegroup acting on a manifold. For instance, recall that CPn can be thought as
CPn = {unit vectors in Cn+1}/(z ∼ λz with |λ| = 1).
This gives us a quotient map
p : S2n+1 → CPn; z 7→ [z]
with fiber p−1([z]) = {λz : λ ∈ S1} ∼= S1. Why is this locally trivial? ConsiderU = {[z] : z0 6= 0} ⊆ CPn. We want to check that p−1(U) ∼= U ×S1 ∼= Cn×S1.The map Cn × S1 → p−1(U) can explicitly be written down as
((z1, . . . , zn), λ) 7→ (λ, λz1, λz2, . . . , λzn)
(1 +∑|zi|2)1/2
.
Other examples include covering spaces with n sheets. Then F are n points.Also, there are examples where E is a group. The group SO(n) acts on
Sn−1 ⊆ Rn. This gives a map
p : SO(n)→ Sn−1; A 7→ Ae1.
The fiber will be SO(n − 1). Likewise, there is a map p : U(n) → S2n−1 withfiber U(n− 1). You can do this for the quaternions too.
Anyways, look at the map p : E → B. Pick a point b0 ∈ B and let F0 =p−1(b0). Pick a base point e0 ∈ F0. Then p : (E,F0, e0) → (B, b0, b0). So weget a map
p∗ : πk(E,F0, e0)→ πk(B, b0).
Proposition 37.2. This map p∗ is an isomorphism.
Recall that there is a long exact sequence
πk(F ) πk(E) πk(E,F ) πk−1(F ) πk−1(E) · · · .
Math 231a Notes 82
Now this proposition tells us that what we have can be also written as
πk(F ) πk(E) πk(B) πk−1(F ) πk−1(E) · · · .
The proof of the proposition will tell us also what the maps are.For example, look at a covering space B → B with F the discrete space. For
k ≥ 1, πk(F ) = 0. This implies that
πk(B) ∼= πk(B)
for all k ≥ 2.Let us look at another example. Take p : S2n+1 → CPn with fiber F = S1.
We have πk(S1) = 0 for k ≥ 2. So again, the long exact sequence gives
πk(CPn) = πk(S2n+1)
for k ≥ 3. If n = 1, then πk(S2) = πk(S3) for k ≥ 3. In particular, π3(S2) =π3(S3) = Z.
Because πk(S∞) = 0 for all k, we get πk(CP∞) = 0 for k ≥ 3. What isπ2(CP∞)? We can use the sequence.
π2(S∞) = 0 π2(CP∞) π2(S1) = Z π1(S∞) = 0
So π2(CP∞) = Z. So all homotopy groups of CP∞ except for π2(CP∞) = Z.The main point is that fiber bundles have the homotopy lifting property (not
to be confused with the homotopy extension property). This says that givenany paracompact space X and a homotopy {gt}, G : I ×X → B, and a lift ofg0, i.e., a f0 : X → E with p ◦ f0 = g0, then there exists a F : I × X → Eextending f0 such that p ◦ F = G.
Proof of proposition 37.2. We fisrt show that p∗ is onto. Given ψ in πk(B, b0),i.e., ψ : (Ik, ∂Ik) → (B, b0), we want to find a Ψ such that p ◦ Ψ = ψ. Now amap (Ik, J, Ik−1)→ X can be thought as a homotopy I × Ik−1 → X. Then thehomotopy lifting property gives that such a Ψ exists.
The injectivity can be thought similarly.
Index
abstract simplicial complex, 7Alexander duality, 69, 71
boundary, 8Brouwer fixed point theorem, 34
chain, 8chain homotopy, 22cochain complex, 45cochain map, 46cohomology, 46cohomology with compact support,
72contractible space, 24cup product, 53CW complex, 38cycle, 8
deformation retract, 33degree, 37direct limit, 41
Eilenberg-Steenrod axioms, 21exact sequence, 16
of chain complexes, 17excision, 31excision axiom, 26Ext, 51
fiber bundle, 81framed cobordism, 78fundamental class, 63fundamental group, 73
graded commutative ring, 60
homology, 13
simplicial homology, 9, 11homotopy, 22, 73homotopy extension property, 75homotopy group, 73homotopy-equivalence, 24
identification space, 36
Jordan curve theorem, 70
manifold, 62
normal framing, 78
orientation, 62local, 62
Poincare duality, 67, 72for manifolds with boundary,
72
reduced homology, 25regular value, 77relative cohomology, 52relative homology, 19retraction, 33
Sard’s theorem, 77simple cover, 67singular cohomology, 47strong deformation retract, 36
taut set, 70
unimodular paring, 66universal coefficients theorem, 47
wedge sum, 36
83