SECTION 1: HOMOTOPY AND THE FUNDAMENTAL GROUPOID

You probably know the fundamental group π1(X,x0) of a space X with a base point x0, definedas the group of homotopy classes of maps S1 → X sending a chosen base point on the circle to x0.In this course we will define and study higher homotopy groups πn(X,x0) by using maps S

n → Xfrom the n-dimensional sphere to X, and see what they tell us about the space X.

Let us begin by fixing some preliminary conventions. By space we will always mean a topologicalspace (later, we may narrow this down to various important classes of spaces, such as compact spaces,locally compact Hausdorff spaces, compactly generated weak Hausdorff spaces, or CW-complexes).A map between two such spaces will always mean a continuous map.

A pointed space is a pair (X,x0) consisting of a space X and a base point x0 ∈ X. A mapbetween pointed spaces f : (X,x0) → (Y, y0) is a map f : X → Y with f(x0) = y0. A pairof spaces is a pair (X,A) consisting of a space X and a subspace A ⊆ X. A map of pairsf : (X,A)→ (Y,B) is a map f : X → Y such that f(A) ⊆ B.

We now recall the central notion of a homotopy. Two maps of spaces f, g : X → Y are calledhomotopic if there is a continuous map H : X × [0, 1]→ Y such that:

H(x, 0) = f(x) and H(x, 1) = g(x), x ∈ X

Such a map H is called a homotopy from f to g. We write f ' g to denote such a situation orH : f ' g if we want to be more precise. The homotopy relation enjoys the following nice properties.

Proposition 1. Let X,Y, and Z be spaces.

(1) The homotopy relation is an equivalence relation on the set of all maps from X to Y .(2) If f, g : X → Y and k, l : Y → Z are homotopic then also kf, lg : X → Z are homotopic.

Proof. Let us prove the first claim. So, let us consider maps f, g, h : X → Y . The homotopy relationis reflexive since we have the following constant homotopy at f :

κf : f ' f given by κf (x, t) = f(x)

Given a homotopy H : f ' g then we obtain an inverse homotopy H−1 as follows:

H−1 : g ' f with H−1(x, t) = H(x, 1− t)

Thus, the homotopy relation is symmetric. Finally, if we have homotopies F : f ' g and G : g ' hthen we obtain a homotopy H : f ' h by the following formula:

H(x, t) =

{F (x, 2t) , 0 ≤ t ≤ 1/2G(x, 2t− 1) , 1/2 ≤ t ≤ 1

Thus we showed that the homotopy relation is an equivalence relation.To prove the second claim let us begin by two special cases. Let us assume that we have a

homotopy H : f ' g. Then we obtain a homotopy from kf to kg simply by post-compositionwith k:

kH : X × [0, 1] H→ Y k→ Z is a homotopy kH : kf ' kg1

2 SECTION 1: HOMOTOPY AND THE FUNDAMENTAL GROUPOID

The next case is slightly more tricky. Given a homotopy K : k ' l then we obtain a homotopyfrom kg to lg as follows:

K ◦ (g × id) : X × [0, 1]→ Y × [0, 1]→ Z defines a homotopy K ◦ (g × id) : kg ' lg

In order to obtain the general case it suffices now to use the transitivity of the homotopy relationsince from the above two special cases we deduce

kf ' kg ' lg

as intended. This concludes the proof. �

The equivalence classes with respect to the homotopy relation are called homotopy classesand will be denoted by [f ]. Given two spaces X and Y then the set of homotopy classes of mapsfrom X to Y is denoted by:

[X,Y ]

This proposition allows us to form a new category where the objects are given by spaces andwhere morphisms are given by homotopy classes of maps: to make this more explicit we begin byrecalling the notion of a category.

Definition 2. A category C consists of the following data:

(1) A collection ob(C) of objects in C.(2) Given two objects X,Y there is a set C(X,Y ) of morphisms (arrows, maps) from X

to Y in C.(3) Associated to three objects X,Y, Z there is a composition map:

◦ : C(Y,Z)× C(X,Y )→ C(X,Z) : (g, f) 7→ g ◦ f

These data have to satisfy the following two properties:

• The composition is associative, i.e., we have (h◦g)◦f = h◦(g◦f) whenever these expressionsmake sense.

• For every object X there is an identity morphism idX ∈ C(X,X) such that for allmorphisms f ∈ C(X,Y ) we have:

idY ◦ f = f = f ◦ idX

We use the following standard notation. Given a category C we write X ∈ C in order to saythat X is an object of C. If f ∈ C(X,Y ) is a morphism from X to Y we write f : X → Y . Moreover,the composition is often just denoted by juxtaposition; i.e., we write gf instead of g ◦ f .

You know already a lot of examples of categories.

Example 3. (1) The category of sets and maps of sets.(2) The category of groups and group homomorphisms.(3) Given a ring R we have the category of R-modules and R-linear maps.(4) The category of fields and field extensions.(5) The category of smooth manifolds and differentiable maps.

Using the conventions introduced above we also have the following examples.

Example 4. (1) The category Top of spaces and maps.(2) The category Top∗ of pointed spaces and maps of pointed spaces.(3) The category Top2 of pairs of spaces and maps of pairs.

SECTION 1: HOMOTOPY AND THE FUNDAMENTAL GROUPOID 3

Now, as a consequence of the above proposition we can form a new category with spaces as objectsand morphisms given by homotopy classes of maps. Two homotopy classes can be composed byforming the composition of representatives and then passing to the corresponding homotopy class.The above proposition guarantees that this is well-defined. It is easy to check that we get a categorythis way.

Corollary 5. Spaces and homotopy classes of maps define a category Ho(Top), the (naive) homo-topy category of spaces.

There are variants of this for the case of Top∗ and Top2. In these cases we are mainly interested

in slightly different variants of the homotopy relation. Two maps f, g : (X,x0) → (Y, y0) in Top∗are called homotopic relative base points, notation

f ' g rel x0,

if there exists a homotopy H : f ' g such that

H(x0, t) = y0, t ∈ [0, 1].

Thus, we are asking for the existence of a homotopy through pointed maps. Again, one checks thatthis is an equivalence relation which is compatible with composition. The equivalences classes [f ]are called pointed homotopy classes and the set of all such is denoted by [(X,x0), (Y, y0)].

These pointed variants are in fact special cases of the following more general notion. Given apair of spaces (X,A), a space Y , and two maps f, g : X → Y such that f(a) = g(a) for all a ∈ A.Then a relative homotopy from f to g is a homotopy H : f ' g such that H(a,−) : [0, 1]→ Y isconstant for all a ∈ A. Thus the additional condition imposed is

H(a, t) = f(a) = g(a), t ∈ [0, 1], a ∈ A.

If for two such maps f and g there is a relative homotopy H, then this will be denoted by

H : f ' g rel A.

This notion specializes to pointed homotopy or homotopy, if A consists of a single point or is emptyrespectively.

Finally, let us consider two maps f, g : (X,A)→ (Y,B) in Top2. A homotopy of pairs from fto g is a homotopy H : f ' g which, in addition, satisfies

H(a, t) ∈ B, t ∈ [0, 1], a ∈ A.

This is of course precisely the condition that for each t ∈ [0, 1] the map H(−, t) : X → Y is actuallya map of pairs (X,A)→ (Y,B). If we want to be precise then we denote such a homotopy by

H : f ' g : (X,A)→ (Y,B).

Also in this case we obtain a well-behaved equivalence relation and the equivalence classes aredenoted as before. We will write [(X,A), (Y,B)] for the set of homotopy classes of pairs. As aconsequence of this discussion we obtain the following result.

Corollary 6. (1) Pointed spaces and pointed homotopy classes define a category Ho(Top∗), thehomotopy category of pointed spaces.

(2) Pairs of spaces and homotopy classes of pairs define a category Ho(Top2), the homotopycategory of pairs of spaces.

4 SECTION 1: HOMOTOPY AND THE FUNDAMENTAL GROUPOID

Definition 7. A map f : X → Y of spaces is a homotopy equivalence if there is a map g : Y → Xsuch that

g ◦ f ' idX and f ◦ g ' idY .

The space X is homotopy equivalent to Y if there is a homotopy equivalence f : X → Y .

It is easy to see that being homotopy equivalent is an equivalence relation. The equivalence classof a space X with respect to this relation is called the homotopy type of X.

Definition 8. A morphism f : X → Y in a category C is an isomorphism if there is a morphismg : Y → X such that

g ◦ f = idX and f ◦ g = idY .

An object X is isomorphic to Y if there is an isomorphism X → Y .

Example 9. (1) A morphism in the category of sets is an isomorphism if and only if it isbijective.

(2) In many categories the objects are given by ‘sets with additional structure’ while morphismsare defined as morphisms of sets ‘respecting this additional structure’. Frequently, it is truethat a morphism in such a category is an isomorphism if and only if the underlying map ofsets is a bijection. This is for example the case for groups, rings, fields, and modules.

(3) In the category Top one has to be careful; a continuous bijection f : X → Y is, in general, notan isomorphism, i.e., a homeomorphism. For this to be true we have to impose additionalconditions on the spaces (like compact and Hausdorff).

(4) A morphism [f ] : X → Y in Ho(Top) is an isomorphism if and only if any map f : X → Yrepresenting this homotopy class is a homotopy equivalence.

Exercise 10. (1) Define a notion of pointed homotopy equivalence (without using theconcept of an isomorphism) and check that the pointed analog of Example 9.(4) holds.

(2) Define a notion of homotopy equivalence of pairs (again, without using the concept ofan isomorphism) and check that the corresponding analog of Example 9.(4) holds.

(3) The notion of being isomorphic is an equivalence relation on the collection of objects in anarbitrary category. The corresponding equivalence classes are called isomorphism classes.

Recall that given a space X and an element x0 ∈ X we have the fundamental group π1(X,x0) ofhomotopy classes of loops at x0. If we take a different point x1 ∈ X we obtain a further such groupπ1(X,x1) which a priori has nothing to do with π1(X,x0). However, if x0 and x1 lie in the same pathcomponent of X then any path joining them induces an isomorphism between the two homotopygroups by conjugation with the given path. It is easy to check that paths homotopic relative tothe boundary induce the same isomorphism. However, in general, the induced isomorphisms maybe different. A convenient way of encoding all these different groups and isomorphisms in a singlestructure is given by the fundamental groupoid of a space. In order to discuss this we have tointroduce one more notion from category theory.

Definition 11. A category C is a groupoid if every morphism in C is an isomorphism.

This terminology reflects the idea that a groupoid is like a group in a certain sense. In fact, forevery object in a groupoid the set of endomorphisms is actually a group. The justification for thisterminology is given by the first of the following examples.

SECTION 1: HOMOTOPY AND THE FUNDAMENTAL GROUPOID 5

Example 12. (1) Every group G gives rise to a groupoid BG as follows. The category BGhas a single object denoted by ∗. Hence, the only set of morphisms we have to specify isthe set of endomorphisms of ∗ and this set is given by:

BG(∗, ∗) = GThe composition is given by the multiplication of the group. It is easy to check that allthe axioms of a groupoid are precisely fulfilled because G is a group. In other words, agroup is essentially the same thing as a groupoid with one object. Similarly, a monoid Mis essentially the same thing as a category with a single object.

