Lecture Notes on Homology Theory

Dr. Thomas Baird (illustrations by Nasser Heydari)

Winter 2014

Contents

1 Introduction 2

2 Review of Point-Set Topology 52.1 New spaces from old. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Connectedness and Path-Connectedness . . . . . . . . . . . . . . . . . . . 82.3 Covers and Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Metric spaces and the Lebesgue number lemma . . . . . . . . . . . . . . . 92.5 Hausdorff spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Singular Homology 103.1 Simplices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Chains, cycles, and boundaries . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2.1 Homology as a functor . . . . . . . . . . . . . . . . . . . . . . . . . 163.3 Homotopy Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.3.1 Chain complexes and chain homotopy . . . . . . . . . . . . . . . . . 173.3.2 The prism operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.4 Relative Homology and the long exact homology sequence . . . . . . . . . 213.4.1 Reduced Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.5 Excision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.5.1 Proof of Excision . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.5.2 Mapping cylinders and cones . . . . . . . . . . . . . . . . . . . . . . 31

3.6 Applications to spheres: the degree of a map . . . . . . . . . . . . . . . . . 333.7 Cellular homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.7.1 Cell complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.7.2 Cell complex propaganda (not to be tested) . . . . . . . . . . . . . 403.7.3 Cellular Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.7.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.8 Mayer-Vietoris Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.9 Homology with coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.9.1 The Universal Coefficient Theorem for Homology . . . . . . . . . . 523.10 Covering spaces and the transfer . . . . . . . . . . . . . . . . . . . . . . . . 54

1

4 Cohomology 564.1 The cup product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.2 The Kunneth formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.3 Manifolds and orientations . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.4 The cap product and Poincare Duality . . . . . . . . . . . . . . . . . . . . 71

1 Introduction

Topology is the study of topological spaces ( e.g. subsets of Rn) and continuous mapsbetween them. The basic idea of algebraic topology is to study functors F fromtopological spaces to groups (or some other type of algebraic category). This means isthat for every topological space X, we assign a group F (X), and to each continuous mapf : X → Y , we assign a group homomorphism F (f) : F (X)→ F (Y ) such that

F (f ◦ g) = F (f) ◦ F (g)

for any pair of composable maps

Xg→ Y

f→ Z

and also that identity maps are sent to identity maps:

F (IdX) = IdF (X)

To see how this sort of thing may be useful, observe that if two spaces X and Yare isomorphic (i.e. homeomorphic), then F (X) and F (Y ) must be isomorphic for everyfunctor F . It turns out that the most powerful way to prove that two spaces X, Y arenot homeomorphic is to find a functor such that F (X) and F (Y ) are not isomorphic.

For another application, we begin with a definition. A subset A ⊆ X is called aretract, if there exists a continuous map r : X → A such that f(a) = a for all a ∈ A. 1.Inclusion of sets defines an injective map i : A ↪→ X. If A is a retract in X, then thereexists r such that

r ◦ i = IdA.

For any functor, this means that

F (r) ◦ F (i) = F (r ◦ i) = F (IdA) = IdF (A).

In particular, this means that F (i) must be injective when A ⊆ X is a retract (if not,F (r) ◦ F (i) = IdF (A) would not be injective, a contradiction). Using this idea, we willprove that the unit circle S1 is not a retract inside the unit disk D2.

The kinds of functors we will learn about in this course are the (singular) homologyand cohomology functors. These functors come in families labelled by non-negativeintegers called the degree (also called dimension): H0, H1, H2, ... for homology andH0, H1, H2, .... for cohomology. Both homology and cohomology take values in abeliangroups, though we will also study variations that take values in vector spaces.

1For example, the inclusion R ↪→ R2 as the x-axis is a retract using the map r(x, y) = (x, 0)

2

The historical motivation for homology theory came from vector calculus.2 Recall thatthere are various versions of the Fundamental Theorem of Calculus ( Stokes’ Theorem,Green’s Theorem, the Divergence Theorem) that equate an integral over a manifold (curve,surface, solid, etc.) with an integral over its boundary (set of points, a curve, surface,respectively). Homology emerged from efforts to understand how many “independent”submanifolds there are with respect to a given domain. Roughly speaking, the 0- homologyH0(X) is generated by points in X, the 1-homology H1(X) is generated by (oriented)closed curves in X, the 2-homology is generated by (oriented) closed surfaces, and so on.The homology class is trivial if the curve, surface, etc. is the boundary of a surface, solid,etc..

To see how this might work, consider the disconnected subset X ⊂ R2 pictured inFigure 1.

Figure 1: A space X with two path components

A point p in one component cannot be joined by a continuous path to a point q inanother component. It follows that p and q determine different elements [p] and [q] inH0(X) . We will show that there is an isomorphism H0(X) ∼= Zn where n is the numberof path-components of X.

Consider now a annulus A in R2 (Figure 2).The closed loop C represents an element in H1(A). It is intuitively clear that S1 is

not the boundary of a surface in A, so C represents a non-trivial element [C] in H1(A).Indeed, we will show that H1(A) ∼= Z and that C represents one of the generators (theother generator is obtained by reversing the orientation on C). On the other hand, ifwe take a union of C with a curve D that winds around the annulus in the oppositedirection, we see that together they form the boundary of a surface (Figure 3). In termsof homology, this will mean that [C+D] = [C]+[D] = 0, or equivalently that [C] = −[D].

2The development of homology theory is usually attributed to Poincare in the 1890’s, though the sub-ject didn’t really come into it’s own until the 1930’s through the work of numerous other mathematicians.

3

Figure 2: Loop in an annulus

Figure 3: Two loop in an annulus bound a surface

4

There are many different kinds of homology that are defined in different ways. Theapproach we will take in this course is called singular homology. Singular homology hassome great theoretical advantages over others (such as simplicial homology and cellularhomology), but has the draw back of being difficult to calculate directly. Indeed, itwill take some time before we establish the fact that oriented submanifolds determinehomology classes which has been the basis of today’s lecture.3 Instead, singular homologyis based on “singular simplices” which we will be introduced in ...

2 Review of Point-Set Topology

In this section we collect some basic facts from general topology that will be required inthis course. Proofs of these results can be found in any introductory textbook on generaltopology (e.g. Munkres’ Topology) or in the point-set topology notes I have posted onD2L.

Definition 1. A topological space (or simply space) (X, τ) is a set X and a collectionτ of subsets of X, called the open sets, satisfying the following conditions:

i) ∅ and X are open,ii) Any union of open sets is open,iii) Any finite intersection of open sets is open.

A set is called closed if its complement is open. Usually, we will denote the topologicalspace (X, τ) simply by X.

Example 1 (Euclidean Topology). An open ball in Rn is a set of the form

B = Bε(p) := {x ∈ Rn| ||x− p|| < ε}

for some p ∈ Rn and ε > 0. A subset U ⊆ Rn is called open if it is a union of openballs. Equivalently, U is open if for every p ∈ U , there exists an open ball B such thatp ∈ B ⊂ U .

In the example above, we say that open balls form a basis for the Euclidean topology.More generally, a collection of open sets B in a topological space X is called a basis ifevery other open set in X is a union of sets in B.

Definition 2. A continuous map f : X → Y between topological spaces is a map ofsets for which pre-images of open sets are open. I.e.

U ⊆ Y is open ⇒ f−1(U) := {x ∈ X|f(x) ∈ U} ⊆ X is open

Definition 3. A homeomorphism is a continuous bijection f : X → Y such that theinverse f−1 is also continuous. This is the notion of isomorphism for topological spaces.

3It is possible to define a kind of homology theory using oriented manifolds directly, called bordism.One of the reasons that this approach is not standard in introductory courses is that the theory ofmanifolds gets complicated in dimensions larger than two.

5

We will only rarely need use the abstract Definition 2 explicitly. More often, we willmake use of certain properties of continuous functions, including the following.

Proposition 2.1. Let X, Y and Z be topological spaces.

• The identity map IdX : X → X is continuous.

• If f : X → Y and g : Y → Z are continuous, then the composition g ◦ f : X → Zis continuous.

• Any constant map f : X → Y is continuous.

The first two conditions above make topological spaces + continuous maps into acategory. We will speak more about categories later.

2.1 New spaces from old.

Most of the topological spaces we encounter in this course are constructed from Rn usingthe operations below.

Definition 4. Let X be a topological space and A ⊂ X a subset. The subspace topol-ogy on A is the topology for which V ⊂ A is open if and only if V = A ∩ U for someopen set U in X.

Example 2. Any subset of Rn acquires a subspace Euclidean topology. Unless otherwisestated, we will always assume subsets of Rn to have this topology.

The inclusion map i : A ↪→ X is continuous (with respect to the subspace topology).In fact, we have the following special property: A map f : Y → A from a topologicalspace Y is continuous if and only if the composition i ◦ f : Y → X is continuous.

Definition 5. The product space X×Y of two spaces X and Y is the Cartesian productof sets X × Y , with a basis of open sets of the form U × V where U ⊂ X and V ⊂ Y areboth open.

The above definition iterates to define products of any finite number of spaces (infiniteproducts require a different definition).

Example 3. The n-fold product R×R×...×R is homeomorphic to Rn with the Euclideantopology.

The key property of product spaces is that a map

F : Z → X × Yis continuous if and only if the coordinate functions F = (F1, F2) are continuous as mapsfrom Z to X and to Y respectively.

Definition 6. Let {Xα} be a (possibly infinite) collection of spaces indexed by α. Thecoproduct space or disconnected union

∐αXα is the disjoint union of the sets Xα

with U ⊆∐

αXα is open if and only if U ∩Xα is open for all α.

6

The inclusions iα0 : Xα0 ↪→∐Xα are all continuous. A map F :

∐Xα → Y is

continuous if and only if the composition F ◦ iα : Xα → Y are continuous for all α.

Definition 7. An equivalence relation on a set X is a relation ∼ satisfying, for allx, y ∈ X

(i) x ∼ x

(ii) x ∼ y implies y ∼ x

(iii) x ∼ y and y ∼ z implies x ∼ z.

Given any relation R on X, we can generate the “smallest” equivalence relation ∼Rsuch that xRy implies x ∼R y. Explicitly, we define x ∼R y if and only if there exists afinite sequence {xi ∈ X}ni=0 for n ≥ 0 satisfying

x0 = x,

xn = y and,

xiRxi−1 or xi−1Rxi for all i = 1, ..., n.

Given x ∈ X, the equivalence class of x is

[x] := {y ∈ X|x ∼ y}

Notice that [x] = [y] if and only if x ∼ y. The equivalence classes determine a partitionof X into disjoint sets. Let E := {[x]|x ∈ X} be the set of equivalence classes (we willsometimes denote E = X/ ∼ ) . There is a canonical map

Q : X → E, x 7→ [x]

called the quotient map.

Definition 8. Let X be a topological space and let ∼ be an equivalence relation on theset underlying X. The quotient topology on E is the topology for which U ⊂ E isopen if and only if Q−1(U) is open in X.

Observe that Q : X → E is continuous and that a map f : E → Y is continuous ifand only if f ◦Q : X → Y is continuous.

Example 4. Suppose X and Y are topological spaces, A ⊆ X is a subspace, and f : A→Y is a continuous map. Define an equivalence relation on the coproduct X

∐Y generated

by f(a) ∼ a for all a ∈ A. We say that the quotient space (X∐Y )/ ∼ is obtained by

attaching X to Y along A using f .

Example 5. Suppose X is a topological space, and G is a group that acts on X viahomeomorphisms. Define two points in X equivalent if they lie in the same orbit of G.The quotient space in this case is called the orbit space and is denoted X/G.

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2.2 Connectedness and Path-Connectedness

Let I denote the unit interval [0, 1] ⊂ R with the Euclidean topology.

Definition 9. A space X is called path-connected if for any two points p, q ∈ X thereexists a continuous map γ : I → X such that γ(0) = p and γ(1) = q.

Definition 10. A space X is called connected if there is no proper subset A ⊂ Xwhich is both open and closed. (proper means other than X or ∅, which are always bothopen and closed).

Observe that if A ⊂ X is both open and closed, then the complement Ac is also bothopen and closed, and there is a natural isomorphism A

∐Ac ∼= X. Thus spaces that are

not connected can be decomposed into a disconnected union of nonempty spaces.

Proposition 2.2. Path-connected spaces are connected.

The converse of Proposition 2.2 is not true in general. However all the connectedspaces we encounter in this course will also be path-connected.

Connectedness and path-connectedness are preserved under the following operations

• A product of (path-)connected spaces is (path-)connected.

• The continuous image of a (path-)connected space is (path-)connected.

• Let {Uα} be a covering of X such that each Uα is (path-)connected and the inter-section ∩αUα is non-empty. Then X is (path-)connected.