(2) Every category has an underlying groupoid given by the same objects and the isomorphismsonly. For example, we have the category of sets and bijections, spaces and homeomorphisms,smooth manifolds and diffeomorphisms, and so on.

We now give a description of the fundamental groupoid π(X) of a space X. The collection ofobjects obj(π(X)) is given by the points of X. Given two points x, y ∈ X the set of morphisms

π(X)(x, y)

is given by the set of homotopy classes of paths from x to y relative to the boundary. To becompletely specific, let us recall that a path α in X from x to y is given by a map

α : [0, 1]→ X such that α(0) = x and α(1) = y.We want to consider two such paths α and β to be equivalent if they are homotopic relative to theboundary, i.e., if there is a map H : [0, 1]× [0, 1]→ X such that

H(−, 0) = α, H(−, 1) = β, H(0, t) = x, and H(1, t) = y for all 0 ≤ t ≤ 1.Now, given a path α from x to y and a path β from y to z then the concatenation β ∗ α is givenby: (

β ∗ α)(t) =

{α(2t) , 0 ≤ t ≤ 1/2β(2t− 1) , 1/2 ≤ t ≤ 1

Exercise 13. (1) The concatenation β ∗α is again continuous and defines a path from x to z.(2) Given three points x, y, z ∈ X then the assignment ([β], [α]) 7→ [β ∗α] defines a well-defined

composition map

◦ : π(X)(y, z)× π(X)(x, y)→ π(X)(x, z).(3) The composition is associative.(4) Prove that the homotopy classes of constant paths give identity morphisms. Thus we

already know that π(X) defines a category.(5) Given a path α from x to y then we the inverse path α−1 from y to x is defined by the

formula

α−1(t) = α(1− t).Show that every morphism [α] in π(X) is an isomorphism by verifying that [α−1] defines atwo-sided inverse of [α].

This exercise shows that π(X) is a groupoid, the fundamental groupoid of X. Let (X,x0)be a pointed space, then the fundamental group π1(X,x0) of X at x0 is defined as the group ofautomorphisms of x0 in π(X), i.e.,

π1(X,x0) = π(X)(x0, x0).

6 SECTION 1: HOMOTOPY AND THE FUNDAMENTAL GROUPOID

Let us introduce the following notation for the unit interval, its boundary, and the 1-sphere:

I = [0, 1], ∂I = {0, 1}, and S1 = {x ∈ R2 | ‖x‖ = 1}Then, the fundamental group is given by

π1(X,x0) = [(I, ∂I), (X,x0)] ∼= [(S1, ∗), (X,x0)]where the latter isomorphism comes from the homeomorphism I/∂I ∼= S1.

Example 14. We assume you have learned about the fundamental group in your undergraduatetopology, and know examples like

π1(S1, ∗) ∼= Z and π1(T, ∗) ∼= Z× Z,

whereT = S1 × S1

is the torus. The latter formula is of course a special case of the following product formula. Given(X,x0), (Y, y0) ∈ Top∗ then the product (X × Y, (x0, y0)) is again a pointed space and the naturalmap

π1(X × Y, (x0, y0))∼=→ π1(X,x0)× π1(Y, y0)

is an isomorphism.

SECTION 2: THE COMPACT-OPEN TOPOLOGY AND LOOP SPACES

In this section we will give the important constructions of loop spaces and reduced suspensionsassociated to pointed spaces. For this purpose there will be a short digression on spaces of mapsbetween (pointed) spaces and the relevant topologies.

To be a bit more specific, one aim is to see that given a pointed space (X,x0), then there is anentire pointed space of loops in X. In order to obtain such a loop space

Ω(X,x0) ∈ Top∗,we have to specify an underlying set, choose a base point, and construct a topology on it. Theunderlying set of Ω(X,x0) is just given by the set of maps

Top∗((S1, ∗), (X,x0)).

A base point is also easily found by considering the constant loop κx0 at x0 defined by:

κx0 : (S1, ∗)→ (X,x0) : t 7→ x0

The topology which we will consider on this set is a special case of the so-called compact-opentopology. We begin by introducing this topology in a more general context.

Let K be a compact Hausdorff space, and let X be an arbitrary space. The set Top(K,X) ofcontinuous maps K → X carries a natural topology, called the compact-open topology. It hasa subbasis formed by the sets of the form

B(T,U) = {f : K → X | f(T ) ⊆ U}where T ⊆ K is compact and U ⊆ X is open. Thus, for a map f : K → X, one can form a typicalbasis open neighborhood by choosing compact subsets T1, . . . , Tn ⊆ K and small open sets Ui ⊆ Xwith f(Ti) ⊆ Ui to get a neighborhood Of of f ,

Of = B(T1, U1) ∩ . . . ∩B(Tn, Un).One can even choose the Ti to cover K, so as to ‘control’ the behavior of functions g ∈ Of on allof K.

The topological space given by this compact-open topology on Top(K,X) will be denoted by:

XK ∈ Top

Proposition 1. Let K be a compact Hausdorff space. Then for any X,Y ∈ Top, there is a bijectivecorrespondence between maps

Yf→ XK and Y ×K g→ X.

Proof. Ignoring continuity for the moment, there is an obvious bijective correspondence betweensuch functions f and g, given by

f(y)(k) = g(y, k)

for all y ∈ Y and k ∈ K. We thus have to show that if f and g correspond to each other in thissense, then f is continuous if and only if g is.

In one direction, suppose g is continuous, and choose an arbitrary subbasic open B(T,U) ⊆ XK .To prove that f−1(B(T,U)) is open, choose y ∈ f−1(B(T,U)), so g({y × T}) ⊆ U . Since T is

1

2 SECTION 2: THE COMPACT-OPEN TOPOLOGY AND LOOP SPACES

compact and g is continuous, there are open V 3 y and W ⊇ T with g(V ×W ) ⊆ U (you shouldcheck this for yourself!). Then surely V is a neighborhood of y with f(V ) ⊆ B(T,U), showing thatf−1(B(T,U)) is open.

Conversely, suppose f is continuous, and take U open in X. To prove that g−1(U) is open,choose (y, k) ∈ g−1(U), i.e., y ∈ f−1(B({k}, U)). Now, in the space XK , if a function K → Xmaps k into U , then it must map a neighborhood Wk of k into U , and if we choose Wk small enoughit will even map the compact set T = W̄k into U . This shows that:

B({k}, U) =⋃{B(T,U) | T is a compact neighborhood of k}

So y ∈ f−1(B(T,U)) for some T = W̄k. By continuity of f , we find a neighborhood V of y withf(V ) ⊆ B(T,U), i.e., g(V × T ) ⊆ U ; and hence surely g(V ×Wk) ⊆ U , showing that g−1(U) isopen. �

As a special case we can consider the compact space K = S1, for which maps out of K are loops.

Definition 2. Let X ∈ Top and let (Y, y0) ∈ Top∗.(1) The space Λ(X) = XS

1 ∈ Top is the free loop space of X.(2) The loop space Ω(Y, y0) ∈ Top∗ of (Y, y0) is the pair consisting of the subspace of Λ(Y )

given by the pointed loops (S1, ∗) → (Y, y0) together with the constant loop κy0 at y0 asbase point.

The above proposition applied to the compact space K = S1 tells us that there is a bijective

correspondence between maps g : X × S1 → Y and maps f : X → Y S1 = Λ(Y ). Let now (X,x0)and (Y, y0) be pointed spaces. We want to make explicit the conditions to be imposed on a mapg : X × S1 → Y such that the corresponding map f actually defines a pointed map:

f : (X,x0)→ Ω(Y, y0) = (Ω(Y, y0), κy0)

Since the correspondence is given by the formula

g(x, t) = f(x)(t)

it is easy to check that these conditions are:

g(x0, t) = y0, t ∈ S1, and g(x, ∗) = y0, x ∈ X

Thus the map g has to send the subspace {x0} × S1 ∪X × {∗} ⊆ X × S1 to the base point y0 ∈ Yand hence factors over the corresponding quotient.

Definition 3. Let (X,x0) be a pointed space. Then the reduced suspension Σ(X,x0) ∈ Top∗ isthe pointed space

Σ(X,x0) = X × S1/({x0} × S1 ∪X × {∗})where the base point is given by the collapsed subspace.

Using the quotient map I → I/∂I ∼= S1, a different description of the reduced suspension Σ(X,x0)is given by

Σ(X,x0) = X × I/({x0} × I ∪X × ∂I)and in either description we will denote the base point by ∗. The quotient map I → I/∂I ∼= S1induces maps of pairs(

X × I, {x0} × I ∪X × ∂I)→(X × S1, {x0} × S1 ∪X × {∗}

)→(Σ(X,x0), ∗

).

SECTION 2: THE COMPACT-OPEN TOPOLOGY AND LOOP SPACES 3

At the level of elements we allow us to commit a minor abuse of notation and simply write

(x, t) 7→ (x, t) 7→ [x, t], x ∈ X, t ∈ I.

Thus, we have the following descriptions of the base point ∗ ∈ Σ(X,x0)

[x0, t] = [x, 0] = [x, 1] = ∗, x ∈ X, t ∈ I,

and similarly if the second factor is an element of S1. Proposition 1 combined with the discussionpreceding the definition of the reduced suspension gives us the following corollary.

Corollary 4. Let (X,x0), (Y, y0) ∈ Top∗. Then there is a bijective correspondence between pointedmaps

g : Σ(X,x0)→ (Y, y0) and f : (X,x0)→ Ω(Y, y0)given by the formula g([x, t]) = f(x)(t) for all x ∈ X and t ∈ S1.

This corollary turns out to be the special case of a pointed analog of Proposition 1. In order tounderstand this, we have to introduce a few constructions of pointed spaces. A pointed analog ofspaces of maps is easily obtained (compare to the difference between the loop space and the freeloop space!).

Definition 5. Let (K, k0), (X,x0) ∈ Top∗ and assume that K is compact. The pointed mappingspace

(X,x0)(K,k0) ⊂ XK

is the subspace of pointed maps (K, k0)→ (X,x0). It has a natural base point given by the constantmap κx0 with value x0, and hence defines an object

(X,x0)(K,k0) ∈ Top∗.

From now on we begin to be a bit sloppy about the notation of base points. If we do not needa special notation for a base point of a pointed space we will simply drop it from notation. Forexample, we will write ‘Let X be a pointed space’, the suspension Σ(X,x0) will be denoted by Σ(X),and similarly. Also the pointed mapping space of the above definition will sometimes simply bedenoted by XK . Moreover, we will sometimes generically denote base points by ∗. Whenever thissimplified notation results in a risk of ambiguity we will stick to the more precise one.

As a next step, let us consider a pair of spaces (X,A). Then we can form the quotient space X/Aby dividing out the equivalence relation ∼A generated by

a ∼A a′, a, a′ ∈ A.

The quotient space X/A is naturally a pointed space with base point given by the equivalence classof any a ∈ A. In the sequel this will always be the way in which we consider a quotient space as apointed space. Note that this was already done in the above definition of the (reduced) suspensionof a pointed space.

Definition 6. Let (X,x0) and (Y, y0) be two pointed spaces, then their wedge X∨Y is the pointedspace

X ∨ Y = X t Y/{x0, y0} ∈ Top∗.

The wedge X ∨ Y comes naturally with pointed maps iX : X → X ∨ Y and iY : Y → X ∨ Y .In fact, this is the universal example of two such maps with a common target in the sense of thefollowing exercise.

4 SECTION 2: THE COMPACT-OPEN TOPOLOGY AND LOOP SPACES

Exercise 7. Let X,Y, and W be pointed spaces and let f : X →W, g : X →W be pointed maps.Then there is a unique pointed map (f, g) : X ∨ Y →W such that:

(f, g) ◦ iX = f : X →W and (f, g) ◦ iY = g : Y →W

Thus, from a more categorical perspective the wedge is the categorical coproduct in the categoryof pointed spaces.

Example 8. (1) The quotient space I/{0, 1/2, 1} is homeomorphic to S1 ∨ S1.(2) For any pointed space X we have X ∨ ∗ ∼= X ∼= ∗ ∨X.(3) For two pointed spaces X and Y we have X ∨ Y ∼= Y ∨X.

The wedge X ∨ Y of two pointed spaces is naturally a subspace of X × Y . In fact, this inclusioncan be obtained by applying the above exercise as follows. For pointed spaces (!), the product(X × Y, (x0, y0)) comes naturally with an inclusion map of (X,x0) given by

(X,x0)→ (X × Y, (x0, y0)) : x 7→ (x, y0).

There is a similar map (Y, y0)→ (X × Y, (x0, y0)). Thus we are in the situation of Exercise 7 andhence obtain a pointed map X ∨ Y → X ∧ Y . This map can be checked to be the inclusion of asubspace. The corresponding quotient construction is so important that it deserves a special name.

Definition 9. Let X and Y be pointed spaces. Then the smash product X ∧ Y of X and Y isthe pointed space

X ∧ Y = X × Y/X ∨ Y ∈ Top∗.

As it is the case for every quotient space, the smash product X ∧ Y naturally comes with aquotient map

q : X × Y → X ∧ Y.We will use the following notation for points in X ∧ Y :

[x, y] = q(x, y), x ∈ X, y ∈ Y

In the next example, we will use the 0-dimensional sphere or 0-sphere S0 which is thetwo-point space:

S0 = {−1,+1} ⊆ [−1,+1]Let us agree that we consider S0 as a pointed space with −1 as base point. Moreover, let I+ be thedisjoint union of I = [0, 1] with a base point, i.e.,

I+ = [0, 1] t ∗ ∈ Top∗with ∗ as base point.

Example 10. (1) For every X ∈ Top∗ we have X ∧ S1 ∼= Σ(X).(2) For every X ∈ Top∗ we have a homeomorphism X ∧ S0 ∼= X ∼= S0 ∧X.(3) For two pointed spaces X and Y we have X ∧ Y ∼= Y ∧X.(4) Let X ∈ Top∗. The reduced cylinder of X is the smash product X ∧ I+. Unraveling the

definition of the smash product, we see that we have

X ∧ I+ ∼= X × I/{x0} × I ∈ Top∗.

Thus, pointed maps out of X ∧ I+ are precisely the pointed homotopies.

SECTION 2: THE COMPACT-OPEN TOPOLOGY AND LOOP SPACES 5

Proposition 11. Let K,X, and Y be pointed spaces and assume that K is compact, Hausdorff.Then there is a bijective correspondence between pointed maps

g : X ∧K → Y and f : X → Y K

given by the formula g([x, k]) = f(x)(k) for all x ∈ X and k ∈ K.

Using the cylinder construction on spaces one can also establish the following result. A proofwill be given in the exercises.

Corollary 12. Let K,X, and Y be pointed spaces and assume that K is compact, Hausdorff. Thenthere is a bijection:

[X ∧K,Y ] ∼= [X,Y K ]

Exercise 13. (1) Give a proof of Proposition 11 using the corresponding result about (‘un-pointed’) spaces. Hint: compare the proof of the special case of K = S1 and realize thatthe smash product is designed so that this proposition becomes true.

(2) Give a proof of Corollary 12. There are some hints on how to attack this on the exercisesheet.

We will now recall the notion of path components which allows us to establish a relation betweenloop spaces and the fundamental group.

Let X be a space and let x0, x1 ∈ X. We say that x0 and x1 are equivalent, notation x0 ' x1,if and only if there there is a path in X connecting them. It is easy to check that this defines anequivalence relation on X with equivalence classes the path components of X. We write π0(X)for the set of path components of X and denote the path component of x ∈ X by [x]. A point in Xcan be identified with a map x : ∗ → X sending the unique point ∗ to x. Under this identification,the above equivalence relation becomes the homotopy relation on maps ∗ → X. Thus, we have abijection:

π0(X) ∼= [∗, X]In the case of a pointed space the set of path components has a naturally distinguished element

given by the path component of the base point. Thus, the set π0(X,x0) of path components of apointed space (X,x0) is a pointed set. Now, a point x ∈ (X,x0) can be identified with a pointedmap S0 → X. In fact, a bijection is obtained by evaluating such a map on 1 ∈ S0 (which is not thebase-point!). Moreover, it is easy to see that we have an isomorphism of pointed sets

π0(X,x0) ∼= [(S0,−1), (X,x0)].

Exercise 14. Verify the details of the above discussion.

As a consequence of our work so far we obtain the following corollary.

Corollary 15. Let (X,x0) be a pointed space. Then there is a canonical bijection of pointed sets:

π1(X,x0) ∼= π0(Ω(X,x0))

Proof. It suffices to assemble our results from above. Since the sphere S1 is a compact, Hausdorffspace we can apply Corollary 12 in order to obtain:

π0(Ω(X,x0)) ∼= [(S0,−1),Ω(X,x0)]∼= [(S0,−1) ∧ (S1, ∗), (X,x0)]∼= [(S1, ∗), (X,x0)]∼= π1(X,x0)

6 SECTION 2: THE COMPACT-OPEN TOPOLOGY AND LOOP SPACES

The third identification is a special case of Example 10.(2). �

Thus, at least for theoretical purposes, in order to calculate the fundamental group of a givenspace it is enough to calculate the path components of the associated loop space. However, thisdoes not really simplify the task since the loop space is, in general, less tractable than the originalspace. This corollary also suggests a definition of higher homotopy groups. Namely, given a pointedspace (X,x0) we could simply set

πn(X,x0) = π0((X,x0)(Sn,∗)), n ≥ 2.

This will be pursued further in the next section where we will see that we actually obtain abeliangroups this way.

The final aim of this section is to introduce the notion of a functor and to remark that many ofthe constructions introduced so far are in fact functorial. Here is the key definition.

Definition 16. Let C and D be categories. A functor F : C→ D from C to D is given by:(1) An object function which assigns to each object X ∈ C an object F (X) ∈ D.(2) For each pair of objects X,Y ∈ C a morphism function C(X,Y )→ D(F (X), F (Y )).

This data is compatible with the composition and the identity morphisms in the sense that:

• For morphisms X f→ Y g→ Z in C we have F (g ◦ f) = F (g) ◦ F (f) : F (X)→ F (Z) in D.• For every object X ∈ C we have F (idX) = idF (X) : F (X)→ F (X) in D.

In order to give some examples of functors, we introduce notation for some prominent categories.

Notation 17. (1) The category of sets and maps of sets will be denoted by Set, the one ofpointed sets and maps preserving the chosen elements by Set∗.

(2) We will write Grp for the category of groups and group homomorphisms.(3) The category of abelian groups and group homomorphisms is denoted by Ab.(4) Given a ring R, we write R-Mod for the category of (left) R-modules and R-linear maps.

You know already many examples of functorial constructions. Let us only give a few examples.

Example 18. (1) The formation of free abelian groups generated by a set defines a functorSet→ Ab. More generally, given a ring then there is a free R-module functor Set→ R-Mod.

(2) Given a group G then we obtain the abelianization of G by dividing out the subgroupgenerated by the commutators aba−1b−1, a, b ∈ G. This quotient is an abelian group andgives us a functor (−)ab : Grp→ Ab : G 7→ Gab.

(3) There are many functors which forget structures or properties. For example there is thefollowing chain of forgetful functors where R is an arbitrary ring:

R-Mod→ Ab→ Grp→ Set∗ → Set

We first forget the action by scalars r ∈ R and only keep the abelian group. We then forgetthe fact that our group is abelian. Next, we drop the group structure and only keep theneutral element. Finally, we also forget the base point.

In this course, many of the constructions on spaces or unpointed spaces turn out to be instancesof suitable functors. Let us mention some of the constructions which were already implicit in thiscourse. More examples will be considered in the exercises.

Example 19. (1) The fundamental group construction defines a functor π1 : Top∗ → Grp.

SECTION 2: THE COMPACT-OPEN TOPOLOGY AND LOOP SPACES 7

(2) The formation of path components defines functors

π0 : Top→ Set and π0 : Top∗ → Set∗.

(3) The first two examples are special cases of the following more general construction. Let Kbe a space, then we can consider the assignment

[K,−] : Top→ Set : X 7→ [K,X].

Given a map f : X → Y of spaces then we obtain an induced map [K,X] → [K,Y ] bysending a homotopy class [g] : K → X to [f ] ◦ [g] : K → X → Y . It is easy to check thatthis defines a functor Top→ Set.

Similarly, if (K, k0) is a pointed space, then the assignment (X,x0) 7→ [(K, k0), (X,x0)]is functorial. Note that the set [(K, k0), (X,x0)] has a natural base point given by thehomotopy class of the constant map κx0 : (K, k0)→ (X,x0). Given a pointed map

f : (X,x0)→ (Y, y0)

then the induced map [(K, k0), (X,x0)]→ [(K, k0), (Y, y0)] : [g] 7→ [f ]◦ [g] preserves the basepoint. Thus, we obtain a functor

[(K, k0),−] : Top∗ → Set∗.

The first two examples are obtained by considering the special cases of K = ∗ ∈ Top andK = S0, S1 ∈ Top∗ respectively.

(4) The construction of the reduced suspension is functorial and similarly for the loop space.Thus, we have two functors:

Σ: Top∗ → Top∗ and Ω: Top∗ → Top∗Let us give some details about the functoriality of Σ (the case of Ω will be treated in theexercises). By definition, Σ(X,x0) is the following quotient space:

X × S1/({x0} × S1 ∪X × {∗})

If we have a pointed map f : (X,x0)→ (Y, y0), we can form the product with the identityto obtain a map

f × idS1 : X × S1 → Y × S1

In order to obtain a well-defined map Σ(f) : Σ(X,x0) → Σ(Y, y0) we want to apply theuniversal property of the construction of quotient spaces. Thus, it suffices to check that:

(f × idS1)({x0} × S1 ∪X × {∗}

)⊆ {y0} × S1 ∪ Y × {∗}

But this is true since f is a pointed map. Thus, we deduce the existence of a unique mapΣ(f) : Σ(X,x0)→ Σ(Y, y0) such that the following square commutes:

X × S1 //

��

Y × S1

��

Σ(X,x0) ∃!//___ Σ(Y, y0)

We leave it as an exercise to check that the uniqueness implies that we indeed get a functorΣ: Top∗ → Top∗. For example, you might want to consider the following diagram to prove

8 SECTION 2: THE COMPACT-OPEN TOPOLOGY AND LOOP SPACES

the compatibility with respect to compositions:

X × S1 //

��

Y × S1

��

// Z × S1

��

Σ(X,x0) ∃!//___ Σ(Y, y0) ∃!