2.3 Covers and Compactness

Definition 11. An open (closed) cover of a topological space X is a collection of open(closed) sets {Uα} such that the union ∪αUα = X.

Proposition 2.3. Let {Uα} be either an open cover or a finite closed cover of X. A mapof sets

f : X → Y

between topological spaces is continuous if and only if the restrictions f |Uα : Uα → Y arecontinuous for all α (where Uα has the subspace topology).

The preceding proposition will be used in two ways: to test if a map f is continuous byconsidering the restrictions, and also to construct a map f by gluing together continuousmaps defined on the Uα that agree on overlaps.

Definition 12. A space X is called compact if every open cover {Uα} of X containsa finite subcover. I.e., there exists a finite collection {U1, ..., Un} ⊆ {Uα} such that∪ni=1Ui = X.

Proposition 2.4. A subspace of Rn is compact if and only if it is closed and bounded.

8

Compactness is preserved under the following:

• A closed subspace of a compact space is compact.

• A finite union of compact spaces is compact.

• A product of compact spaces is compact.

• If f : X → Y is continuous and X is compact, then the image f(X) ⊆ Y is compact.

2.4 Metric spaces and the Lebesgue number lemma

Definition 13. Let X be a set. A metric on X is a function

d : X ×X → R≥0

called the distance or metric function, satisfying

1. d(x, x′) = 0 ⇔ x = x′ (d separates points)

2. d(x, x′) = d(x′, x) (d is symmetric)

3. d(x, x′′) ≤ d(x, x′) + d(x′, x′′) (the triangle inequality)

A metric space (X, d) determines a metric topology on X, which is generated bythe basis of open balls Bε(p) = {x ∈ X|d(x, p) < ε}. If A is a subset of a metric space Xthen A becomes a metric space by restriction. The metric topology on A is the same asthe subspace topology on A.

The following result will come up repeatedly.

Lemma 2.5 (Lebesgue number Lemma). Let A be an open covering of a compact metricspace X. There exists δ > 0, called the Lebesgue number, such that for all p ∈ X, theopen ball Bδ(p) is contained in some U ∈ A.

2.5 Hausdorff spaces

Definition 14. A space X is called Hausdorff if for any pair of distinct points p, q ∈ X,there exist open sets U, V such that p ∈ U , q ∈ V and U ∩ V = ∅.

Proposition 2.6. Any metric space is Hausdorff. In particular, any subset of Rn isHausdorff.

The Hausdorff property is preserved under the following:

• Products of Hausdorff spaces are Hausdorff.

• Subspaces of Hausdorff spaces are Hausdorff.

• Coproducts of Hausdorff spaces are Hausdorff.

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Definition 15. A space X is locally compact, if every point p ∈ X is contained in anopen neighbourhood p ∈ U such that the closure U is compact.

For example, compact spaces are locally compact. Also, Rn is locally compact.

Proposition 2.7. Let X be a locally compact space and let Y be a Hausdorff space. Ifφ : X → Y is a continuous map that is also a bijection of sets, then φ is a homeomorphism.

3 Singular Homology

3.1 Simplices

The standard q-simplex ∆q is the simplex spanned by the zero vector e0 = ~0 and thestandard basis vectors e1, ..., eq in Rq (Figure 4). Thus,

∆q := {(t1, ..., tn)| ≥ 0 for all i = 1, ..., q and t1 + ...+ tq ≤ 1 }

Figure 4: The standard simplices

If X is a topological space, a singular q-simplex (or simply simplex)in X is a(continuous) map

σ : ∆q → X.

A singular 0-simplex in X is simply a point in X, a singular 1-simplex in X is acontinuous path in X, etc.. We can think of singular simplices as probes used to studythe space X.

Example 6. Let v0, ..., vq be a set of q + 1-vectors in Rn for some n. Define

[v0, ..., vq] : ∆q → Rn, (t1, ..., tn) 7→ (1− t1 − ...− tq)v0 + t1v1 + ...+ tqvq.

10

We call [v0, ..., vq] the affine simplex defined by v0, ..., vq. Slightly abusing notation,define the face maps for 0 ≤ i ≤ q, define by

F iq : ∆q−1 → ∆q

by F iq = [e0, ..., ei, ...., en] where the ei means “omit ei”.

Figure 5: Faces of the standard 2-simplex

The ith face of a singular q-simplex σ : ∆q → X is the q − 1-simplex

σ(i) : ∆q−1 → X

defined be composition with the face map:

σ(i) := σ ◦ F iq .

3.2 Chains, cycles, and boundaries

Define Sq(X) to be the free Abelian group generated by singular q-simplices. The elementsof Sq(X) are called singular chains and are formal linear combinations of the form∑

σ

aσσ

where the coefficients aσ ∈ Z and the sum is over a finite number of singular q-simplicesσ. By convention, Sq(X) = 0 for q < 0.

The boundary map ∂q : Sq(X) → Sq−1 is a homomorphism, defined on singular sim-plices by

∂q(σ) =

q∑i=0

(−1)iσ(i)

and extended linearly to all of Sq(X) by the rule

∂q(∑σ

aσσ) =∑σ

aσ∂q(σ).

We will often drop the subscript and write ∂ = ∂q when it is unlikely to cause confusion.

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Example 7. Let σ1 and σ2 be singular 2-simplices in X. Then −2σ1 + 3σ2 ∈ S2(X) is a2-chain and

∂2(−2σ1 + 3σ2) = −2∂2(σ1) + 3∂2(σ2)

= −2(σ(0)1 − σ

(1)1 + σ

(2)1 ) + 3(σ

(0)2 − σ

(1)2 + σ

(2)2 )

= −2σ(0)1 + 2σ

(1)1 − 2σ

(2)1 + 3σ

(0)2 − 3σ

(1)2 + 3σ

(2)2

is a 1-chain in S1(X).

The boundary map can be understood schematically from Figure 6, but be careful notto confuse singular simplices (which are maps) with their images (which are sets).

Figure 6: Boundary of simplices - intuition

Proposition 3.1. The composition ∂q−1 ◦ ∂q : Sq(X)→ Sq−2(X) is the zero map. Drop-ping subscripts, we write this

∂2 = 0.

Proof. Since Sq(X) is generated by simplices, it suffices to check that ∂q−1 ◦ ∂q(σ) = 0 forall q-simplices σ.

It is an easy check that if 0 ≤ j < i ≤ q, the face maps satisfy

F iq ◦ F

jq−1 = F j

q ◦ F i−1q−1.

Thus

12

∂q−1 ◦ ∂q(σ) = ∂q−1(

q∑i=0

(−1)iσ(i))

=

q∑i=0

(−1)i∂q−1(σ ◦ F iq)

=

q∑i=0

(−1)iq−1∑j=0

(−1)j(σ ◦ F iq ◦ F

jq−1)

=

q∑i=0

q−1∑j=0

(−1)i+j(σ ◦ F iq ◦ F

jq−1)

=∑

0≤i≤j≤q−1

(−1)i+j(σ ◦ F iq ◦ F

jq−1) +

∑0≤j<i≤q

(−1)i+j(σ ◦ F iq ◦ F

jq−1)

=∑

0≤i≤j≤q−1

(−1)i+j(σ ◦ F iq ◦ F

jq−1)−

∑0≤j≤i−1≤q−1

(−1)i−1+j(σ ◦ F jq ◦ F i−1

q−1)

= 0

as these two sums cancel term by term by changing the index of the first sum by i 7→ j,j 7→ i− 1.

A more geometric illustration of ∂ ◦ ∂ = 0 is provided in Figure 7.

Figure 7: ∂2 = 0

The group of q-cycles Zq(X) is the kernel of ∂q:

Zq(X) := {α ∈ Sq(X) | ∂(α) = 0}

The group of q-boundaries Bq(X) is the image of ∂q+1:

Bq(X) := {∂(β) | β ∈ Sq+1(X)}.

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by Proposition 3.1, Bq(X) is a subgroup of Zq(X). The qth degree singular homology ofX is the quotient group:

Hq(X) := Zq(X)/Bq(X).

Example 8. The homology of a point. If X = {pt} is a single point, then there isonly one singular simplex in each degree, which is the constant map σq : ∆q → {pt}. Thechain groups are

Cq({pt}) = Zσq ∼= Z.4

The boundary map satisfies (for q ≥ 1)

∂q(σq) =

q∑i=0

(−1)iσ(i)q

=

q∑i=0

(−1)iσq−1

=

{σq−1 if q is even

0 if q is odd

while ∂0(σ0) = 0.It follows that for q ≥ 1

Zq({pt}) = Bq({pt}) =

{0 if q is even

Cq({pt}) if q is odd

while Z0({pt}) ∼= Z and B0({pt}) = 0. Thus

Hq({pt}) = Zq({pt})/Bq({pt}) =

{0 if q ≥ 1

Z if q = 0

A space X for which Hq(X) ∼= Hq({pt}) for all q is called acyclic meaning no cyclesthat are not also boundaries.

Proposition 3.2. Let {Xk} be the set of path components of a space X (indexed by k).Then

Hq(X) = ⊕kHq(Xk)

for all q ≥ 0.

Proof. Because the standard q-simplex is path connected (indeed convex), the image ofa singular q-simplex σ : ∆q → X must be path connected and in particular must liewithin one of the path components of X. It follows that for all q we have a canonicaldecomposition,

Cq(X) = ⊕qCq(Xk).

4The case of a point is highly unusual in this respect. For most spaces Y , Cq(Y ) has uncountablerank.

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Moreover, it is clear that the boundary map ∂ respects this decomposition, so that

Zq(X) = ⊕kZq(Xk)

andBq(X) = ⊕kBq(Xk)

and finally that

Hq(X) = Zq(X)/Bq(X)

= (⊕kZq(Xk))/(⊕kBq(Xk))

= ⊕k(Zq(Xk)/Bq(Xk))

= ⊕kHq(Xk)

We denote by π0(X) the set of path components of of a space X.

Proposition 3.3. There is a canonical isomorphism

H0(X) = Zπ0(X).

Thus H0(X) ∼= Zn where n is the number of path components of X.

Proof. By Proposition 3.2, it suffices to show that if X is path connected, then there is acanonical isomorphism

H0(X) = Z.Recall that a singular 0-simplex is the same thing as a point in X. Thus

S0(X) = Z0(X) =⊕p∈X

Zp.

The standard one simplex ∆1 is equal to the unit interval [0, 1] ⊆ R, so a singular 1-simplex is a continuous path in σ : [0, 1] → X. Since X is path-connected, for any twopoints p, q ∈ X, there exists a path σ such that σ(0) = p and σ(1) = q. Consequently,the boundary satisfies

∂(σ) = σ(1)− σ(0) = p− q ∈ S0(X),

and thusB0(X) = SpanZ{p− q | p, q ∈ X} ⊂

⊕p∈Z

Zp.

Observe that B0(X) is equal to the kernel of the homomorphism

ε :⊕p∈Z

Zp→ Z

defined on generators by ε(p) = 1. It follows that ε descends to a homomorphism

H0(X) = Z0(X)/B0(X) ∼= Z.

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3.2.1 Homology as a functor

Suppose that f : X → Y is a continuous map. If σ is a q simplex for X, then thecomposition f ◦ σ is a q-simplex for Y . This defines a homomomorphism

Sq(f) : Sq(X)→ Sq(Y ), Sq(f)(∑σ

aσσ) =∑σ

aσf ◦ σ.

Clearly Sq(IdX) = IdSq(X) and Sq(f ◦ g) = Sq(f) ◦Sq(g) for composable continuous mapsf and g. Thus Sq is functor from topological spaces to abelian groups. It allows commuteswith the boundary map.

Lemma 3.4. ∂q ◦ Sq(f) = Sq−1(f) ◦ ∂q.

Proof. It is enough to check for simplices.

∂Sq(f)(σ) = ∂(f ◦ σ)

=

q∑i=0

(−1)i(f ◦ σ)(i)

=

q∑i=0

(−1)if ◦ σ ◦ F iq

=

q∑i=0

(−1)if ◦ σ(i)

= Sq(f)(

q∑i=0

(−1)iσ(i))

= Sq(f)(∂σ)

using associativity of composition.

It follows then that Sq(f) sends Zq(X) to Zq(Y ) and Bq(X) to Bq(Y ) and thus inducesa homomorphism between the quotient groups

Hq(f) : Hq(X)→ Hq(Y ).

It follows easily from the fact that Sq is a functor that Hq is a functor from topologicalspaces to abelian groups. It is common to use short hand

f∗ = Hq(f)

though we will try to avoid doing so.

16

3.3 Homotopy Invariance

Two continuous maps f, g : X → Y are said to be homotopic if there exists a continuousmap

h : X × I → Y

where I = [0, 1] is the unit inverval and both h(x, 0) = f(x) and h(x, 1) = g(x). Intu-itively, two maps are homotopic if one can be continuously deformed into the other.

The goal of this section is to prove the following theorem.

Theorem 3.5. Let f and g be homotopic maps from X to Y . For all q ≥ 0, the inducedmaps on homology are equal: Hq(f) = Hq(g).

Two spaces X and Y are called homotopy equivalent if there exist maps f : X → Yand g : Y → X such that f ◦ g is homotopic to IdY and g ◦ f is homotopic IdX .

Corollary 3.6. If X and Y are homotopy equivalent, then H∗q (X) ∼= H∗q (Y ) for all q ≥ 0.

Proof. By Theorem 3.5 and functoriality, we have

Hq(f) ◦Hq(g) = Hq(f ◦ g) = Hq(IdY ) = IdHq(Y )

and similarly Hq(g) ◦ Hq(f) = IdHq(X). The Hq(f) and Hq(g) are inverse isomorphismsbetween Hq(X) and Hq(Y ).

A space is called contractible if it is homotopy equivalent to a point. Examples ofcontractible spaces include all convex subspaces of Rn (exercise). By Corollary 3.6, acontractible space X satisfies Hq(X) = 0 for q ≥ 1 and H0(X) = Z (i.e. contractiblespaces are acyclic).

Before proving Theorem 3.5, it will be helpful to introduce some abstract ideas aboutchain complexes.

3.3.1 Chain complexes and chain homotopy

A chain complex (of abelian groups)

C := (Cq, ∂q)q∈Z

is a sequence of abelian groups (Cq)q∈Z and homomorphisms ∂q : Cq → Cq−1 such that∂q∂q+1 = 0 for all q ∈ Z.

· · · ∂q+3−→ Cq+2∂q+2−→ Cq+1

∂q+1−→ Cq∂q−→ Cq−1

∂q−1−→ Cq−2∂q−2−→ . . .

Typically Cq = 0 for q < 0.

Example 9. The singular chain complex S(X) = (Sq(X), ∂q)q∈Z is a chain complex.

17

We define Zq(C) = ker(∂q) and Bq(C) = im(∂q+1) and Hq(C) = Zq(C)/Bq(C), calledrespectively the q-chains, q-boundaries, and q-homology groups of the chain complex. Ifz ∈ Zq(C), denote by [z] ∈ Hq(C) the coset represented by z.

A morphism of chain complexes f : C → C ′ is a sequence of homomorphisms(fq : Cq → C ′q)q∈Z that commutes with boundary maps: fq−1∂q = ∂′qfq for all q. In otherwords, the following diagram commutes:

Cq∂q //

fq��

Cq−1

fq−1

��C ′q

∂′q // C ′q−1

.

Example 10. Given a continuous map φ : X → Y , the morphisms Sq(φ) : Sq(X) →Sq(Y ) determine a chain morphism S(φ) : S(X)→ S(Y ).

A chain map f : C → C ′ induces a homomorphism in homology Hq(f) : Hq(C) →Hq(C

′) for all q ∈ Z by the same reasoning as in §3.2.1 by the rule

Hq(f)([z]) = [fq(z)].

Each Hq is a functor from chain complexes to abelian groups.Let f, g : C → C ′ be two chain maps. A chain homotopy between f and g is a

sequence of homomorphism (Pq : Cq → C ′q+1)q∈Z such that

∂′q+1Pq + Pq−1∂q = fq − gq.

Cq+1∂q+1 //

��

Cq∂q //

fq−gq

��

Pq

~~}}}}

}}}}

}}}}

}}}}

}Cq−1

��

Pq−1

~~}}}}

}}}}

}}}}

}}}}

}

C ′q+1∂′q+1

// C ′q∂′q

// C ′q−1

.

The chain maps f and g are called chain homotopic if there exists a chain homotopybetween them.

Proposition 3.7. If chain maps f, g : C → C ′ are chain homotopic, then the inducedmaps on homology are equal: Hq(f) = Hq(g) as maps from Hq(C) to Hq(C

′).

Proof. Let [z] ∈ Hq(C) be represented by z ∈ Zq(C). Then

fq(z)− gq(z) = Pq−1∂q(z) + ∂q+1Pq(z) = ∂q+1Pq(z)

is a boundary. Thus

Hq(f)([z])−Hq(g)([z]) = [fq(z)]− [gq(z)] = [fq(z)− gq(z)] = 0

so Hq(f)([z]) = Hq(g)([z]).

18

3.3.2 The prism operator

For t ∈ I = [0, 1], define

it : X ↪→ X × I, it(x) = (x, t).

Lemma 3.8. The two maps i0, i1 : X → X × I, determine chain homotopic chain mor-phisms S(i0) and S(i1).

Proof. The first step is to define a decomposition of ∆q× I ⊂ Rn+1 into (q+ 1)-simplices.Denote the vertices lying in ∆q×{0} by v0, ..., vq and those lying in ∆q×{1} by w0, ..., wq.For each i, the image of the affine simplex

[v0, ..., vi, wi+1, ...., wn] : ∆q → ∆q × I

can be thought of as the graph of a map from ∆q to I, because composing the projection∆q × I → ∆q is the identity map. These graphs slice ∆q × I into the images of affine(q + 1)-simplices

[v0, ..., vi, wi, ..., wq] : ∆q+1 → ∆q × I, i ∈ {0, ..., q}.

Figure 8: Prism decomposition

For arbitrary X, define the prism operator Pq : Sq(X)→ Sq+1(X × I) on simplicesby

Pq(σ) :=

q∑i=0

(−1)i(σ × IdI) ◦ [v0, ..., vi, wi, ..., wq].

19

This will be our chain homotopy.

∂q+1Pq(σ) = ∂q+1

q∑i=0

(−1)i(σ × IdI) ◦ [v0, ..., vi, wi, ..., wq]

=

q∑i=0

∑j≤i

(−1)i+j(σ × IdI) ◦ [v0, .., vj, ..., vi, wi, ..., wq]

+

q∑i=0

∑j≥i

(−1)i+j+1(σ × IdI) ◦ [v0, ..., vi, wi, ..., wj..., wq]

On the other hand,

Pq−1∂q(σ) = Pq−1(

q∑j=0

(−1)jσ ◦ [e0, ..., ej, ..., eq])

=

q∑i=0

∑j<i

(−1)i+j+1(σ × IdI) ◦ [v0, ..., vj, ..., vi, wi, ..., wq])

+

q∑i=0

∑j>i

(−1)i+j(σ × IdI) ◦ [v0, .., vi, wi, ..., wj, ..., wq].

Adding together, we get

∂q+1Pq(σ) + Pq−1∂q(σ) =

q∑i=0

(σ × IdI) ◦ [v0, ..., vi, wi, ..., wq]

−q∑i=0

(σ × IdI) ◦ [v0, ..., vi, wi, ..., wq]

= σ × IdI ◦ [w0, ..., wq]− σ × IdI ◦ [v0, ..., vq]

= i1 ◦ σ − i0 ◦ σ.

It follows that∂q+1Pq + Pq−1∂q = Sq(i1)− Sq(i0).

Proof of Theorem 3.5. A homotopy between two maps f, g : X → Y is a map h : X×I →Y such that f = h◦i0 and g = h◦i1. By Lemma 3.8, Sq(i0) and Sq(i1) are chain homotopic,so Proposition 3.7 implies Hq(i0) = Hq(i1). Finally we see that

Hq(f) = Hq(h) ◦Hq(i0) = Hq(h) ◦Hq(i0) = Hq(g).

In fact, it is not hard to show that S(f) and S(g) are chain homotopic via the chainhomotopy Sq+1(H) ◦ Pq.

20

3.4 Relative Homology and the long exact homology sequence

A topological pair (X,A) consists of a space X and a subspace A ⊆ X. A pair (X,A)gives rise to an inclusion of chain groups Sq(A) ≤ Sq(X) (technically the inclusion mapi : A ↪→ X determines injective homomorphism Sq(i)). Define the relative chain groupof the pair to be the quotient group

Sq(X,A) := Sq(X)/Sq(A).

The relative chain groups combine to form the relative chain complex

. . . Sq+1(X,A)∂q+1−→ Sq(X,A)

∂q−→ Sq−1(X,A)∂q−1−→ . . .

where the boundary map is defined by the following commutative diagram

Sq(X)

��

∂q // Sq−1(X)

��Sq(X,A)

∂q // Sq−1(X,A)

where the vertical arrows are quotient maps. Note that ∂q is well defined because ∂q sends

Sq(A) to Sq−1(A) and that ∂2

= 0 because ∂2 = 0. It follows that we can define relativecycles, relative boundaries, and relative homology as described in 3.3.1, which are denoted

Zq(X,A), Bq(X,A), Hq(X,A)

respectively. Geometrically, a relative cycle in Zq(X,A) is represented by a chain in Zq(X)whose boundary lands in Sq−1(A).

Remark 1. Observe that if A = ∅, then Sq(A) = 0 for all q. It follows that Sq(X, ∅) =Sq(X) and that Hq(X, ∅) = Hq(X). Thus it is possible to think of homology as just aspecial case of relative homology.

A map of topological pairs

f : (X,A)→ (X ′, A′)

is a continuous map f : X → X ′ such that f(A) ⊆ A′. Such a map determines amorphism of chain complexes S(f) : S(X,A)→ S(X ′, A′) and thus also a homomorphismon homology

Hq(f) : Hq(X,A)→ Hq(X′, A′).

The following properties are proven similarly to their counterparts for Hq(X).

• Hq is functor from topological pairs to Abelian groups. I.e. Hq(f)◦Hq(g) = Hq(f◦g)and Hq(Id(X,A)) = IdHq(X,A).

• If {Xk} is the set of path components of X and Ak = A ∩ Xk, then there is acanonical isomorphism Hq(X,A) =

⊕kHq(Xk, Ak).

21

• Let h : X×I → X ′ be a homotopy between maps map such that ht(a) := h(a, t) ∈ A′for all a ∈ A and t ∈ I. Then Hq(h0) = Hq(h1) as homomorphisms from Hq(X,A)to Hq(X

′, A′).

The quotient morphisms Sq(X) → Sq(X,A) fit together into a morphism of chaincomplexes j : S(X) → S(X,A). Combined with inclusion chain morphism i : S(A) →S(X) we get a commutative diagram

. . . // Sq+1(A) //

i��

Sq(A) //

i��

Sq−1(A) //

i��

. . .

. . . // Sq+1(X) //

j

��

Sq(X) //

j

��

Sq−1(X) //

j

��

. . .

. . . // Sq+1(X,A) // Sq(X,A) // Sq−1(X,A) // . . .

. (1)

By functoriality, these chain morphisms give rise to homology homomorphismsHq(A)→Hq(X) → Hq(X,A) for all q ≥ 0. The most important property of relative homology isthe existence of a connecting homomorphism

Hq(X,A)∂→ Hq−1(A),

(denoted by ∂ using abuse of notation). Let η ∈ Hq(X,A), be represented by a cyclej(z) ∈ Sq(X,A) which is the image of a chain z ∈ Sq(X). Then ∂(z) must land inSq(A) ≤ Sq(X), because ∂(j(z)) = j(∂(z)) = 0 ∈ Sq(X,A) = Sq(X)/Sq(A). Define

∂([z]) = [∂(z)] ∈ Hq−1(A).

(We leave it as an exercise prove that this homomorphism is well defined and does notdepend on the representatives z or z.). The connecting homomorphism permits us toextend to a long sequence of homomorphisms

· · · → Hq+1(X,A)∂→ Hq(A)

Hq(i)→ Hq(X)Hq(j)→ Hq(X,A)

∂→ Hq−1(A)→ . . . (2)

By convention, this sequence ends at · · · → H0(X)→ H0(X,A)→ 0.

Definition 16. A sequence of abelian groups and homomorphisms

Af→ B

g→ C

is called exact at B if ker(g) = im(f).

Theorem 3.9. The sequence (3) is exact (meaning exact at all groups in the sequence).It is called the long exact homology sequence associated to the pair (X,A).

22

Proof. We will only prove exactness at Hq(X,A) and leave the rest as an exercise. Con-sider

Hq(X)Hq(j)→ Hq(X,A)

∂→ Hq−1(A)

If η ∈ im(Hq(j)), that means we can choose a representative cycle z ∈ Cq(X) suchthat [j(z)] = η. Since z is a cycle, it follows that ∂(z) = 0, so

∂(η) = [∂(z)] = [0] = 0 ∈ Hq−1(A).

Therefore im(Hq(j)) ≤ ker(∂).Conversely, suppose that η ∈ ker(∂). This means that we can choose a chain z ∈ Sq(X)

such that [j(z)] = η and ∂(z) = ∂(α) for some α ∈ Sq(A). Consequently, z − α is a cyclein Sq(X) and

Hq(j)([z − α]) = [j(z − α)] = [j(z)]− [j(α)] = [j(z)] = η.