//___ Σ(Z, z0)

(5) By adding disjoint base points to spaces we obtain a functor (−)+ : Top→ Top∗. There isalso a forgetful functor Top∗ → Top in the opposite direction which forgets the base points.

SECTION 3: HIGHER HOMOTOPY GROUPS

In this section we will introduce the main objects of study of this course, the homotopy groups

πn(X,x0)

of a pointed space (X,x0), for each natural number n ≥ 2. (Recall that the (pointed) set ofcomponents π0(X,x0) and the fundamental group π1(X,x0) have already been defined.) One goalof this course is to develop some techniques which will allow us to calculate these homotopy groupsin interesting examples.

We begin by introducing some notation for important spaces. Let us denote by

In = [0, 1]× . . .× [0, 1] ⊆ Rn

the n-cube and ∂In ⊆ In for its boundary. Thus,

∂In = {(t1, . . . , tn) ∈ In | at least one of the ti ∈ {0, 1}}.

Let us agree on the convention that ∂I0 = ∅ is empty. Note that the boundary satisfies (and iscompletely determined by ∂I and) the Leibniz formula

∂(In × Im) = (∂In)× Im ∪ In × (∂Im).

The n-sphere is denoted by

Sn = {x ∈ Rn+1 | ‖x‖ = 1}.Note that there are homeomorphisms In/∂In ∼= Sn; we will write [t1, . . . , tn] ∈ Sn for the image of(t1, . . . , tn) ∈ In under the composition In → In/∂In ∼= Sn.

The definition of the underlying (pointed) set of πn(X,x0) is simple enough:

πn(X,x0) = [(In, ∂In), (X,x0)]

Thus, an element [α] of πn(X,x0) is represented by a map α : In → X sending the entire bound-

ary ∂In to the base point x0; and two such α and α′ represent the same element of πn(X,x0) if

and only if there is a homotopy H : In × I → X such that

H(∂In × I) = x0, H(−, 0) = α, and H(−, 1) = α′.

Obviously, a map f : (X,x0)→ (Y, y0) induces a function

f∗ : πn(X,x0)→ πn(Y, y0)

which in fact only depends on the homotopy class of f . Just like for the fundamental group, wehave

Proposition 1. (1) For each pointed space (X,x0) and n ≥ 1, the set πn(X,x0) is a group,the n-th homotopy group of (X,x0).

(2) For each map (X,x0) → (Y, y0), the induced operation πn(X,x0) → πn(Y, y0) is a grouphomomorphism, defining a functor πn : Top∗ → Grp. The functor πn is homotopy invariant,i.e., πn(f) = πn(g) for homotopic maps f ' g.

1

2 SECTION 3: HIGHER HOMOTOPY GROUPS

Proof. For two elements [α] and [β] in πn(X,x0), the product [β] ◦ [α] is represented by the mapβ ∗ α : In → X defined by:

(β ∗ α)(t1, . . . , tn) ={α(2t1, t2, . . . , tn) , 0 ≤ t1 ≤ 1/2β(2t1 − 1, t2, . . . , tn) , 1/2 ≤ t1 ≤ 1

Notice that the definition agrees with the known group structure on the fundamental group for n =1. The proof that ◦ is well-defined and associative, that the constant map κx0 : In → X representsa neutral element, and that each element [α] has an inverse represented by

α−1(t1, . . . , tn) = α(1− t1, t2, . . . , tn)is exactly the same as for π1, and we leave the details as an exercise. Also the functoriality is anexercise. �

Exercise 2. Carry out the details of the proof.

Warning 3. Let X be a space and let x0, x1 ∈ X. In general, πn(X,x0) and πn(X,x1) can bevery different. In fact, homotopy groups only ‘see the path-component of the base point’. Moreprecisely, let (X,x0) be a pointed space and let X

′ = [x0] be the path-component of x0. Then theinclusion i : (X ′, x0) → (X,x0) induces an isomorphism i∗ : πn(X ′, x0) → πn(X,x0) for all n ≥ 1.This follows immediately from the fact that Sn is path-connected for n ≥ 1. We will see later thatany path between two points x0, x1 ∈ X induces an isomorphism πn(X,x0) ∼= πn(X,x1).

Remark 4. (1) Of course it is not only the validity of the proposition which is important,but also the explicit description of the product. However, it can be shown that this groupstructure is unique for n ≥ 2.

(2) We said that the proof of the group structure is analogous to the argument for π1. In fact,there is a more formal way to see this, as we will see below.

One may object that the definition of the group structure is a bit unnatural, because the firstcoordinate t1 is given a preferred rôle in the definition of the group structure. We could also definea product as follows:

(β ∗i α)(t1, . . . , tn) ={α(t1, . . . , 2ti, . . . , tn) , 0 ≤ ti ≤ 1/2β(t1, . . . , 2ti − 1, . . . , tn) , 1/2 ≤ ti ≤ 1

The explanation is that these two products induce the same operation on homotopy classes. Theproof of this fact is given by the following observation (Lemma 5) together with the so-calledEckmann-Hilton argument (Proposition 6).

Lemma 5. The operation ∗ distributes over the operation ∗i in the sense that(α ∗i β) ∗ (γ ∗i δ) = (α ∗ γ) ∗i (β ∗ δ)

for all maps α, β, γ, δ : (In, ∂In)→ (X,x0).

Proof. We only have to look at the case n = 2, i = 2. Then the expressions on the left and rightcorrespond to the same subdivisions of the square so define identical maps (draw the picture!). �

Proposition 6. (‘Eckmann-Hilton trick’)Let S be a set with two associative operations •, ◦ : S × S → S having a common unit e ∈ S.Suppose • and ◦ distribute over each other, in the sense that

(α • β) ◦ (γ • δ) = (α ◦ γ) • (β ◦ δ)Then • and ◦ coincide, and define a commutative operation on S.

SECTION 3: HIGHER HOMOTOPY GROUPS 3

Proof. Taking β = e = γ in the distributive law yields α ◦ δ = α • δ, while taking α = e = δyields β ◦ γ = γ • β. �

Applying this proposition to ∗ and ∗i shows that these define the same operation on πn(X,x0)for n ≥ 2. The proposition also shows:

Corollary 7. The groups πn(X,x0) are abelian for n ≥ 2.

Remark 8. For this reason, one often employs additive notation for the group structure on πn(X,x0),writing:

[β] + [α] = [β ∗ α]0 = [κx0 ]−[α] = [α−1]

There is yet another way of describing πn(X,x0).

Proposition 9. (1) There is a bijection of sets natural in the pointed space (X,x0):

[(Sn, ∗), (X,x0)]∼=→ πn(X,x0)

(2) The group structure on [(Sn, ∗), (X,x0)] induced by this bijection coincides with the oneobtained by composition with the ‘pinch map’

∇ : Sn → Sn ∨ Sn

defined by collapsing the equator in Sn to a single point.

Proof. (1) This follows immediately from the isomorphism

(In/∂In, ∗)→ (Sn, ∗).

(2) Recall from the exercises that the wedge ∨ defines a coproduct in the category of pointed spaces,so that two maps α, β : (Sn, ∗)→ (X,x0) together uniquely define a map

α ∨ β : (Sn ∨ Sn, ∗)→ (X,x0).

Thus we get an induced operation on [(Sn, ∗), (X,x0)] defined by

β ∗ α = (α ∨ β) ◦ ∇

It is easy to check that this corresponds to the operation ∗ on maps from (In, ∂In), once one takesthe equator in Sn to be the image of {t1 = 1/2} ⊆ In under the map In → In/∂In ∼= Sn. �

Example 10. For the one-point space ∗ ∈ Top∗ there is precisely one pointed map Sn → ∗ foreach n ≥ 0. Thus we have

π0(∗) ∼= ∗, π1(∗) ∼= 1, and πn(∗) ∼= 0, n ≥ 2.

We will refer to this by saying that πn(∗) is trivial for all n ≥ 0.

This example together with the homotopy invariance immediately gives us the following.

Corollary 11. Let f : (X,x0)→ (Y, y0) be a pointed homotopy-equivalence. Then the induced map

f∗ : πn(X,x0)→ πn(Y, y0)

is an isomorphism. In particular, πn(X,x0) is trivial for all choices of base points in a contractiblespace X and all n ≥ 0.

4 SECTION 3: HIGHER HOMOTOPY GROUPS

Proof. Let g : (Y, y0)→ (X,x0) be an inverse pointed homotopy equivalence so that we have:

g ◦ f ' idX rel x0 and f ◦ g ' idY rel y0Homotopy invariance gives g∗ ◦ f∗ = (g ◦ f)∗ = id, and similarly f∗ ◦ g∗ = id. For the second claim,given a contractible space X and x0 ∈ X, it suffices to consider the pointed homotopy equivalence(X,x0)→ ∗. �

We will later see that these groups πn(X,x0) are non-trivial and highly informative, but we needto develop (or know) a little more theory before we can make this precise. However, assuming a bitof background knowledge, we observe the following.

Example 12. (Preview of examples)

(1) The identity map id : Sn → Sn defines an element of πn(X,x0). If you know somethingabout degrees, you know that the constant map has degree zero while the identity hasdegree 1. It is fact that the degrees of homotopic maps coincide, we conclude that 0 6= [id]in πn(S

n, ∗). In fact, we will show that there is an isomorphism πn(Sn, ∗) ∼= Z and that[id] is a generator. This could be proved, e.g., using singular homology but we will obtainthis calculation as a consequence of the homotopy excision theorem.

(2) If you know a bit of differential topology then you know that any map Sk → Sn is homotopicto a smooth map, and that a smooth map f : Sk → Sn cannot be surjective if k < n. Sosuch a map f factors as a composition

Sk → Sn − {x} ∼= Rn → Sn

for some point x ∈ Sn not in the image of f . The contractibility of Rn implies that f ishomotopic to a constant map. Thus,

πk(Sn, ∗) ∼= 0, k < n.

We will later deduce this result from the cellular approximation theorem.(3) Consider the scalar multiplication on the complex vector space of dimension 2,

µ : C× C2 → C2.

When we restrict to the complex numbers, respectively vectors of norm 1, we obtain a map

µ : S1 × S3 → S3,

an action of the circle-group on the 3-sphere. It can be shown that the orbit space of thisaction is S2. The quotient map S3 → S2 is the famous Hopf fibration, and defines anon-zero element in π3(S

2, ∗).

Let us now examine loop spaces in some more detail. Recall the construction of the loop spaceΩ(X,x0) associated to a pointed space (X,x0), and the isomorphism

π1(X,x0) ∼= π0(Ω(X,x0)).