So ker(∂) ≤ im(i∗).

Proofs like the one above are called a diagram chase, because they amount to chasingelements through a diagram like (1).

Denote by π0(X,A) the set of path components of X that do not intersect A.

Proposition 3.10. There is a canonical isomorphism

H0(X,A) ∼= Zπ0(X,A).

In particular, H0(X,A) ∼= Zm where m is the number of path components of X that donot intersect A.

Proof. Let i : A ↪→ X denote the inclusion of A into X. Then we have an exact sequence

H0(A)H0(i)→ H0(X)

H0(j)→ H0(X,A)→ 0.

Exactness implies that H0(X,A) ∼= H0(X)/im(H0(i)). We know (Prop 3.3) that H0(X) =Zπ0(X) and the image of H0(i) is generated by those path components of X that containa path component of A. The result follows.

One of the great merits of the long exact homology sequence is that it is functorialwith respect to maps of pairs.

Proposition 3.11. Let f : (X,A) → (Y,B) be a map of pairs. This induces homomor-phisms on homology such that the following diagram commutes.

. . . // Hq+1(X,A)

f∗��

∂ // Hq(A) //

f∗��

Hq(X)

f∗��

// Hq(X,A) ∂ //

f∗��

Hq−1(A) //

f∗��

. . .

. . . // Hq+1(Y,B) ∂ // Hq(B) // Hq(Y ) // Hq(Y,B) ∂ // Hq−1(B) // . . .

Proof. We leave verification up to the reader.

23

The following lemma is helpful for calculations:

Lemma 3.12 (The Five Lemma). Consider a commutative diagram of abelian groups

A1//

α

��

A2//

β

��

A3//

γ

��

A4//

�

A5

ε

��B1

// B2// B3

// B4// B5

where the rows are exact. If α, β, δ and ε are all isomorphisms, then γ is also an isomor-phism.

Proof. Diagram chase.

Corollary 3.13. If f : (X,A) → (Y,B) is a map of pairs such that two out of threefamilies of induced maps on homology

{Hq(A)→ Hq(B)}q∈Z,

{Hq(X)→ Hq(Y )}q∈Z.

{Hq(X,A)→ Hq(Y,B)}q∈Z,

are isomorphisms in all degree, then the remaining family is isomorphisms in all degrees.

Proof. Simply apply the Five Lemma to the diagram:

. . . // Hq+1(X,A)

f∗��

∂ // Hq(A) //

f∗��

Hq(X)

f∗��

// Hq(X,A) ∂ //

f∗��

Hq−1(A) //

f∗��

. . .

. . . // Hq+1(Y,B) ∂ // Hq(B) // Hq(Y ) // Hq(Y,B) ∂ // Hq−1(B) // . . .

Example 11. If a map of pairs f : (X,A) → (Y,B) restricts to homotopy equivalencesbetween X and Y and between A and B, then f∗ : Hq(X,A) → Hq(Y,B) is an isomor-phism in all degrees.

3.4.1 Reduced Homology

It is sometimes convenient to use a modified version of singular homology called reduced homology.For any space X, there exists a unique map to a point ε : X → {pt}. Define the reducedhomology

Hq(X) := kerHq(ε).

It is an easy consequence of (8) that if X has n path components,

Hq(X) ∼=

{Hq(X) if q ≥ 1

Zn−1 if q = 0

24

More canonically, H0(X) is the kernel of the map Zπ0(X)→ Z that sends each generatorto 1. For relative homology we define

Hq(X,A) = Hq(X,A)

if A 6= ∅ is non-empty. Basically, reduced homology is designed so that Hq({pt}) = 0 forall degrees without exception and this sometimes makes calculations less clumsy.

Functoriality, homotopy invariance, and the long exact sequence all work for relativehomology. In particular, if A is non-empty, then we have a long exact sequence

· · · → Hq+1(X,A)∂→ Hq(A)→Hq(X)→Hq(X,A)

∂→ Hq−1(A)→ . . . (3)

Remark 2. If X is a path-connected space and p ∈ X, then the long exact sequencedefines a natural isomorphism

Hq(X, p) = Hq(X, p) ∼= Hq(X)

in all degrees.

3.5 Excision

The last property we need before we can do calculations is called excision. Given anordered pair (X,A) we say a subspace of B ⊆ A can be excised if the inclusion (X −B,A−B) ↪→ (X,A) induces isomorphisms

Hq(X −B,A−B) ∼= Hq(X,A)

in all degrees q.

Theorem 3.14. If the closure of B is contained in the interior of A, then B can beexcised.

Corollary 3.15. If V ⊆ B ⊂ A and

1. V can be excised, and

2. the inclusion (X −B,A−B) ↪→ (X − V,A− V ) determine homotopy equivalencesX −B ↪→ X − V and A−B ↪→ A− V ,

then B can be excised.

Proof. We want to prove that Hq(X − B,A − B) → Hq(X,A) is an isomorphism. Byfunctoriality, it is enough to show that the homomorphisms Hq(X−B,A−B)→ Hq(X−V,A−V ) andHq(X−V,A−V )→ Hq(X,A) are isomorphisms. The first is an isomorphismby homotopy invariance (Example 11) and the second is an isomorphism because V canbe excised.

We postpone the proof so that we can (finally!) do some actual calculations.

25

Proposition 3.16. The homology groups of the unit sphere Sn for n ≥ 1 satisfy

Hq(Sn) =

{Z if q = 0 or q = n

0 otherwise

Proof. It will be more convenient to work with reduced homology, so our goal is to provethat

Hq(Sn) =

{Z if q = n

0 otherwise.

Let En+ and En

− denote the upper and lower closed hemispheres of Sn. Note that for n ≥ 1,En

+ ∩ En−∼= Sn−1. I claim that by an excision that

Hq(Sn, En

−) ∼= Hq(En+, S

n−1), ∀q ∈ Z, n ≥ 1.

Here I am excising the interior of the lower hemisphere En−. This does not satisfy the

hypotheses of Theorem 3.14, but a slightly smaller open disk does and then I can applyCorollary 3.15.

Now consider the long exact sequences (LES) associated to these pairs. Because En+

is contractible, the LES of the pair (En+, S

n−1) breaks into isomorphisms

0→ Hq(En+, S

n−1)∼→ Hq−1(Sn−1)→ 0

for all n ≥ 1 and all q ∈ Z. Likewise, the LES of (Sn, En−) gives rise to isomorphisms

0→ Hq(Sn)∼→ Hq(S

n, En−)→ 0

for all n ≥ 1 and all q ∈ Z. Combined, we obtain isomorphisms

Hq(Sn) ∼= Hq−1(Sn−1)

for all n ≥ 1 and all q ∈ Z.Since S0 is a disconnected union of two points, it follows that

Hq(S0) =

{Z if q = 0

0 otherwise.

The result now follows by induction.

Proposition 3.16 hints at the special role that spheres play in homology. Later, wewill consider a special class of spaces built out of spheres called cell-complexes that areparticularly well suited to algebraic topology.

Theorem 3.17 (Brouwer Fixed Point Theorem). Let f : Dn → Dn be a continuous mapfrom the closed n-disk Dn to itself. There exists p ∈ Dn such that f(p) = p.

Proof. For the sake of contradiction, suppose that no such p exists. Then f(x) 6= x forall x ∈ Dn and we can define a continuous map r : Dn → Sn−1 as illustrated in Figure 9

Notice that for points x ∈ Sn−1, r(x) = x. This implies that r is retract. In par-ticular, this means r : Hn(Dn) → Hn(Sn−1) is surjective which contradicts the fact thatHn−1(Sn−1) ∼= Z and Hn−1(Dn) = 0.

26

Figure 9: Brouwer retraction

3.5.1 Proof of Excision

We begin with some notation. Let V := {Vi} be an open cover of a space X. Definethe chain group SVq (X) to be the subgroup of Sq(X) generated by simplices whose imagelands entirely within some Vi ∈ V . The proof of excision boils down to showing that everychain in Sq(X) can be replaced - up to a boundary - by one in SVq (X), where the relevant

open cover is {X \B, int(A)}. 5

The idea is to subdivide simplices into unions of simplices through a process calledbarycentric subdivision. Iterating the process results in smaller and smaller simplicesuntil each one is contained in some set in the cover V via a Lebesgue number argument.

First we define barycentric subdivision for affine simplices. If [v0, ..., vq] is an affinesimplex in Rn, the barycentre is the point

b =1

q + 1v0 + ...+

1

q + 1vq.

The (geometric) barycentric subdivision of [v0, ..., vq] is defined inductively to be the unionof simplices [b, w0, ..., wq−1] where [w0, ..., wq−1] arises in the barycentric subdivision of aface [v0, ..., vi, ..., vq].

Recall that the diameter of a set A ⊂ Rn equals sup{|x− y| | x, y ∈ A}.Lemma 3.18. The diameter of (the image of) any simplex arising in the barycentricsubdivision of [v0, ..., vq] is no greater than q

q+1times the diameter of [v0, ..., vq].

Proof. The distance between any two points v and∑tivi in (the image of) the simplex

[v0, ..., vq] satisfies

|v −∑

tivi| = |∑

ti(v − vi)| ≤∑

ti|v − vi| ≤ max{|vi − v|}.

5In fact, since the boundary operator ∂ sends SVq (X) to SVq−1(X), it actually forms a sub-chain complexSV(X). Hatcher proves that the inclusion SV(X) ↪→ S(X) is a chain homotopy equivalence and thusinduces an isomorphism on homology.

27

Figure 10: Barycentric subdivision

because ti ≥ 0 and t0 +...+tq = 1. Therefore the diameter of an affine simplex [v0, ..., vq] isequal to the maximum distance between vertices max{|vi − vj|}. Now let [b, w0, ..., wq−1]be a simplex in the subdivision of [v0, ..., vq]. By induction on q, we may assume that forany i, j ∈ 0, ..., q − 1, that

|wi − wj| ≤ diam([w0, ..., wq−1]) ≤ q − 1

qdiam[v0, ..., vk, ..., vq]) ≤

q

q + 1diam([v0, ..., vq]).

On the other hand, if one of the vertices is b we have a bound

|b− wj| ≤ max{|b− vi| | i = 0, ..., q}.

Notice that

b =1

q + 1vi +

q

q + 1bi

where bi is the barycentre of the face [v0, ..., vi, ..., vq]. Thus

|vi − b| =q

q + 1|vi − bi| ≤ diam([v0, ..., vn]).

completing the calculation.

28

A consequence of Lemma 3.18 is that repeatedly iterating barycentric subdivisionallows us to decompose a given simplex into simplices of arbitrarily small diameter.

We now turn this geometric construction into an algebraic one. We will define a chainmap:

B : S(X)→ S(X)

which is chain homotopic to the identity map. This means B induces the identity mapon homology. For any chain c ∈ Sq(X) we will show that Bk(c) ∈ SVq (X) for sufficientlylarge k and the theorem will follow pretty easily from that.

To begin, assume X is a convex subset of Rn and consider only chains lying in thesubcomplex Saff (X) ⊂ S(X) generated by affine chains.

Let [v0, ..., vq] be an affine simplex in (a subset of) Rn. Given b ∈ Rn, define

b[v0, ..., vq] = [b, v0, ...., vn]

and extend linearly to a map Saffq (X)→ Saffq+1(X).Observe that

∂b[v0, ..., vq] = [v0, ..., vq] +

q∑i=0

(−1)i+1[b, ..., vi, ..., vq] = [v0, ..., vq]− b∂[v0, ..., vq]

or more succinctly

∂bσ = σ − b∂σ (4)

Next we define the chain map B on Saff (X)→ Saff (X) by induction on degree. Foran affine simplex σ, denote bσ it’s barycentre. For zero simplex B0(σ) = σ. For q ≥ 1,define

Bq(σ) = bσBq−1(∂σ).

A little thought confirms that the affine simplices arising in the chain Bq(σ) are exactlythose arising in the geometric barycentric subdivision above.

We wish to show that Bq−1∂ = ∂Bq to establish that B is a chain map. By induction,suppose that Bq−2∂ = ∂Bq−1. For an affine q-simplex σ we have

∂Bq(σ) = ∂bσBq−1∂σ

= Bq−1∂σ − bσ∂Bq−1∂σ

= Bq−1∂σ − bσBq−2∂2σ

= Bq−1∂σ

using induction, (4), and ∂2 = 0.Next, we construct a chain homotopy

{Tq : Saffq (X)→ Saffq+1(X)}q∈Z.

such thatT∂ + ∂T = Id−B.

29

Figure 11: Chain homotopy geometrically

We define Tq inductively, by the T−1 = 0 and

Tqσ = bσ(σ − Tq−1∂σ)

Assume inductively that Tq−2∂ + ∂Tq−1 = Id−Bq−1. Then

∂Tqσ = ∂bσ(σ − Tq−1∂σ)

= σ − Tq−1∂σ − bσ∂(σ − Tq−1∂σ)

= σ − Tq−1∂σ − bσ(∂σ − ∂Tq−1∂σ)

= σ − Tq−1∂σ − bσ(Bq−1(∂σ)− Tq−1∂2σ)

= σ − Tq−1∂σ − bσBq−1(∂σ)

= σ − Tq−1∂σ −Bqσ

completing the induction.Finally, we extend the definition of B and of T to the singular chain complex S(X)

of an arbitrary topological space. Let δq : ∆q → ∆q be the identity map on the standardsimplex. For singular q-simplex σ define

Bσ = Sq(σ)B(δq), Tσ = Sq+1(σ)T (δq).

Since ∂ commutes both with Sq(σ) and with B for affine chains, we get the formulas

∂Bσ = ∂Sq(σ)B(δq) = Sq(σ)B(∂δq) = B∂σ

and similarly

∂Tσ + T∂σ = S(σ)(∂Tδq + T∂δq) = S(σ)(δq −Bδq) = σ −Bσ.

Remark 3. The construction of B is totally natural and independent of the space X.Thus, if A is a subspace of X, then B sends S(A) to S(A) and thus determines a chain mapB : S(X,A) → S(X,A). Because T is natural the relative chain map is also homotopicto the identity and B induces identity maps on Hq(X,A) for any pair.

30

Lemma 3.19. Given an open cover V of X, and a chain c ∈ S(X), there exists k ≥ 0such that Bk(c) ∈ SV(X).

Proof. A chain is a finite formal sum of simplices and B distributes over sums. Thus itis enough to prove that some k works for each simplex and then the largest of these willwork for the chain.

Let σ : ∆q → X be a singular simplex (a continuous map). The preimages {σ−1(Vi) | Vi ∈V} is an open cover of ∆q. Therefore, there exists a Lebesgue number d > 0 such thatsubset of ∆q of diameter less than d is contained in some open set σ−1(Vi) in the cover.Applying Lemma 3.18, by iterating barycentric subdivision a finite number k times, allsimplices in the decomposition are made to have diameter less than d. It follows thatBk(σ) ∈ SV(X).

Proof of Theorem 3.14. We are attempting to show that the map Hq(X \ B,A \ B) →Hq(X,A) is an isomorphism.

To prove surjectivity we must show that every element of Hq(X,A) is represented bya chain Sq(X) contained in the subgroup Sq(X \ B). Let η ∈ Hq(X,A) be representedby z ∈ Sq(X). By Lemma 3.19, we can choose k such that Bk(z) also represents η, andevery simplex occurring in Bk(z) maps into either X \ B or A. Discard those simpliceslanding in A to get a chain c ∈ Sq(X \B) representing η.

To prove injectivity, suppose ξ ∈ Hq(X \B,A\B) maps to 0 ∈ Hq(X,A). This meansthat ξ is represented by z ∈ Sq(X \B) and there exists β ∈ Sq+1(X) such that

∂(β) = z + α

where α ∈ Sq(A). Choose k so that the simplices of Bk(β) are contained in one of X \Bor A. Decompose Bk(β) = β1 + β2 where β1 ∈ S(X \B) and β2 ∈ S(A). Then

∂β1 + ∂β2 = ∂Bk(β) = Bk∂β = Bk(z) +Bk(α)

thus∂(β1)−Bk(z) = −∂(β2) +Bk(α)

lies in both Sq(X\B) and Sq(A), so must lie in Sq(A\B). It follows that Bk(z) ∈ Sk(X\B)represents ξ and that Bk(z) = ∂(β1) + γ where γ = (−∂(β2) + Bk(α)) ∈ Sq(A \ B), soBk(z) is a relative boundary in Sq(X \B,A \B), and we conclude that ξ = 0.

3.5.2 Mapping cylinders and cones

A subspace A ⊆ X is called a deformation retract if there is a homotopy h : X×I → Asuch that h(x, 0) = x and h(x, 1) ∈ A for all x ∈ X and h(a, t) = a for all a ∈ A and t ∈ I.The inclusion i : A ↪→ X is a homotopy equivalence (this was a homework problem).

A closed subspace A ⊆ X is called a neighbourhood deformation retract if thereexists an open neighbourhood A ⊆ U ⊆ X such that A is a deformation retract of U .

Example 12. The inclusion of S1 ⊂ R2 is a neighbourhood deformation retract becauseit includes as a deformation retract into an open annulus.

31

Proposition 3.20. Let A ⊆ X be a neighbourhood deformation retract that intersectsevery path component of X. Then there is a canonical isomorphism

Hq(X,A) ∼= Hq(X/A)

where X/A is quotient space of X obtained by identifying all of A to a point.

Proof. Since A ↪→ U is a homotopy equivalence, we know by Example 11 that

Hq(X,A) ∼= Hq(X,U)

for all q. By excision (Theorem 3.14)

Hq(X − A,U − A) ∼= Hq(X,U).

On the other hand, if we denote by A/A the point in X/A that A is collapsed to, it is nothard to see that U/A deformation retracts onto A/A. Thus

Hq(X/A) ∼= Hq(X/A,A/A) ∼= Hq((X/A)− (A/A), (U/A)− (A/A)) ∼= Hq(X −A,U −A),

where the first isomorphism follows from Remark 2 since X/A is path connected.

Example 13. The quotient space Dn/∂Dn is homeomorphic to the sphere Sn. It followsthat

Hq(Dn, ∂Dn) ∼= Hq(S

n)

which was basically what we used in the proof of Proposition 3.16.

Example 14. If X contains a contractible neighbourhood deformation retraction A,then Hq(X) ∼= Hq(X/A) for all q. (Indeed, one may show that X → X/A is a homotopyequivalence.

The hypotheses of Proposition 3.20 hold in many situations, but not always, so it isconvenient to have a construction that works in general. Let f : Y → X be a continuousmap. The mapping cylinder associated to f is the quotient space (or adjunction)

Cyl(f) := (Y × I) tf X = ((Y × I) tX)/ ∼

where the relation is generated by (y, 1) ∼ f(y) for all y ∈ Y . The inclusion X ↪→ Cyl(f)is a homotopy equivalence with homotopy inverse Cyl(f) 7→ X x 7→ x and (y, t) 7→ f(x).

The mapping cone of f : Y → X is the quotient space

Cone(f) := Cyl(f)/(Y × {0})

Proposition 3.21. Given any map of spaces f : Y → X such that the image of fintersects every path component of X, we can define a long exact sequence in homology

· · · → Hq+1(Cone(f))→ Hq(Y )→ Hq(X)→ Hq(Cone(f))→ Hq−1(Y )→ ...

in case i : Y ↪→ X is a subspace inclusion, this is canonically isomorphic to the long exactsequence of the pair

· · · → Hq+1(X, Y )→ Hq(Y )→ Hq(X)→ Hq(X, Y )→ Hq−1(Y )→ ...

32

Proof. The subspace Y ×{0} is a closed subset of Cyl(f) and is a deformation retract ofthe open subset Y × [0, 1), so Y ×{0} is a neighbourhood deformation retract in Cyl(f).Thus by ... we have canonical isomorphisms

Hq(Cone(f)) ∼= Hq(Cyl(f), Y × {0}).

Since Y ×{0} is homotopy equivalent to Y and Cyl(f) is homotopy equivalent to X, thelong exact sequence ... can be obtained from the LES of the pair (Cyl(f), Y × {0}) byreplacing groups with isomorphic groups.

In case f : Y → X is subspace inclusion, then the homotopy equivalence Cyl(f)→ Xsending (y, t) ∈ Y × I to f(y) restricts to a homeomorphism from Y × {0} to Y . Theresulting morphism of long exact sequences

Hq(Y × {0}) //

��

Hq(Cyl(f)) //

��

Hq(Cyl(f), Y × {0}) //

��

Hq−1(Y × {0}) //

��

Hq−1(Cyl(f))

��Hq(Y ) // Hq(X) // Hq(X, Y ) // Hq−1(Y ) // Hq−1(X)

which must be an isomorphism by the Five Lemma.

Example 15. The wedge sum Let Xk be a collection of spaces containing base pointspk ∈ Xk . The wedge product is the space

∨kXk = (tkXk)/ ∼

where we identify basepoints pi ∼ pj for all i, j. If the base points neighbourhood defor-mation retracts then

H(∨kXk) ∼= ⊕kH(Xk)

by Proposition 3.20.

3.6 Applications to spheres: the degree of a map

Recall that our calculation of Hn(Sn) relied on the following sequence of isomorphisms

Hn(Sn)∼= // Hn(Sn, En

−) Hn(En+, S

n−1)∼=oo

∼= // Hn(Sn−1)

We can use this to construct a cycle representing the generator of H1(S1) by the Figure12

Indeed, the cycle we have constructed is the barycentric subdivision of a simplexσ : ∆1 → S1 that winds once around the circle. It is not hard to show that (exercise) thatany ‘’chain” of 1-simplices that wraps once around the circle also represents the generatorof H1(S1).

Recall that Hn(Sn) ∼= Z. Given a continuous map f : Sn → Sn, the induced mapf∗ : Hn(Sn) → Hn(Sn) must be of the form f∗(α) = dα for some integer d ∈ Z. We calld = deg(f) the degree of the map f , .

33

Figure 12: Constructing a cycle generating H1(S1)

Figure 13: Cycles generating H1(S1)

Since Hn is a functor, we see immediately that deg(IdSn) = 1, that deg(f ◦ g) =deg(f) deg(g) for two maps f, g : Sn → Sn, and that homotopic maps are have the samedegree.6

Proposition 3.22. A map f : Sn → Sn that is not surjective has degree zero.

Proof. Suppose p ∈ Sn is not in the image of f . Then f factors through the inclusionmap Sn → Sn \{p} → Sn so by functoriality Hq(f) factors through Hq(S

n \{p}) ∼= 0 andthus must be zero.

Given a space X, define the suspension SX := X × I/ ∼ to be the quotient of X × Iwhere ∼ collapses X ×{0} and X ×{1} to distinct points. If f : X → Y is a map, definethe suspension of f

Sf : SX → SY, Sf(x, t) = (f(x), t).

This defines the suspension functor from spaces to spaces.

Lemma 3.23. The suspension of a sphere satisfies SSn = Sn+1. Given a map f : Sn →Sn, the suspension Sf : Sn+1 → Sn+1 satisfies deg(f) = deg(Sf).

Proof. The homeomorphism SSn = Sn+1 is pretty clear; this is the picture where Sn

includes into Sn+1 as the equator. Because the long exact homology sequence is functorialwith respect to pairs and the excision isomorphism is canonical, we obtain a commutative

6It is also true that two maps from Sn to Sn are homotopic if and only if they have the same degree.The proof of this is beyond the scope of this course.

34

diagram of with horizontal arrows isomorphisms

Hn(Sn) //

(Sf)∗��

Hn(Sn, En−)

(Sf)∗��

// Hn(En+, S

n−1)

(Sf)∗��

// Hn(Sn−1)

f∗��

Hn(Sn) // Hn(Sn, En−) // Hn(En

+, Sn−1) // Hn(Sn−1)

so S(f) and f have the same degree.

Proposition 3.24. Let rn : Sn → Sn be a restriction of a reflection on Rn+1 → Rn+1.Then deg(f) = −1.

Proof. For n ≥ 1, we can identify rn = Srn−1, so by induction it suffices to prove the casen = 0. In this case, S0 = {N,S} is a pair of points and r0 transposes them. The pointsrepresent 0-simplices and H0(S0) is generated by [N ]− [S]. We have

(r0)∗([N ]− [S]) = [r0(N)]− [r0(S)] = [S]− [N ] = −([N ]− [S])

so deg(r0) = −1.

We define the antipodal map on Sn by x 7→ −x.

Proposition 3.25. If f : Sn → Sn is a map with no fixed points (i.e. there is no pointp ∈ Sn such that f(p) = p), then f is homotopic to the antipodal map. In particular,deg(f) = (−1)n+1.

Proof. If f has no fixed points, then the path tf(x)− (1− t)x does not pass through theorigin. It follows that

h : Sn × I → Sn, ht(x) =tf(x)− (1− t)x|tf(x)− (1− t)x|

is a homotopy joining the antipodal map h0 to f = h1. Finally, note that the antipodalmap is equal to a composition of (n+1) reflections on Sn ⊂ Rn+1 so it has degree (−1)n+1

by Proposition 3.24.

Theorem 3.26 (Hairy Ball Theorem). Every continuous vector field on an even dimen-sional sphere has a zero.