The group structure on π1(X,x0) comes from a ‘group structure up to homotopy’ on Ω(X,x0).Explicitly, writing

H = Ω(X,x0)

and e = κx0 ∈ H for the constant loop at x0, there is a multiplication map on H given by theconcatenation:

µ : H ×H → H : (β, α) 7→ β ∗ α

SECTION 3: HIGHER HOMOTOPY GROUPS 5

This multiplication is associative up to homotopy in the sense that the following two maps arehomotopic:

H ×H ×Hid×µ

//

µ×id��

H ×H

µ

��

(γ, β, α) � // (γ, β ∗ α)_

��

(γ, β, α)_

��

H ×H µ // H γ ∗ (β ∗ α) (γ ∗ β, α)� // (γ ∗ β) ∗ α

Moreover, this multiplication is unital up to homotopy, i.e., we have homotopies from id to µ◦(e×id)and µ ◦ (id× e):

H

id##F

FFFF

FFFFe×id

// H ×Hµ

��

Hid×e

oo

id{{xx

xxxx

xxx

α � // (e, α)_

��

(α, e)_

��

α�oo

H e ∗ α α ∗ e

Indeed, these homotopies are given by the usual reparametrization homotopies. A pointed space(H, e) with such an additional structure is called an (associative) H-space.

Given any such H-space (H, e), composition with µ defines an associative multiplication on theset of homotopy classes of maps

[(Y, y0), (H, e)]

for an arbitrary pointed space (Y, y0). Moreover, this ‘multiplicative structure’ is natural in (Y, y0).(If you do not know what we mean by this naturality, then see the exercise sheet.)

Moreover, the associative multiplication defines a group structure on this set if H has a homo-topy inverse, i.e., if there is a map i : H → H such that the following diagram commutes up tohomotopies:

H

e""E

EEEE

EEEEi×id

// H ×H

µ

��

Hid×i

oo

e||yy

yyyy

yyy

α � // (i(α), α)_

��

(α, i(α))_

��

α�oo

H i(α) ∗ α α ∗ i(α)

In this case (H, e) is called an H-group. Thus, Ω(X,x0) is an H-group, and for each pointedspace (Y, y0) the set [(Y, y0),Ω(X,x0)] carries a natural group structure. For the one-point space∗ ∈ Top∗, this defines the usual group structure on:

[∗,Ω(X,x0)] ∼= π0(Ω(X,x0)) ∼= π1(X,x0)

Exercise 13. Define the notion of a commutative H-space and a commutative H-group. Isthe loop space of a pointed space with the concatenation pairing a commutative H-group?

Theorem 14. For every n ≥ 1, there is a natural isomorphism of groups:

πn(X,x0) ∼= πn−1(Ω(X,x0))

Let us begin with a lemma. Recall our notation [x, y] ∈ X ∧ Y for points in a smash product.Moreover, let us use a similar notation for elements in a quotient space, i.e., we will write [x] ∈ X/A.For convenience, we will drop base points from notation in the next lemma and we use In/∂In asour model for the n-sphere.

6 SECTION 3: HIGHER HOMOTOPY GROUPS

Lemma 15. The following map is a pointed homeomorphism:

Sn ∧ Sm → Sn+m :[[t1, . . . , tn], [t

′1, . . . , t

′m]

]7→ [t1, . . . , tn, t′1, . . . , t′m]

Proof. One checks directly that this is a well-defined continuous bijection between compact Haus-dorff spaces and hence a homeomorphism. �

This map can be described as:

Sn ∧ Sm = (In/∂In) ∧ (Im/∂Im)= (In/∂In × Im/∂Im)/(∗ × Im/∂Im ∪ In/∂In × ∗)∼= In × Im/(∂In × Im ∪ In × ∂Im)∼= In+m/∂In+m

= Sn+m

Proof. (of Theorem 11). The multiplication on πn−1(Ω(X,x0)) is the ‘loop multiplication’. Wealready know from an earlier lecture that there is a natural isomorphism:

[(Sn−1, ∗),Ω(X,x0)] ∼= [(Sn−1 ∧ S1, ∗), (X,x0)]Using the homeomorphism of the above lemma, it is immediate that the pairing on [(Sn, ∗), (X,x0)]induced by the loop multiplication is ∗n, the ‘concatenation with respect to the last coordinate’.But we already know that this is identical to the group structure on πn(X,x0). �

If we look at the theorem for n ≥ 2, we see that πn−1(Ω(X,x0)) has two group structures: one isthe group structure on πn−1 for any pointed space, and the other is an instance of the group structureon [(Y, y0), (H, e)] for any H-group H, in this case for Y = S

n−1 and H = Ω(X,x0). Moreover, thesegroup structures distribute over each other. Indeed, the multiplication µ : H ×H → H induces agroup homomorphism

µ∗ : πn−1(H, e)× πn−1(H, e)→ πn−1(H, e)and this precisely means that the multiplication coming from the H-group distributes over the onecoming from πn−1. So, by Eckmann-Hilton (Proposition 6), the two multiplications coincide andare commutative.

Corollary 16. The fundamental group π1(H, e) of an H-group (H, e) is abelian.

We now come to a different model for the loop space. We have seen that Ω(X,x0) has amultiplication which is associative and unital ‘up to homotopy’. One may wonder whether there isa way to make this multiplication strictly associative and unital. For a general H-space this neednot be possible. But in this special case there is an easy way to do this. Let M(X,x0) be thespace of Moore loops (named after J. C. Moore). Its points are pairs (t, α) where t ∈ R, t ≥ 0,and α : [0, t] → X is a loop at x0 of length t (i.e., α(0) = x0 = α(t)). We can topologize this setas a subspace of R ×X [0,∞), identifying a path α : [0, t] → X with the map [0,∞) → X which isconstant on [t,∞), and the resulting space is the Moore loop space. Then there is a continuous andstrictly associative multiplication on M(X,x0), given by

(t, β) · (s, α) = (t+ s, β ∗M α)where:

(β ∗M α)(r) ={α(r) , 0 ≤ r ≤ sβ(r − s) , s ≤ r ≤ t+ s

A strict unit for this multiplication is (0, κx0).

SECTION 3: HIGHER HOMOTOPY GROUPS 7

The space M(X,x0) is homotopy equivalent to Ω(X,x0). Indeed there are maps

Ω(X,x0)ψ

//M(X,x0),

φoo

ψ is simply the inclusion, while φ is defined by

φ(t, α)(r) = α(t · r), 0 ≤ r ≤ 1.Then obviously φ ◦ ψ is the identity, while

Hs(t, α)(r) = α(((1− s)t+ s)r

)defines a homotopy from H0 = ψ ◦ φ to the identity H1.

Perspective 17. We just observed that the loop space and the Moore loop space are homotopyequivalent spaces. Note that the respective H-group structures correspond to each other under thesehomotopy equivalences. However the multiplications have different formal properties: the Mooreloop space is strictly associative while the loop space is only associative up to homotopy. Thuswe see that a space X homotopy equivalent to a space with a strictly associative multiplicationdoes not necessarily inherit the same structure. But it is easy to see that X can be turned into anH-space that way. To put it as a slogan:

‘strictly associative multiplications do not live in homotopy theory’

As we already mentioned not all H-spaces can be rectified in the sense that they would be homotopyequivalent to spaces with a strictly associative multiplication. One might wonder what additionalstructure would be needed for this to become true. There is an answer to this question lying beyondthe scope of these lectures. Nevertheless, these questions and the more general search for homotopyinvariant algebraic structures initiated the development of a good deal of mathematics.

Let us formalize the notion of a homotopy invariant functor. Let C be an arbitrary category.Then a functor F : Top∗ → C is homotopy invariant if pointed maps which are homotopic relativeto the base point always have the same image under F :

f ' g implies F (f) = F (g)Now, note that there is a canonical functor

γ : Top∗ → Ho(Top∗)which is the identity on objects and which sends a pointed map to its pointed homotopy class.

Exercise 18. (1) The above assignments, in fact, define a functor γ : Top∗ → Ho(Top∗) andthis functor is homotopy invariant.

(2) Let C be a category. A functor F : Top∗ → C is homotopy invariant if and only if there is afunctor F ′ : Ho(Top∗) → C such that F = F ′ ◦ γ : Top∗ → C. In this case the functor F ′ isunique.

(3) Redo a similar reasoning for the categories Top and Top2.

Thus a homotopy invariant functor ‘is the same thing’ as a functor defined on the homotopy categoryof (pointed or pairs of) spaces. In particular, we have:

π0 : Ho(Top∗)→ Set∗, π1 : Ho(Top∗)→ Grp, and πn : Ho(Top∗)→ Ab

SECTION 4: RELATIVE HOMOTOPY GROUPS AND THE ACTION OF π1

In this section we will introduce relative homotopy groups of a (pointed) pair of spaces. Associatedto such a pair we obtain a long exact sequence in homotopy relating the absolute and the relativegroups. This and related long exact sequences are useful in calculations as we will see later.Moreover, we want to clarify the rôle played by the choice of base points. Expressed in a fancyway, we will show that the assignment x0 7→ πn(X,x0) defines a functor on the fundamentalgroupoid π(X) of X. This encodes, in particular, an action of the fundamental group on higherhomotopy groups.

To begin with let us consider a pointed space (X,x0) and a subspace A ⊆ X containing thebase point x0. Thus we have an inclusion of pointed spaces i : (A, x0) → (X,x0) and we referto (X,A, x0) as a pointed pair of spaces. The inclusion induces a map at the level of homotopygroups (or sets)

i∗ : πn(A, x0)→ πn(X,x0), n ≥ 0.

which, in general, is not injective. A homotopy class α ∈ πn(A, x0) lies in the kernel of i∗ if forany map f : (In, ∂In) → (A, x0) representing it the induced map i ◦ f : (In, ∂In) → (X,x0) ishomotopic to the constant map κx0 . Such a homotopy is a map H : I

n × I → X satisfying thefollowing relations:

H(−, 1) = f, H(−, 0) = κx0 , and H |∂In×I= κx0

Thus, if we denote by Jn the subspace of the boundary ∂In+1 = In × ∂I ∪ ∂In × I given by

Jn = In × {0} ∪ ∂In × I

then such a homotopy is a map of triples of spaces (in the obvious sense):

H : (In+1, ∂In+1, Jn)→ (X,A, x0)

There is also an adapted notion of homotopies of maps of triples which we want to introduce infull generality. Let X2 ⊆ X1 ⊆ X0 and Y2 ⊆ Y1 ⊆ Y0 be triples of spaces and let

f, g : (X0, X1, X2)→ (Y0, Y1, Y2)

be maps of triples. Then a homotopy H : f ' g is a map of triples

H : (X0, X1, X2)× I = (X0 × I,X1 × I,X2 × I)→ (Y0, Y1, Y2)

which satisfies H(−, 0) = f and H(−, 1) = g. Thus, we are asking for a homotopy H : X0× I → Y0which has the property that each map H(−, t) respects the subspace inclusions, i.e., is a map oftriples H(−, t) : (X0, X1, X2)→ (Y0, Y1, Y2). In the special case that X2 and Y2 are just base points,this gives us the notion of homotopies of maps of pointed pairs.

Exercise 1. This homotopy relation is an equivalence relation which is well-behaved with respectto maps of triples. Similarly, we get such a result for pointed pairs of spaces. There are homotopycategories of triples of spaces and pointed pairs of spaces.