Proof. A continuous vector field on a Sn is equivalent to a map V : Sn → Rn+1 such thatV (x) is orthogonal to x for all x ∈ Sn. If a non-vanishing vector field V exists, then wecan define an associated map f : Sn → Sn by f(x) = V (x)/|V (x)| which has no fixedpoints. By Proposition ..., this implies that deg(f) = (−1)n+1

One the other hand, since f(x) and x are always orthogonal, we can build a homotopy

h : Sn × I → Sn, ht(x) = cos(tπ/2)x+ sin(tπ/2)f(x)

between the identity map and f , from with we conclude that deg(f) = 1. If n is even,this leads to a contradiction.

35

Remark 4. In contrast with Theorem 3.26, if n is odd Sn always admits a non-vanishingvector field. This is because S2m−1 ⊆ R2m = Cm and we can use complex scalar multica-tion to rotate each vector by 90 degrees. Explicitly, V (x1, y1, ..., xn, yn) = (−y1, x1, ...,−yn, xn).

Alternative approach:For n ≥ 0, the homology group Hn(Sn) is isomorphic to Z. There are two possible

isomorphisms Hn(Sn) ∼= Z depending on a choice of generator. A choice of this generatoris called an (global) orientation of Sn.

Given a point p ∈ Sn and an open neighbourhood p ∈ U ⊆ Sn, we have canonicalisomorphisms

Hn(Sn)∼=−→ Hn(Sn, Sn − {p})

∼=←− Hn(U,U − {p}). (5)

composing the long exact sequence of the pair (Sn, Sn − {p}) with excision. A choice oforientation for Hn(U,U − {p}) ∼= Z is called a local orientation of Sn at p. Becausethe isomorphism (5) is natural, an orientation of Sn determines local orientations at allpoints p ∈ Sn, and vice versa.

Now suppose that f : Sn → Sn is a map and for some point p ∈ Sn the preimagef−1(p) is a finite set of points {q1, ..., qk} ⊂ Sn.7 Suppose further that for some openneighbourhood p ∈ U the preimage f−1(U) is a disjoint union of open sets V1 ∪ ... ∪ Vkfor which qi ∈ Vi. For each i, the restriction of f induces homomorphism

Hn(Vi, Vi \ {qi})→ Hn(U,U \ {p}).

Since both groups are isomorphic to Z, the homomorphism must be multiplication by aninteger di which we call the local degree.

Proposition 3.27. Under the conditions above, the degree of f is the sum of the localdegrees: deg(f) =

∑ki=1 di.

Proof. Fix an orientation Z = Hn(Sn) and use this to impose local orientations at allpoints. We have a commutative diagram where natural and orientation isomorphisms areindicated by double lines.

Hn(Sn)f∗ //

��

Hn(Sn)

Hn(Sn, Sn − f−1(p))f∗ // Hn(Sn, Sn − {p})

Z

A

..

⊕ki=1Hn(Vi, Vi − {qi})

f∗ // Hn(U,U − {p})

Zk B // Z7Such a point always exists if f is differentiable (Sard’s Theorem)

36

In matrix notation, we have

A =

11...1

B =

(d1 d2 ... dk

)So the composition is d1 + ...+ dk is the degree of f .

In the simplest case, p and U can be chosen so that f restricts to local homemorphismsVi → U . In this case the local degrees are all ±1, so the degree is obtained by countingpoints q1, ..., qk signs according to whether f is locally orientation preserving or reversing.

Example 16. We can construct a map Sn → Sn of degree d ≥ 2 as follows. Let A ⊂ Sn

be the complement of d disjoint open disks Bi in Sn. Let

q : Sn → Sn = X/A ∼= ∨dSn

be the quotient map. The orientation on Sn induces local orientations and hence globalorientations on each sphere in the wedge sum. Let

p : ∨dSn → Sn

map each sphere by a degree 1 homeomorphism to Sn.The preimage (pq)−1(y) of a generic point y ∈ Sn consists of a single point in each

disk Bi each with local degree is 1 because pq is a local homeomorphism. Thereforedeg(pq) = d. By precomposing pq with a reflection, we can construct a map of degree −d.

Consider the map given d ∈ Z

wd : S1 → S1, wd(eiθ) = eidθ

for d ≥ 1 we can see by Figure 14 that deg(wd) = d.

Figure 14: the winding map w4

37

Note that w−d is equal to the composition of wd with a reflection, so deg(w−d) = −d.By suspension, we construct maps

Snwd : Sn+1 → Sn+1

of degree d for any integer.

Theorem 3.28 (Fundamental Theorem of Algebra). A complex polynomial function f(z)of degree d ≥ 1 has a complex root.

Proof. The case d = 1 is obvious, so suppose that d ≥ 2. We assume f is monic forsimplicity so f(z) = zd + O(zd−1). Assume that f(z) has no complex roots. Then thereis a well-defined, continuous map

g : C→ S1, g(z) =f(z)

|f(z)|

(=

zd +O(zd−1)

|zd +O(zd−1)|

).

Define a homotopy h : S1 × I → S1 by

ht(eiθ) = g(

t

1− teiθ).

for t < 1 and extend by continuity for t = 1. We have h0(eiθ) = g(0) is a constant andthus deg(h0) = 0. On the other hand, for large values of z, g(z) becomes dominated bythe highest degree terms in the numerator and denominator, so in the limit t → 1, wehave

h1(eiθ) = eidθ

so deg(h1) = d, which contradicts degree being a homotopy invariant.

3.7 Cellular homology

3.7.1 Cell complexes

LetDn := {x ∈ Rn||x| ≤ 1}

denote the unit disk or closed n-cell with boundary

Sn−1 = ∂Dn := {x ∈ Rn||x| = 1}.

Given a topological space X and a continuous map f : Sn−1 → X, we may construct anew space

Y := (X qDn)/ ∼

where we quotient by the equivalence relation generated by p ∼ f(p) for all p ∈ Sn−1.We say that Y is obtained from X by attaching an n-cell; the map f is called theattaching map. More generally, if we have a collection of maps fα : Sn−1 → X, then weconstruct

Y = (X q (∐α

Dnα))/ ∼

38

where p ∼ fα(p) for all p ∈ Sn−1α and α.

A cell complex (also called CW-complex) is a space that is constructed inductivelyby attaching cells. For instance,

• A 0-dimensional complex X0 is a discrete set of points (i.e. a disconnected union of0-cells).

• A 1-dimensional cell complex X1 is a space constructed by attaching a collection of1-cells to X0.

• A 2-dimensional cell complex X2 is constructed by attaching 2-cells to X1.

• and so on ...

In general, a cell complex X may have cells in arbitrarily high dimensions, in which caseit is called ∞-dimensional. Each n-cell determines a characteristic map φα : Dn → X.A subset S ⊆ X is open/closed if and only if φ−1

α (S) ⊆ Dn is open/closed for all cells.

Example 17. A wedge of n-spheres ∨ISn is constructed by attaching I many n-cells ontoa point X0 = {p} by the only possible attaching map f : Sn−1 → {p}.

Example 18. The torus S1 × S1 can be constructed by attaching a 2-cell onto a wedgeof two circles X = S1 ∨ S1. If we denote by a and b the loops defined by the two circlesin X, then the attaching map f : S1 → X is the loop a · b · a−1 · b−1.

Example 19. More generally, the genus g surface Σg is constructed by gluing a 2-cell toa wedge of 2g circles. If the loops defined by the circles are called a1, b1, ..., ag, bg, thenthe attaching map sends S1 to the concatenation

∏gi=1[ai, bi], where [ai, bi] = aibia

−1i b−1

i

is the commutator.

A subspace A ⊆ X is called a subcomplex if it is a closed union of cells (that is,of images of characteristic maps). Given a subcomplex A ⊆ X, the quotient space X/Adefined by identifying all points in A with each other, is naturally a cell complex called aquotient complex of X.

Proposition 3.29. A subcomplex A of a cell complex X is a neighbourhood deformationretract. Thus, Hq(X,A) ∼= Hq(X/A) we have a long exact sequence in homology

. . . Hq+1(X/A)→ Hq(A)→ Hq(X)→ Hq(X/A)→ Hq−1(A)→ . . .

Proof. Skipped.

The subcomplex Xn ⊆ X consisting of all cells of dimension ≤ n is called the n-skeleton of X (by convention X−1 = ∅). If X is infinite dimensional, the topology onX satisfies S ⊂ X is open (resp. closed) if and only if S ∩ Xn is open (resp. closed) inXn for all n. In particular, a map f : X → Y is continuous if and only if the restrictionsfn : Xn → Y are continuous for all n. We say X has the direct limit topology with respectto Xn.

39

Figure 15: orientable surfaces

Lemma 3.30. Let X be a cell complex and C ⊂ X a compact subspace. Then C iscontained within finitely many cells of X.

Proof. Choose a sequence of points xi ∈ C lying in distinct cells. We will show that theset S := {xi} is finite. We begin by showing S is closed.

First observe that

S ⊆ X is closed⇔ S ∩Xn is closed in X, ∀ n

We use induction on n.Clearly S∩X0 is closed in X0 hence in X, because every subset of X0 is closed. Assume

by induction that S ∩Xn−1 is closed in X. Thus for any characteristic map

φα : Dk → X

the pre-image φ−1α (S ∩ Xn−1) is closed in Dk. For k < n, the pre-image φ−1

α (S ∩ Xn) =φ−1α )(S ∩ Xn−1) is closed in Dk. For k = n, the pre-image φ−1

α (S ∩ Xn) ⊆ Dn equalsφ−1α (S∩Xn−1) plus at most one point, thus it is a union of two closed sets, hence is closed

in Dn. We deduce that S ∩Xn is closed in Xn hence also in X. By induction, this holdsfor all n so S is closed in X.

The same argument shows that every subset of S is also closed, so S has the discretetopology. But S is a closed subset of the compact set C, so it is compact. We concludethat S is finite.

3.7.2 Cell complex propaganda (not to be tested)

We present some results showing that many interesting spaces are either homeomorphicor homotopy equivalent to cell complexes. The material in this section will not be tested.

40

Definition 17. A real analytic function f : Rn → R is a infinitely differentiablefunction such that at every point p ∈ Rn, f equals its Taylor series at p on some positiveradius. A real analytic set X ⊂ Rn is the solution set finite collection of equationsf1(x) = ... = fn(x) = 0, for fi real analytic.

Example 20. Polynomial functions, exponential functions, trigonometric functions, etc.are real analytic.

Theorem 3.31 (Lojasiewicz 1964). Every real analytic set X ⊆ Rn is homeomorphic toa cell complex.

Let X and Y be two topological spaces. Let Cont(X, Y ) be the set of all continuousmaps from X to Y with the compact-open topology.

Example 21. The space LY = Cont(S1, Y ) is called the free loop space of Y .

Theorem 3.32 (Milnor 1959). If X and Y are cell complexes and X is compact, thenCont(X, Y ) is homotopy equivalent to a cell complex.

Definition 18. A topological space X is called a (topological) n-manifold if it is Haus-dorff and if every point p is contained in an open neighbourhood p ∈ U ⊂ X that ishomeomorphic to Rn.

• Every open set in Rn is an n-manifold.

• The sphere Sn is an n-manifold.

• Surfaces of any genus are 2-manifolds.

• The product of an m-manifold and an n-manifold is an m+ n-manifold.

An example of a space that is locally Euclidean but is not a manifold is constructedby taking two copies of the real line R q R = R × {a, b} and forming the quotient by(t, a) ∼ (t, b) if t 6= 0. This space looks locally like R, but the points (0, a) and (0, b)cannot be separated by open sets.

Theorem 3.33. Every compact n-manifold is homotopy equivalent to a cell complex. Itremains an open question whether or not every compact n-manifold is homeomorphic toa cell complex.

3.7.3 Cellular Homology

Lemma 3.34. If X is a cell complex, then:

(a) Hq(Xn, Xn−1) is zero if q 6= n and is an free abelian group with generators corre-sponding to the n-cells when q = n.

(b) Hq(Xn) = 0 for q > n. Thus Hq(X) = 0 for q > dim(X).

41

(c) The inclusion i : Xn ↪→ X induces an isomorphism Hq(i) : Hq(Xn) → Hq(X) forq < n.

Proof. By Proposition 3.29, we have isomorphism Hq(Xn, Xn−1) ∼= Hq(Xn/Xn−1) andXn/Xn−1 is a wedge of spheres indexed by the n-cells of X. Property (a) follows.

Property (b) is proven by induction. Clearly true for n = 0. Now suppose it has beenproven for n− 1. The long exact sequence of the pair contains

→ Hq(Xn−1)→ Hq(Xn)→ Hq(Xn, Xn−1)→

where both Hq(Xn−1) = Hq(Xn, Xn−1) = 0 for q > n by induction and property (a). ThusHq(Xn) = 0 as well.