1

2 SECTION 4: RELATIVE HOMOTOPY GROUPS AND THE ACTION OF π1

Maybe you should not carry out this exercise in detail but only play a bit with the notions inorder to convince yourself that they behave as expected.

Now, back to our pointed pair (X,A, x0). The above discussion motivates the following definition:

πn(X,A, x0) = [(In, ∂In, Jn−1), (X,A, x0)], n ≥ 1

(Note that in the case of A = {x0} we have πn(X,x0, x0) = πn(X,x0).) A priori, the πn(X,A, x0)are only pointed sets, the base point being given by the homotopy class of the constant map κx0 .However, it turns out that we get groups for n ≥ 2 which are abelian for n ≥ 3. To this end let usconsider maps

α, β : (In, ∂In, Jn−1)→ (X,A, x0), n ≥ 2Then we can define the concatenation β ∗ α : (In, ∂In, Jn−1)→ (X,A, x0) by the ‘usual formula’:

(β ∗ α)(t1, . . . , tn) ={α(2t1, t2, . . . , tn) , 0 ≤ t1 ≤ 1/2β(2t1 − 1, t2, . . . , tn) , 1/2 ≤ t1 ≤ 1

It follows immediately that β ∗α again is a map of triples. As in earlier lectures one checks that thisconcatenation is well-defined on homotopy classes and defines a group structure on πn(X,A, x0)with neutral element given by the homotopy class of the constant map.

Definition 2. Let (X,A, x0) be a pointed pair of spaces. Then the group

πn(X,A, x0) = [(In, ∂In, Jn−1), (X,A, x0)], n ≥ 2,

is the n-th relative homotopy group of (X,A, x0). The pointed set

π1(X,A, x0) = [(I1, ∂I1, 0), (X,A, x0)]

is the first relative homotopy set of (X,A, x0).

To avoid awkward notation we will simply write πn(X,A) instead of πn(X,A, x0) unless thereis a risk of ambiguity. Now, if n ≥ 3 one could again object that the above definition for theconcatenation is not very natural. In fact, one could also define pairings ∗i, where 1 ≤ i ≤ n − 1,given by the formula:

(β ∗i α)(t1, . . . , tn) ={α(t1, . . . , 2ti, . . . , tn) , 0 ≤ ti ≤ 1/2β(t1, . . . , 2ti − 1, . . . , tn) , 1/2 ≤ ti ≤ 1

(Note that there is no ∗n unless A = {x0} and this is why π1(X,A) is only a pointed set ingeneral.) Following the lines of the last lecture (‘Eckmann-Hilton trick’) one checks that thesedifferent pairings induce the same group structure and that πn(X,A) is abelian for n ≥ 3. If wedenote by Top2∗ the category of pointed pairs of spaces, then our discussion gives us the following:

Corollary 3. The assignments (X,A, x0) 7→ πn(X,A) can be extended to define functors:

π1 : Top2∗ → Set∗, π2 : Top

2∗ → Grp, and πn : Top

2∗ → Ab, n ≥ 3

Exercise 4. Convince yourself that (X,A, x0) 7→ π2(X,A) really defines a functor taking valuesin groups by drawing some diagrams. If you are ambitious, then do similarly in order to seethat π3(X,A) always is an abelian group.

A different way of proving this corollary is sketched in the exercises. There, you will showthat πn+1(X,A) is naturally isomorphic to the n-th homotopy group of a certain space P (X;x0, A).

SECTION 4: RELATIVE HOMOTOPY GROUPS AND THE ACTION OF π1 3

The motivation for this discussion was the observation that an inclusion i : (A, x0) → (X,x0)induces a morphism of homotopy groups which is not necessarily injective. The relative homo-topy groups are designed to measure the deviation from this. In fact, if j denotes the inclu-sion j : (X,x0)→ (X,A) then there is the following result.

Proposition 5. Given a pointed pair of spaces (X,A, x0), there are connecting homomorphisms∂ : πn(X,A)→ πn−1(A, x0), n ≥ 1, such that the following sequence is exact:

. . .→ πn+1(X,A)∂→ πn(A, x0)

i∗→ πn(X,x0)j∗→ πn(X,A)

∂→ . . . ∂→ π0(A, x0)i∗→ π0(X,x0)

This is the long exact homotopy sequence of the pointed pair (X,A, x0). Moreover, this sequenceis natural in the pointed pair.

Before we attack the proof let us be a bit more precise about the statement. Recall that a diagram

of groups and group homomorphisms G1f→ G2

g→ G3 is exact at G2 if we have the equalityim(f) = ker(g) of subgroups of G2. In particular, the composition g ◦ f sends everything to theneutral element of G3, but we also have a converse inclusion. Namely, if x2 ∈ G2 lies in ker(g),then it already comes from G1, i.e., there is an element x1 ∈ G1 such that f(x1) = x2.

More generally, a diagram of groups and group homomorphisms

G1 → G2 → . . .→ Gn−1 → Gn

is exact if it is exact at Gi for all 2 ≤ i ≤ n− 1. A special case is a short exact sequence whichis an exact diagram of the form:

1→ G1 → G2 → G3 → 1

Example 6. Let G and H be groups.

(1) A homomorphism G→ H is injective if and only if 1→ G→ H is exact.(2) A homomorphism G→ H is surjective if and only if G→ H → 1 is exact.(3) A homomorphism G→ H is an isomorphism if and only if 1→ G→ H → 1 is exact.(4) A group G is trivial if and only if 1→ G→ 1 is exact.

In particular, a short exact sequence basically encodes a surjective homomorphism G2 → G3 to-gether with the inclusion of the kernel N = G1 ⊆ G2.

Now, in the diagram we consider in the above proposition not all maps are homomorphisms ofgroups. In fact, the last three entries π1(X,A), π0(A, x0), and π0(X,x0) are only pointed sets. Thenotion of exactness is extended to the context of maps of pointed sets by defining the kernel of sucha map to be the preimage of the base point.

Finally, let us make precise the meaning of the naturality in the above proposition. If we havea map of pointed pairs f : (X,A, x0) → (Y,B, y0), then we have a connecting homomorphism foreach of the pointed pairs. The naturality means that the following square commutes:

πn+1(X,A)∂ //

f∗

��

πn(A, x0)

f∗

��

πn+1(Y,B)∂

// πn(B, y0)

4 SECTION 4: RELATIVE HOMOTOPY GROUPS AND THE ACTION OF π1

It is easy to check that from this we actually get a commutative ladder of the form:

. . . // π1(X,x0) //

��

π1(X,A) //

��

π0(A, x0) //

��

π0(X,x0)

��

. . . // π1(Y, y0) // π1(Y,B) // π0(B, y0) // π0(Y, y0)

For the purpose of the following lemma let us introduce some notation. Recall that Jn is obtainedfrom In+1 by removing the ‘interior of the cube and the interior of the top face’. From a differentperspective Jn is obtained from In = In × {0} by gluing a further copy of In on each face of ∂In.Now, if F is a collection of faces of In, then let JnF ⊆ Jn be obtained from In by gluing onlythose copies of In which correspond to faces in F . More formally, the cube In ⊆ Rn has 2n faces.These can be parametrized by the set {1, . . . , n} × {0, 1} in a way that the first component of sucha pair (j, ij) tells us which coordinate is constant while the second coordinate is the value of that

coordinate. Thus the face In−1f ⊆ In corresponding to an index f = (j, ij) is given by:

In−1f = {(t1, . . . , tn) ∈ In | tj = ij}

With this notation the space JnF ⊆ Jn ⊆ ∂In+1 associated to a set F of faces is given by:

JnF = In × {0} ∪ (

⋃f∈F

In−1f × I)

Lemma 7. (1) The map i : Jn−1 → In is the inclusion of a strong deformation retract,i.e., there is a map r : In → Jn−1 which satisfies r ◦ i = idJn−1 and i ◦ r ' idIn (rel Jn−1).

(2) Given a set F of faces of In−1 then Jn−1F ⊆ In is the inclusion of a strong deformationretract.

Proof. We will only give the proof of the first claim, the second one is an exercise. If we considerthe space In ⊆ Rn as the unit cube of length one, then let s be the point s = (1/2, . . . , 1/2, 2)sitting ‘above the center of the cube’. For each point x ∈ In let l(x) be the unique line in Rnpassing through s and x. This line l(x) intersects Jn−1 in a unique point which we take as thedefinition of r(x). It is easy to see that the resulting map r : In → Jn−1 is continuous and that wehave r ◦ i = id. The homotopy i ◦ r ' id (rel Jn−1) is obtained by ‘collapsing the line segmentsbetween x and r(x)’ to r(x). We leave it to the reader to write down an explicit formula for thisand to check that this gives us the intended relative homotopy. �

With this preparation we can now turn to the proof of the proposition.

Proof. (of Proposition 5) Let us begin by defining the connecting homomorphism. Given a class ωin πn(X,A) it can be represented by a map of triples H : (I

n, ∂In, Jn−1) → (X,A, x0). This mapcan be restricted to the top face In−1 × {1} to give a map h = H| : (In−1, ∂In−1) → (A, x0). Weset:

∂ : πn(X,A)→ πn−1(A, x0) : [H] 7→ [h] = [H|]We leave it to the reader to check that this defines a group homomorphism or a map of pointedsets depending on the value of n. The naturality of ∂ follows immediately from the definition.

Let us prove that the sequence is exact. Thus, we have to establish exactness at three differentpositions, one of which we will leave as an exercise. So, we will content ourselves showing exactnessat πn(A, x0) and at πn(X,A). So, we have to show that there are four inclusions:

SECTION 4: RELATIVE HOMOTOPY GROUPS AND THE ACTION OF π1 5

(1) im(∂) ⊆ ker(i∗): This inclusion is immediate; given the homotopy class of a map

H : (In, ∂In, Jn−1)→ (X,A, x0),we have to show that the map i ◦ h : (In−1, ∂In−1)→ (X,x0) is homotopic to the constantmap (relative to the boundary). But such a homotopy is given by H itself.

(2) ker(i∗) ⊆ im(∂): This follows by definition of the relative homotopy groups and the con-necting homomorphism (see the motivational discussion!).

(3) im(j∗) ⊆ ker(∂): Given an arbitrary α ∈ πn(X,x0) it is easy to see that ∂◦j∗ is by definitionrepresented by the constant map κx0 : I

n−1 → X.(4) ker(∂) ⊆ im(j∗): Let us consider a map H : (In, ∂In, Jn−1) → (X,A, x0) which lies in the

kernel of ∂. By definition this means that the restriction

h = H| : (In−1 × {1}, ∂In−1 × {1})→ (A, x0)is homotopic to the constant map κx0 relative to the boundary. Choose an arbitrary suchhomotopy H ′ : h ' κx0 (rel x0). Then we obtain a map

H ′′ : Jn = In × {0} ∪ ∂In × I → Xwhich is H on In × {0}, the homotopy H ′ on In−1 × {1} × I and which takes the constantvalue x0 on the rest of ∂I

n× I.1 An application of Lemma 7 gives us a map K : In+1 → Xwhich restricts to H ′′ along Jn ⊆ In+1. By construction, K is a homotopy of maps oftriples (In, ∂In, Jn−1) → (X,A, x0) from H to K(−, 1) : (In, ∂In) → (X,x0). Thus, wehave [H] = j∗([K(−, 1)]) as intended.

�

Exercise 8. Conclude the proof of Proposition 5 by showing that the sequence is exact at πn(X,x0).

Corollary 9. (1) Given a pointed pair of spaces (X,A, x0) such that there is a pointed homo-topy equivalence X ' ∗ then there are isomorphisms πn(X,A) ∼= πn−1(A), n ≥ 1.

(2) Let i : (A, x0) → (X,x0) be the inclusion of a retract, i.e., we have r ◦ i = id for somepointed map r : (X,x0)→ (A, x0). Then there are split short exact sequences

1→ πn(A, x0)→ πn(X,x0)→ πn(X,A)→ 1, n ≥ 1,i.e., short exact sequences such that πn(A, x0)→ πn(X,x0) admits a retraction.

We can apply the first part to the special case of the reduced cone CA of a pointed space (A, ∗).The reduced cone comes naturally with an inclusion (A, ∗) → (CA, ∗) so that we have a pointedpair (CA,A, ∗). By the corollary, the connecting homomorphism ∂ : πn+1(CA,A)→ πn(A, ∗) is anisomorphism. We can combine this with the map induced by the quotient map q : (CA,A)→ (ΣA, ∗)in order to obtain the suspension homomorphism:

S : πn(A, ∗)δ−1→ πn+1(CA,A)

q∗→ πn+1(ΣA, ∗)As opposed to the context of singular homology, this suspension homomorphism is not an iso-morphism (even not for nice spaces as –say– CW-complexes). However, this map can be iteratedand we will later show that the suspension homomorphisms S : πn+k(Σ

kA, ∗)→ πn+k+1(Σk+1A, ∗)eventually are isomorphisms. Thus, the groups πn+k(Σ

kA, ∗) stabilize for large values of k.

1In the notation introduced before Lemma 7 we thus put the homotopy H on In−1(n,1)

× I and constant maps κx0on In−1f × I ⊆ ∂I

n × I, f 6= (n, 1).

6 SECTION 4: RELATIVE HOMOTOPY GROUPS AND THE ACTION OF π1

We will now turn to the action of the fundamental group on higher homotopy groups. Thiswill also allow us to understand more precisely the difference between πn(X,x0) and [S

n, X]. Tobegin with let us collect some basic facts about free homotopies. Given a space X and a homotopyH : Sn × I → X we obtain a path u in X by setting

u = H(∗,−) : I → Xwhere ∗ is the base point of Sn. If H is a homotopy from f to g and if u is the path of the basepoint, then this will be denoted by:

H : f 'u gThe fact that the homotopy relation is an equivalence relation takes the following form if we keeptrack of the paths of the base point.

Lemma 10. (1) For every map f : Sn → X we have f 'κf(∗) f .(2) If for two maps f, g : Sn → X there is a homotopy f 'u g then we also have g 'u−1 f .(3) Let f, g, h : Sn → X be maps such that f 'u g and g 'v h. Then there is a homotopy

f 'v∗u h.

Lemma 11. For every map f : Sn → X and every path u : I → X such that u(0) = f(∗) there isa map g : Sn → X such that f 'u g.

Proof. Let q : In → In/∂In ∼= Sn be the quotient maps. The maps f ◦ q : In × {0} → X andu ◦ pr : ∂In × I → I → X together define a map as follows:

(f ◦ q, u ◦ pr) : Jn = In × {0} ∪ ∂In × I //

��

X

In+1∃H

44iiiiiiiiiii

It follows from Lemma 7 that we can find an extension H : In+1 → X as indicated in the diagram.By construction, H(−, t) : In → X takes the constant value u(t) on the boundary ∂In and hencefactors as In × I → Sn × I → X. The induced map Sn × I → X defines a homotopy f 'u g. �

Thus g is obtained from f by ‘stacking a copy of the path on top of each point of ∂In’ and thenchoosing a certain reparametrization. In the special case of n = 1 it is easy to see that this way weobtain g = u ∗ f ∗ u−1. In the notation of the lemma, we want to show that the assignment

([u], [f ]) 7→ [g]is well-defined.

Lemma 12. Let f, f0, f1, g, g0, g1 : Sn → X be maps and let u, v : I → X be paths in X.

(1) If f 'u g and u ' v (rel ∂I) then also f 'v g.(2) Let us assume that f0(∗) = f1(∗) = x0 and g0(∗) = g1(∗) = x1. If f0 ' f1 (rel x0),

g0 ' g1 (rel x1) and f0 'u g0 then also f1 'u g1.

Proof. Let us begin with a proof of the first claim. We recommend that you draw a picture in thecase of n = 1 to see what is happening. Now, let H : In × I → X be a homotopy f 'u g andsimilarly G : I × I → X a homotopy u ' v (rel ∂I) which both exist by assumption. From this weconstruct a new map K : Jn+1 → X as follows. Note that Jn+1 ⊆ ∂In+2 can be written as a unionof three subspaces (use the Leibniz rule!):

Jn+1 = In × I × {0} ∪ ∂In × I × I ∪ In × ∂I × I

SECTION 4: RELATIVE HOMOTOPY GROUPS AND THE ACTION OF π1 7

On the first subspace we take the homotopy H, on the second subspace ∂In × I × I pr→ I × I G→ X,and on the remaining one the constant homotopies of f and g, i.e., we take:

In × ∂I × I pr→ In × ∂I ∼= In t In(f,g)→ X

We leave it to the reader to check that these maps fit together in the sense that they definea map K : Jn+1 → X. Now, an application of Lemma 7 shows that K can be extended to amap L = K◦r : In+2 → Jn+1 → X. By construction it follows that the restriction of L to In×I×{1}gives us the desired homotopy f 'v g.

The second claim is now easy. By assumption we have a chain of homotopies:

f1 'κx0 f0 'u g0 'κx1' g1But since κx1 ∗ u ∗ κx0 ' u (rel ∂I) we can conclude f1 'u g1 (by the first part of this lemma). �

Recall that given a space X we denote its fundamental groupoid by π(X). The objects in π(X)are the points inX while morphisms are given by homotopy classes of paths relative to the boundary.

Corollary 13. Let f : (Sn, ∗)→ (X,x0), let u : I → X be a path from x0 to x1, and let f 'u g forsome g : (Sn, ∗)→ (X,x1). Then the homotopy class [g] ∈ πn(X,x1) only depends on the homotopyclasses [f ] ∈ πn(X,x0) and [u] ∈ π(X)(x0, x1).

Proof. Let us assume we were also given f 'κx0 f′, u ' v (rel ∂I), and f ′ 'v g′. Then in order to

show that g 'κx1 g′ we observe that:

g 'u−1 f 'κx0 f′ 'v g′

But since v ∗ κx0 ∗ u−1 ' κx1 (rel ∂I) we can conclude by Lemma 10 and Lemma 12. �

Thus, we obtain a well-defined pairing

π(X)(x0, x1)× πn(X,x0)→ πn(X,x1) : ([u], [f ]) 7→ [u][f ] = [g]

for f 'u g as in the notation of Lemma 11.

Proposition 14. Given a space X then we have a functor πn(X,−) : π(X)→ Grp which sends anobject x0 ∈ π(X) to πn(X,x0) and a map [u] ∈ π(X)(x0, x1) to [u](−) : πn(X,x0)→ πn(X,x1).

Proof. We know already that πn(X,x0) is a group for all x0 ∈ X and that we have a well-definedmap of sets [u](−) : πn(X,x0)→ πn(X,x1). To check that the assignment [u] 7→ [u](−) is compatiblewith compositions and identities it suffices to recall the definition of this action. In fact, since itwas obtained from ‘stacking copies of u on top of ∂In’ it is easy to see that this is true. It remainsto show that the maps [u](−) : πn(X,x0) → πn(X,x1) are group homomorphisms. But this is leftas an exercise. �

Exercise 15. Given a path u : I → X with u(0) = x0 and u(1) = x1 show that [f ] 7→ [u][f ] definesa group homomorphism

[u](−) : πn(X,x0)→ πn(X,x1).

Thus, we have isomorphisms πn(X,x0) ∼= πn(X,x1) whenever x0, x1 ∈ X lie in the same path-component. Note that such an isomorphism is, in general, not canonical, since it depends on thechoice of a homotopy class of paths from x0 to x1. However, if π1(X,x0) ∼= 1 then there is only aunique such homotopy class so that the identification πn(X,x0) ∼= πn(X,x1) is canonical.

8 SECTION 4: RELATIVE HOMOTOPY GROUPS AND THE ACTION OF π1

Corollary 16. Given a pointed space (X,x0) then there is an action of π1(X,x0) on πn(X,x0).For n = 1 this specializes to the conjugation action, i.e., we have:

[u][f ] = [u][f ][u]−1, [u], [f ] ∈ π1(X,x0)

Proof. Since we have a functor πn(X,−) : π(X)→ Grp, it is completely formal that we get an actionof π1(X,x0) on πn(X,x0). In the context of Lemma 11, we already observed that our constructionsends (u, f) to u ∗ f ∗ u−1. Thus, at the level of homotopy classes we obtain the conjugation. �

Instead of using the actual construction of Lemma 11 to deduce this corollary, we can also argueusing the essential uniqueness of the construction (Corollary 13): we just have to observe that thereis a homotopy:

f 'u u ∗ f ∗ u−1

Whenever we have a group acting on a set we can pass to the set of orbits. In the case of theaction of the fundamental group on higher homotopy groups we obtain the following convenientresult.

Corollary 17. Let X be a path-connected space and let x0 ∈ X. Then the forgetful mapπn(X,x0) = [(S

n, ∗), (X,x0)]→ [Sn, X]exhibits [Sn, X] as the set of orbits of the action of π1(X,x0) on πn(X,x0).

Exercise 18. Give a proof of this corollary, i.e., show that the forgetful map is surjective andthat two elements [f ] and [g] have the same image if and only if there is a loop u at x0 suchthat [u][f ] = [g].

SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS

In this section we will introduce two important classes of maps of spaces, namely the Hurewiczfibrations and the more general Serre fibrations, which are both obtained by imposing certainhomotopy lifting properties. We will see that up to homotopy equivalence every map is a Hurewiczfibration. Moreover, associated to a Serre fibration we obtain a long exact sequence in homotopywhich relates the homotopy groups of the fibre, the total space, and the base space. This sequencespecializes to the long exact sequence of a pair which we already discussed in the previous lecture.

Definition 1. (1) A map p : E → X of spaces is said to have the right lifting property(RLP) with respect to a map i : A → B if for any two maps f : A → E and g : B → Xwith pf = gi, there exists a map h : B → E with ph = g and hi = f :

Af

//

i

��

E

p

��

B g//

h

>>~~

~~

X

(So h at the same time ‘extends’ f and ‘lifts’ g.)(2) A map p : E → B of spaces is a Serre fibration if it has the RLP with respect to all

inclusions of the form

In × {0} → In × I = In+1, n ≥ 0,and a Hurewicz fibration if it has the RLP with respect to all maps of the form

A× {0} → A× Ifor any space A. (So evidently, every Hurewicz fibration is a Serre fibration.)