To prove property (c), consider the exact sequence

Hq+1(Xn+1, Xn)→ Hq(Xn)→ Hq(Xn+1)→ Hq(Xn+1, Xn).

By (a), the two groups on the end vanish if q < n so Hq(Xn) ∼= Hq(Xn+1). Repeating thisargument, we get

Hq(Xn) ∼= Hq(Xn+1) ∼= Hq(Xn+2) ∼= ...

which suffices if X is finite dimensional. To take care of the infinite dimensional case,observe that Lemma 3.30 implies that every chain in Sq(X) must be in the image ofSq(Xn) for some n (since the union of images of simplices occurring in the chain is acompact subset of X). Thus every cycle Zq(X) arises as the image of a cycle in Zq(Xn)for some n, and every boundary in Bq(Xn) arises as the image of a boundary in Bq(Xn)for some n. The result follows (details left to reader).

Define a homomorphism dn : Hn(Xn, Xn−1)→ Hn−1(Xn−1, Xn−2) by the commutativediagram

0

0

((PPPPPPPPPPPPPPP Hn(Xn+1) ∼= Hn(X)

55jjjjjjjjjjjjjjjjjj

Hn(Xn)jn

((RRRRRRRRRRRRR

66lllllllllllll

Hn+1(Xn+1, Xn)dn+1 //

∂n+1

66nnnnnnnnnnnnHn(Xn, Xn−1)

dn //

∂n ))SSSSSSSSSSSSSSHn−1(Xn−1, Xn−2)

Hn−1(Xn−1)

jn−1

55llllllllllllll

0

55jjjjjjjjjjjjjjjjjj

42

where the diagonal maps occur in the long exact sequences of pairs. Notice that dn◦dn+1 =0 because it factors through ∂n ◦ jn = 0. Thus (Hn(Xn, Xn−1), dn)n∈Z forms a chaincomplex, called the cellular chain complex. The homology of the cellular chain complexis called the cellular homology.

Theorem 3.35. The cellular homology groups are naturally isomorphic to the singularhomology groups.

Proof. From diagram, we may identify Hn(X) ∼= Hn(Xn)/im(∂n+1). Since jn is injective,this is isomorphic to im(jn)/im(dn+1). By exactness, this is the same as ker(∂n)/im(dn+1).Finally, because jn−1 is injective, this is equal to ker(dn)/im(dn+1).

Theorem 3.35 is very useful for calculations, because it allows us to replace the usuallyuncountably infinite rank Sq(X) by the often finite rank Hn(Xn, Xn−1). Before gettingstarted with examples, we want a more direct understanding of the boundary maps dn.

Denote by {enα}α the set of n-cells of a cell complex X, so that Hn(Xn, Xn−1) is thefree abelian group generated by {enα}α.

Proposition 3.36. For n > 1, the cellular boundary map satisfies

dn(enα) =∑β

dα,βen−1β

where dα,β is the degree of the map

Sn−1α → Xn−1 → Sn−1

β

defined by composing the attaching map of enα with the quotient map Xn−1 → Sn−1β =

Xn−1/(Xn−1 − enβ).

Proof. The proof is based on the following commutative diagram

Hn(Dnα, ∂D

nα) ∂

∼=//

Φα∗��

Hn−1(∂Dnα)

fα∗��

(∆α,β)∗ // Hn−1(Sn−1β )

Hn(Xn, Xn−1)∂n //

dn

))SSSSSSSSSSSSSS Hn−1(Xn−1)q∗ //

jn−1

��

Hn−1(Xn−1/Xn−2)

∼=��

qβ∗

OO

Hn−1(Xn−1, Xn−2)∼= // Hn−1(Xn−1/Xn−2, Xn−2/Xn−2)

where

• Φα is the characteristic map for enα and fα the attaching map.

• q : Xn−1 → Xn−1/Xn−2 is the quotient map.

• qβ : Xn−1/Xn−2 → Sn−1β is the quotient map obtained by collapsing everything be

the β sphere to a point.

43

• ∆α,β = qβ ◦ q ◦ fα is the map whose degree is dα,β.

The map Φα∗ sends the generator of Hn(Dnα, ∂D

nα) to enα and qβ∗ can be understood as

projection onto the β-summand of Hn−1(Xn−1/Xn−2) ∼= Hn−1(Xn−1, Xn−2). The resultfollows by a diagram chase.

3.7.4 Examples

Example 22. Recall that a genus g surface Σg is constructed by attaching a 2-cell toa wedge of 2g circles using an attaching map

∏gi=1[ai, bi]. The cellular complex is:

0→ Z d2−→ Z2g d1−→ Z

The boundary map d1 is zero, because each 1-cell meets the 0-cell twice and the attachingmap S0 → {pt} sends H0(S0) to zero by definition. The boundary map d2 is also zero,because the attaching map winds around each loop twice, but in opposite directions,giving total degree zero. It follows that

Hq(Σg) =

Z if q = 0, 2

Z2g if q = 1

0 otherwise

Example 23. The non-orientable surface Ng of genus g ≥ 0 is constructed by attach-ing a single 2-cell to a wedge sum of g + 1 circles a0, ..., ag by the attaching map a2

0...a2g.

The surface N0 in the real projective plane and N1 is the Klein bottle. The cellularchain complex is

Z d2−→ Zg+1 d1−→ Z.As before, d1 = 0. The attaching map for the 2-cell winds twice around each circle in thesame direction and thus has degree 2 for each 1-cell. Consequently, d2(n) = (2n, ..., 2n). Ifwe do a change of basis for Z2g using generators (1, 1, ..., 1), (0, 1, 0, ..., 0), ..., (0, 0, ..., 0, 1)then with respect to the new basis, d2(n) = (2, 0, ..., 0).

Hq(Ng) =

Z if q = 0

Z2 ⊕ Zg if q = 1

0 otherwise

Example 24. A product of spheres Sm × Sn with m,n ≥ 1 has the structure of acell complex with four cells, in dimensions 0, m, n, and m+n. To see this observe thatDm+n ∼= Dm ×Dn and

Sm × Sn ∼= (Dm ×Dn)/ ∼ ∼= (Dm/ ∼)× (Dn/ ∼)

where ∼ is generated by the relations (x, y) ∼ (x, y′) if y, y′ ∈ ∂Dn and (x, y) ∼ (x′, y)if x, x′ ∈ ∂Dm. Since the relations occur only in the boundary of Dm ×Dn, this can beunderstood as attaching a (m+ n)-cell to the quotient of

∂(Dm ×Dn) = ∂Dm ×Dn ∪Dm × ∂Dn

44

Figure 16: non-orientable surfaces

which is identified as the wedge sum Sm ∨ Sn, which is a cell complex with cells indimension 0, m, and n (compare the case of a torus S1 × S1).

Suppose that n ≥ m and n > 1. Then the (n+ 1)-skeleton is equal to Sm ∨ Sn so byLemma 3.34, Hq(S

m × Sn) = Hq(Sm ∨ Sn) for q ≤ n. It follows that the boundary map

in the cellular chain complex is trivial and that

Hq(Sm × Sn) =

{Z if q = 0,m, n, or m+ n

0 otherwise

Example 25. Real projective space RP n is the set of 1-dimensional vector subspacesof Rn+1. Topologically, RP n is the quotient space of Rn − {0} by the relation v ∼ λv forλ ∈ R∗ a non-zero scalar. Equivalently, RP n is identified with quotient space Sn/ ∼ bythe relation v ∼ −v.

By restricting to a closed hemisphere of Sn, we see that RP n can be obtained byattaching an n-cell to RP n−1 using the quotient Sn−1 → RP n as the attaching map. Byinduction, RP n = e0 ∪ e1 ∪ ... ∪ en is a cell complex with one cell in each dimension ≤ n.Thus, the cellular chain complex looks like

0→ Z dn−→ Z dn−1−→ · · · d1−→ Z→ 0

To understand the boundary maps, we must determine the degrees of the composition ofthe attaching map and the quotient map

Sq−1 φ→ RP q−1 q→ RP q−1/RP q−2 ∼= Sq−1.

Observe that q ◦ φ restricts to local homeomorphisms from each of the two open hemi-spheres Sq−1 − Sq−2 onto Sq−1 − {pt} so by ... the degree is either 0 or ±2. Notice thatthese two restrictions are interchanged by precomposing with the antipodal map whichhas degree (−1)q. Thus the local degrees have the same sign if q is even and differentsigns if q is odd. Thus dq is multiplication by 0 if q is odd and ±2 if q is even. The chaincomplex is

0→ Z 0−→ Z 2−→ Z 0−→ · · · 2−→ Z 0−→ Z→ 0

45

if n is odd and0→ Z 2−→ Z 0−→ Z 2−→ · · · 2−→ Z 0−→ Z→ 0

if n is even. We obtain homology groups,

Hq(RP n) =

Z if q = 0 or q = n and n is odd

Z2 if 0 < q < (n− 1) and is odd

0 otherwise

Example 26. Complex projective space CP n is the set of one dimensional vectorsubspaces of Cn+1. It may also be construct as the quotient of the sphere S2n+1 by therelation v ∼ λv where λ ∈ S1 is a unit scalar. Similar to the real projective spaces, wecan construct CP n inductively by attaching a 2n-dimensional cell to CP n−1. To see this,consider the embedding from D2n ⊆ Cn to Cn × C by

w 7→ (w,√

1− |w|2).

The boundary of D2n is sent to the unit sphere in S2n−1 ⊂ Cn × {0} and there is a one-to-one correspondence between the interior of Dn and the one-dimensional subspaces ofCn×C not contained in Cn×{0}. Thus CP n is obtained by attaching D2n to CP n−1 bythe quotient map S2n−1 → CP n−1.

Since CP n = e0∪e2∪...∪e2n only has cells in even dimension, this means the boundarymaps in the cellular chain complex are necessarily zero and we obtain

Hq(CP n) =

{Z if 0 ≤ q ≤ 2n and is even

0 otherwise

3.8 Mayer-Vietoris Sequence

The Mayer-Vietoris sequence is an alternative to the long exact sequence of a pair (X,A)that sometimes more convenient to use.

Let X be a topological space and let A,B ⊆ X be a pair of subspaces such thatA ∪B = X (i.e. a covering of X).

Theorem 3.37. Let X be a topological space and let A,B ⊆ X be a pair of subspacessuch that A ∪B = X. If we have an excision isomorphism

Hq(B,A ∩B) = Hq(X \Bc, A \Bc) ∼= Hq(X,A)

then there is a long exact sequence in homology

· · · → Hn(A ∩B)I→ Hn(A)⊕Hn(B)

J→ Hn(X)D→ Hn−1(A ∩B)→ . . .

where I, J , and D are defined below. A similar sequence exists for reduced homology

· · · → Hn(A ∩B)I→ Hn(A)⊕ Hn(B)

J→ Hn(X)D→ Hn−1(A ∩B)→ . . . .

46

Example 27. The excision hypothesis of Theorem 3.37 is satisified if

int(A) ∪ int(B) = X

or if X is a cell complex and A, B are sub-cell complexes.

Proof of Theorem 3.37. consider that inclusion determines of pairs (X \ Bc, A \ Bc) =(B,A ∩B)→ (X,A) and thus a morphism of long exact sequences

. . . // Hq+1(B,A ∩B) //

∼=��

Hq(A ∩B) //

��

Hq(B) //

��

Hq(B,A ∩B) //

∼=��

. . .

. . . // Hq+1(X,A) // Hq(A) // Hq(X) // Hq(X,A) // . . .

where every third map between the sequences is an isomorphism. The results follows bythe following lemma.

Lemma 3.38. Given a commuting diagram

// Ci+1

γi+1∼=��

hi+1 // Aifi //

αi��

Bi

βi��

gi // Cihi //

γi∼=��

Ai−1

αi−1

��

//

// C ′i+1

h′i+1 // A′if ′i // B′i

g′i // C ′ih′i // A′i−1

//

in which the rows are exact and the γi are isomorphisms, then there is a long exactsequence

→ AiI→ A′i ⊕Bi

J→ B′iD→ Ai−1 →

where

I(a) = (αi(a), fi(a)),

J(a′, b) = f ′i(a)− βi(b),D(b′) = hiγ

−1i g′i(b

′).

Proof. Diagram chase.

Corollary 3.39. For a path connected space X, there is a canonical isomorphism

Hq+1(SX) ∼= Hq(X)

for all q ∈ Z, where SX denotes the suspension of X.

Proof. Homework exercise.

47

3.9 Homology with coefficients

So far we have developed singular homology theory for integer coefficients, meaning thatour chains are finite formal sums

∑σ aσσ with coefficients aσ ∈ Z. More generally, it is

possible (and useful) to work with coefficients in any commutative ring R with identity1 ∈ R; in particular, this means we have a canonical ring homomorphism Z → R. Themost interesting cases are when R = Zn is the ring of integers modulo n, or when R is afield, such as Q, R, C or Zp for prime p.