(3) If p : E → X is a map of spaces (but typically one of the two kinds of fibrations) and x ∈ X,then p−1(x) ⊆ E is called the fiber of p over x. If x = x0 is a base point specified earlier,we just say the fiber of p for the fiber over x0.

Thus, Hurewicz fibrations are those maps p : E → X which have the homotopy lifting prop-erty with respect to all spaces: given a homotopy H : A × I → X of maps with target X and alift G0 : A→ E of H0 = H(−, 0) : A→ X against the fibration p : E → X then this partial lift canbe extended to a lift of the entire homotopy G : A× I → E, i.e., G satisfies pG = H and Gi = G0:

A× {0} G0 //

i

��

E

p

��

A× IH

//

G

;;ww

ww

wX

By definition, the class of Serre fibrations is given by those maps which have the homotopy liftingproperty with respect to all cubes In.

Example 2. (1) Any projection X × F → X is a Hurewicz fibration, as you can easily check.1

2 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS

(2) The evaluation map

� = (�0, �1) : XI → X ×X

at both end points is a Hurewicz fibration. Indeed, suppose we are given a commutativesquare

A× {0}f

//

i

��

XI

�

��

A× I g //h

99ss

ss

sX ×X

Or equivalently, we are given a map

φ : (A× {0} × I) ∪ (A× I × {0, 1})→ X

which we wish to extend to A× I × I. But ({0} × I) ∪ (I × {0, 1}) = J1 ⊆ I2 is a retractof I2, and hence so is A × J1 ⊆ A × I2. Therefore we can simply precompose φ with theretraction r : A× I2 → A× J1 to find the required extension.

(3) Let p : E → X and f : X ′ → X be arbitrary maps. Form the fibered product or pullback

E ×X X ′ = {(e, x′) | p(e) = f(x′)}

topologized as a subspace of the product E ×X ′. Then if E → X is a Hurewicz (or Serre)fibration, so is the induced projection E×XX ′ → X ′. This follows easily from the universalproperty of the pullback (see Exercise 1).

(4) If E → D → X are two Hurewicz (or Serre) fibrations, then so is their composition E → X(see Exercise 2).

(5) Let (X,x0) be a pointed space. The path space P (X) (or more precisely, P (X,x0) ifnecessary) is the subspace of XI (always with the compact-open topology) given by paths αwith α(0) = x0. The map �1 : P (X)→ X given by evaluation at 1 is a Hurewicz fibration.This follows by combining the previous examples (2) and (3).

(6) If f : Y → X is any map, the mapping fibration of f is the map

p : P (f)→ X

constructed as follows. The space P (f) is the fibered product

P (f) = XI ×X Y = {(α, y) | α(1) = f(y)}

and the map p is given by p(α, y) = α(0). We claim that p is a Hurewicz fibration. Indeed,suppose we are given a commutative diagram

A× {0}

i

��

u // P (f)

p

��

A× I v // X

SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS 3

Denoting by π1 : P (f)→ XI and π2 : P (f)→ Y the two projection maps belonging to thepullback P (f), we can first extend π2 ◦ u as in:

A× {0} π2◦u //

i

��

Y

A× Iũ

;;ww

ww

w

and then use example (2) to find a diagonal as in

A× {0} π1◦u //

��

XI

�=(�0,�1)

��

A× Iw

99ss

ss

s

(v,f◦ũ)// X ×X

Then (w, ũ) : A × I → P (f) is a diagonal filling in the original diagram we are lookingfor. In fact, it lands in P (f) because �1w = fũ, and makes the diagram commute becausepw = �0w = v and (w, ũ)i = (π1u, π2u) = u.

Here is a more abstract way of proving this: we showed in example (2) that � : XI → X×X is a Hurewicz fibration. Hence, by Exercise 1, so is the pullback Q = (X×Y )×(X×X)XIin:

Q //

��

XI

�

��

X × Yid×f

// X ×X

But then, by example (1) and Exercise 2, the composition Q→ X × Y → X is a Hurewiczfibration as well. Now note that

Q→ P (f) : (α, x, y) 7→ (α, y)

defines a homeomorphism and that under this homeomorphism the two maps Q→ X andP (f)→ X are identified. Thus also the map P (f)→ X is a Hurewicz fibration.

Exercise 3. Prove that the map φ in

Y

f!!D

DDDD

DDDφ

// P (f)

p

��

X

given by φ(y) = (κf(y), y) is a homotopy equivalence which makes the diagram commute. Here, asusual, we denote by κf(y) the constant path at f(y).

This exercise thus shows that every map is ‘homotopy equivalent’ to a Hurewicz fibration. Motivatedby this, one calls the fiber of p over x,

p−1(x) = {(α, y) | α(1) = f(y), α(0) = x}

4 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS

the homotopy fiber of f over x. Note that there is a canonical map from the fiber of f over x tothe homotopy fiber of f over x given by a restriction of φ, namely:

f−1(x)→ p−1(x) : y 7→ (κx, y)

Thus, the fiber ‘sits in’ the homotopy fiber while the homotopy fiber can be thought of as a ‘relaxed’version of the fiber: the condition imposed on a point y ∈ Y to lie in the fiber over x is that it hasto be mapped to x by f , i.e., f(y) = x, while a point of the homotopy fiber is a pair (α, y) consistingof y ∈ Y together with a path α in X ‘witnessing’ that y ‘lies in the fiber up to homotopy’.

So far, all our examples are examples of Hurewicz fibrations. However, we will see in the nextlecture that the weaker property of being a Serre fibration is a local property, and hence that allfiber bundles are examples of Serre fibrations. Moreover, this weaker notion suffices to establishthe following theorem.

Theorem 4. (The long exact sequence of a Serre fibration)Let p : (E, e0)→ (X,x0) be a map of pointed spaces with i : (F, e0)→ (E, e0) being the fiber. Supposethat p is a Serre fibration. Then there is a long exact sequence of the form:

. . .→ πn+1(X,x0)δ→ πn(F, e0)

i∗→ πn(E, e0)p∗→ πn(X,x0)

δ→ . . . i∗→ π0(E, e0)p∗→ π0(X,x0)

The ‘connecting homomorphism’ δ will be constructed explicitly in the proof. Before turning tothe proof, let us deduce an immediate corollary. By considering the homotopy fiber Hf instead ofthe actual fiber, we see that we can obtain a long exact sequence for an arbitrary map f of pointedspaces.

Corollary 5. Let f : (Y, y0)→ (X,x0) be a map of pointed spaces and let Hf be its homotopy fiber.Then there is a long exact sequence of the form:

. . .→ πn+1(X,x0)→ πn(Hf , ∗)→ πn(Y, y0)f∗→ πn(X,x0)→ . . .→ π0(Y, y0)

f∗→ π0(X,x0)

Proof. Apply Theorem 4 to the mapping fibration (Example 2.(6)) and use Exercise 3. �

Before entering in the proof of the theorem, we recall from the previous lecture the definition ofthe subspace Jn ⊆ In+1,

Jn = (In × {0}) ∪ (∂In × I) ⊆ ∂In+1 ⊆ In+1.

Note that, by ‘flattening’ the sides of the cube, one can construct a homeomorphism of pairs

(In+1, Jn)∼=→ (In+1, In × {0}).

Thus, any Serre fibration also has the RLP with respect to the inclusion Jn ⊆ In+1. We will usethis repeatedly in the proof.

Proof. (of Theorem 4) The main part of the proof consists in the construction of the operation δ.Let α : (In, ∂In)→ (X,x0) represent an element of πn(X,x0) = πn(X). Let ē0 : Jn−1 → E be theconstant map with value e0. Then the square

Jn−1ē0 //

��

E

p

��

In α//

β

SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS 5

commutes, so by the definition of a Serre fibration we find a diagonal β. Then δ[α] is the elementof πn−1(F ) represented by the map

β(−, 1) : In−1 → F, t 7→ β(t, 1).

Note that this indeed represents an element of πn−1(F ), because the boundary of In−1 × {1} is

contained in Jn−1, and β maps the top face In−1 × {1} into F since p ◦ β = α maps it to x0.The first thing to check is that δ is well defined on homotopy classes. Suppose [α0] = [α1],

as witnessed by a homotopy h : In × I → X from α0 to α1. Suppose also that we have chosenliftings β0 and β1 of α0 and α1 as above. Then we can define a map k making the solid square

J̃n

��

k // E

p

��

In × Ih

//

l

6 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS

we have that β(−, 1) is homotopic to the constant map by a homotopy h relative to ∂In−1 whichmaps into the fiber F . But then, stacking this homotopy h on top of β (i.e., by forming h ◦n β),we obtain a map representing an element β′ of πn(E). The image p ◦ β′ is obviously homotopic top ◦ β = α because p ◦ h is constant, showing that [α] lies in the image of p∗.

Exactness at πn−1(F ). For α : In → X representing an element of πn(X), the map β in the

construction of δ[α] = [β(−, 1)] shows that β(−, 1) ' β(−, 0) = ē0 in E, so i∗δ[α] = 0. Forthe other inclusion, suppose γ : In−1 → F represents an element of πn−1(F ) with i∗[γ] = 0, asrepresented by a homotopy h : In−1 × I → E with h(−, 1) = γ and h(−, 0) = ē0. Then α = p ◦ hrepresents an element of πn(X), and in the construction of δ[α] we can choose the diagonal filling βto be identical to h, in which case δ[α] is represented by γ. This shows that ker(i∗) ⊆ im(δ), andcompletes the proof of the theorem. �

Exercise 6. Show that the long exact sequence of a pointed pair (X,A), constructed in the previouslecture, can be obtained from this long exact sequence, by considering the mapping fibration of theinclusion A→ X (see also the last exercise sheet).

Exercise 7. Prove that the connecting homomorphism δ : πn(X,x0)→ πn−1(F, e0) is a homomor-phism of groups for n ≥ 2.

Exercise 8. Let p : E → X be a Hurewicz fibration, and let α : I → X be a path from x to y. Usethe lifting property of E → X with respect to p−1(x)× {0} → p−1(x)× I to show that α inducesa map α∗ : p

−1(x) → p−1(y). Show that the homotopy class of α∗ only depends on the homotopyclass of α, and that this construction in fact defines a functor on the fundamental groupoid,

π(X)→ Ho(Top).

This last exercise shows in particular that the homotopy type of the fiber of a Hurewicz fibrationis constant on path components. More precisely, if p : E → X is a Hurewicz fibration, then anypath α : I → X induces a homotopy equivalence between the fiber over α(0) and the fiber over α(1).

SECTION 6: FIBER BUNDLES

In this section we will introduce the interesting class of fibrations given by fiber bundles. Fiberbundles play an important role in many geometric contexts. For example, the Grassmaniannvarieties and certain fiber bundles associated to Stiefel varieties are central in the classificationof vector bundles over (nice) spaces. The fact that fiber bundles are examples of Serre fibrationsfollows from Theorem 11 which states that being a Serre fibration is a local property.

Definition 1. A fiber bundle with fiber F is a map p : E → X with the following property: everypoint x ∈ X has a neighborhood U ⊆ X for which there is a