This generalization is pretty straight forward. Define Sq(X;R) to be the free R-modulewhose chains are finite sums the form

∑σ aσσ where the coefficients aσ are elements

of R. So far we have been studying the case Sq(X) = Sq(X; Z). The boundary map∂q : Sq(X;R)→ Sq−1(X;R) is defined as before

∂q(∑σ

aσσ) =∑σ

∂(aσσ) =∑σ

q∑i=0

(−1)iaσσ(i)

and satisfies ∂2 = 0 giving rise to a chain complex S(X;R) = (Sq(X;R); ∂q). Thisdetermines cycles Zq(X;R) := ker(∂q), boundaries Bq(X;R) := im(∂q+1) and homology

Hq(X;R) := Zq(X;R)/Bq(X;R),

all of which are now R-modules. We call Hq(X;R) singular homology with R-coefficients.8

The proofs of all the main theorems carry over unchanged. In particular

• Hq(−;R) is a functor for topological spaces to the category of R-modules. Thus forevery continuous map f : X → Y , there is a homomorphism of R-modules

Hq(f ;R) = f∗ : Hq(X;R)→ Hq(Y ;R)

such that (f ◦ g)∗ = f∗ ◦ g∗ and (IdX)∗ = IdHq(X;R).

• If f, g : X → Y are homotopic, then f∗ = g∗.

• A topological pair (X,A) gives rise to a long exact sequence of R-modules

→ Hq(A;R)→ Hq(X;R)→ Hq(X,A;R)∂→ Hq−1(A;R)→ ...

where the connecting homomorphism is defined using the same diagram chase asbefore.

• The excision isomorphism Hq(X \B,A\B : R) ∼= Hq(X,A;R) hold under the samehypotheses as before.

• H0(X;R) is a free R-module generated by the path components of X.

8Note that an abelian group is the same thing as a Z-module. If R is a field, an R-module is just anR-vector space.

48

• H0(pt;R) = R and Hq(pt;R) = 0 for q 6= 0. Define Hq(X;R) to be the kernel of thenatural map Hq(X;R)→ Hq(pt;R).

• Hn(Sn;R) ∼= R and Hq(Sn;R) = 0 for q 6= n. If f : Sn → Sn is a degree d map,

then f∗ is simply multiplication by d ∈ R.

• If X is a cell complex, then Hn(Xn, Xn−1;R) is a free R-module with genera-tors corresponding to n-cells. We can form a cellular chain complex Cn(X;R) =Hn(Xn, Xn−1;R) and boundary map defined by ... and cellular homology is natu-rally isomorphic to singular homology.

• The Mayer-Vietoris Sequence is still exact and defined in the same way.

One of the big advantages in working with coefficients in a field F is that homologyare now vector spaces. This means for instance that short exact sequences always splitand this can simplify a lot of calculations.

Example 28. Recall that the cellular chain complex C(RP 2) (Example 25) is either of

0→ Z 0−→ Z 2−→ Z 0−→ · · · 2−→ Z 0−→ Z→ 0

0→ Z 2−→ Z 0−→ Z 2−→ · · · 2−→ Z 0−→ Z→ 0

depending on whether n is odd or even.Now consider the cellular chain complex C(RP n; Z2). The chain groups become free

Z2-modules and the boundary maps are all zero because 0 = 2 in Z2

0→ Z20=2−→ Z2

0=2−→ Z20=2−→ · · · 0=2−→ Z2

0=2−→ Z2 → 0.

Thus

Hq(RP n; Z2) =

{Z2 for q = 0, ..., n

0 otherwise.

On the other hand, if F is a field of characteristic other than 2, then 2 is a unit. Thecellular chain complexes becomes one of

0→ F0−→ F

∼=−→ F0−→ · · ·

∼=−→ F0−→ F → 0

0→ F∼=−→ F

0−→ F∼=−→ · · ·

∼=−→ F0−→ F → 0

and we get homology groups

Hq(RP n;F ) =

{F if q = 0 or q = n and n is odd

0 otherwise.

49

The only isomorphism invariant of a vector space is the dimension. Given a field F thedimensions dim(Hq(X;F )) are called Betti numbers of X. These are basic invariantsof a space X that are often easier to compute than the homology groups Hq(X; Z).

Suppose that X is a space for which dim(Hq(X);F ) is finite in all degrees and is zeroin all but finitely many degrees. Define the Euler characteristic

χ(X) :=∞∑q=0

(−1)q dim(Hq(X;F ))

to be the alternating sum of the Betti numbers.

Proposition 3.40. If X is finite cell complex, the Euler characteristic is well defined andsatisfies

χ(X) :=∞∑n=0

(−1)ncn

where cn is the number of cells of dimension n. In particular, χ(X) is a homotopy invari-ant of X and is independent of the field F .

Proof. We will simplify notation to prove this result in a more abstract context. Supposewe have a finite length chain complex of vector spaces

0→ CN∂n−→ ...

∂1−→ C0∂0−→ 0

for which each Cq is finite dimension cq. For each q we have short exact sequences ofvector spaces

0→ Zq → Cq → Bq−1 → 0

and0→ Bq → Zq → Hq → 0.

Denote by zq, bq, hq the dimensions of Zq, Bq, Hq respectively. The rank nullity theoremgives us equations

cq = zq + bq−1, hq = zq − bqfor all q ∈ Z. In case Cq = Cq(X;F ) we get

χ(X) =∑q

(−1)qcq

=∑q

(−1)q(zq + bq−1)

=∑q

(−1)qzq − (−1)q−1bq−1

=∑q

(−1)q(zq − bq)

=∑q

(−1)qhq

50

• A surface of genus g has Euler characteristic 1− 2g + 1 = 2− 2g.

• CP n has Euler characteristic 1 + 0 + 1 + 0 + ...+ 1 = n+ 1.

• A sphere has Euler characteristic 1 + 1 = 2 if n is even and 1− 1 = 0 if n is odd.

A well known example of the Euler characteristic comes up for the boundaries of convexpolytopes, like a tetrahedron or a cube. The Euler characteristic is then V −E+F whereV is the number of vertices, E is the number of edges, and F is the number of faces. Sincethe boundary of a convex polytope must be homeomorphic to S2, it follows that

V − E + F = 2

Theorem 3.41 (Lefshetz Fixed Point Theorem). Let X be a finite cell complex and letf : X → X be a continuous map. The Lefshetz number for f is the number

L(f) :=∞∑q=0

(−1)qTr(Hq(f ; Q))

where Tr(Hq(f ; Q)) is the trace of the linear map Hq(f ; Q) : Hq(X; Q) → Hq(X; Q). IfL(f) 6= 0, then f has a fixed point.

Some examples

• (Generalization of Brower Fixed Point Theorem) If X is a finite cell complex suchthat H0(X; Q) = Q and Hq(X; Q) = 0 for q > 0, then L(f) = 1 for any mapf : X → X. Consequently every f must have a fixed point.

• Note that it is a necessary condition that X is finite, because the translation mapt : R→ R, t(x) = x+ 1 has L(t) = 1 but no fixed points.

• A map of spheres f : Sn → Sn with degree d has Lefshetz number L(f) = 1+(−1)ndwhich vanishes if and only if d = (−1)n+1, recovering ...

• If f is homotopic to the identity map, then L(f) = χ(X). Thus if χ(X) 6= 0 thenany f homotopic to the identity must have a fixed point.

Proof. We will only briefly sketch the argument.Suppose that the map f : X → X restricts to maps f : Xk → Xk for each skeleton.

If this happens, it is not hard to show that f determines a morphism from the cellularchain complex of X to itself:

. . . // Cq+1(X) //

Cq+1(f)

��

Cq(X) //

Cq(f)

��

Cq−1(X) //

Cq−1(f)

��

. . .

. . . // Cq+1(X) // Cq(X) // Cq−1(X) // . . .

51

Using an argument similar to the proof of Proposition ..., one shows that

L(f) =∞∑q=0

(−1)qTr(Cq(f ; Q)).

In particular, this means that if L(f) 6= 0, then Tr(Cq(f ; Q)) 6= 0 for some q. This impliesin particular that some cell ek ∈ X must map onto itself, i.e. ek ⊆ f(ek).

To go further, we must impose more assumptions on X and f . First, we suppose thatX has the special property that the attaching maps of cells are all injective and hencethat the characteristic map Dk → X are injective. Next, we suppose that f sends eachk-cell either in to the k− 1-skeleton or homeomorphically onto a k-cell. These conditionsare satisfied if X is a simplical complex and f is a simplicial map.

With these assumptions imposed, we deduce that L(f) 6= 0 then f sends some k-cellto itself. By the Brouwer Fixed Point Theorem, we deduce that there must be a fixedpoint.

To make this argument rigorous, we would have to show that any continuous mapf : X → X without fixed points can be replaced “up to homotopy” by a simplical mapf ′ : X ′ → X ′ without fixed points. This is true (see 2.C of Hatcher), but the proof israther long and technical.

3.9.1 The Universal Coefficient Theorem for Homology

The Universal Coefficient Theorem for Homology allows Hq(X;R) to be calculated fromHq(X; Z) and Hq−1(X; Z). We need some definitions before stating the formula.

Every Abelian group G can be fit into a short exact sequence of the form

0→ Zr A→ Zg → G→ 0 (6)

where Zr and Zg are (possibly infinitely generated) free abelian groups on r and g gen-erators respectively. We call (6) a free resolution of G. We think of the basis of Zg asthe generators of G and the basis of Zr as relations. The homomorphism A is defined interms of an g × r matrix.

Example 29. For example, every finitely generated abelian group is isomorphic to

G = Zf ⊕ Zn1 ⊕ ...⊕ Znt

for some choice f , n1,..., nt. The integer f is called the rank of G. This fits into ashort exact sequence

0→ Zt A→ Zf ⊕ Zt → G→ 0 (7)

where A = (0, A′) and A′ is the diagonal matrix with entries n1, ..., nt.

Now let R be another abelian group. Given a free resolution (6) of G, we obtain amap

Rr AR−→ Rg

52

where AR is given by the same integral matrix as A. We define

Tor0Z(G,R) := cok(AR) = Rg/im(AR).

T or1Z(G,R) := ker(AR).

Equivalently, these are the homology groups of the chain complex 0→ Rr → Rg → 0.The group Tor0

Z(G,R) is known as the tensor product of G and R and is more com-monly denoted

G⊗Z R := TorZ(G,R).

If R is a ring, then both Tor0Z(G,R) and Tor1

Z(G,R) are R-modules.

Theorem 3.42 (Universal Coefficient Theorem). For any space X and coefficient ringR, there is a non-canonical isomorphism

Hq(X;R) = Tor0Z(Hq(X), R)⊕ Tor1

Z(Hq−1(X), R).

Proof. Skipped. See Hatcher 3.A for a proof.

The moral of Theorem 3.42 is that homology groups with coefficents in R contain nomore information than homology groups with integer coefficients. Here are a couple ofeasy consequences.

Corollary 3.43. If Hq(X; Z) is free abelian in every degree, then Hq(X;R) is a freeR-module of the same rank in each degree.

Proof. For a free abelian group G = Zg, we form a resolution

0→ 0→ Zg ∼=→ G→ 0.

Thus Tor0Z(Zg, R) = Rg and Tor1

Z(Zg, R) = 0 for any R for any R. It follows that

Hq(X;R) = Tor0Z(Hq(X), R)⊕ Tor1

Z(Hq−1(X), R) = Hq(X)⊗Z R = Rn

where n is the rank of Hq(X).

Corollary 3.44. Suppose that Hq(X;F ) is finitely generated in all degrees. If F is afield, then

dimF (Hq(X;F )) ≥ rank(Hq(X; Z))

with equality if F has characteristic zero (e.g. F = Q, R, C).

Proof. For a finitely generated abelian group G, we have resolution (7) above. If F is afield then Tor0

Z(G,F ) and Tor1Z(G,F ) are vector spaces of dimension f + t − rank(A′F )

and t− rank(A′F ). Consequently

Hq(X;F ) = Tor0Z(Hq(X), F )⊕ Tor1

Z(Hq−1(X), F )

has dimension at least as large as the rank ofHq(X) with equality if and only if rank(A′F ) =t respectively. If F has characteristic zero, then equality holds.

53

3.10 Covering spaces and the transfer

A covering map f : X → Y is a continuous map such that for every point p ∈ Y , thereis an open neighbourhood p ∈ U ⊆ Y such that the preimage f−1(U) is a disjoint unionof open sets ∪Vi ⊂ X each mapping homeomorphically to U by f (we call U a trivializingopen set for the cover). We call X a covering space of Y . The Vi are called “sheets” ofthe cover. If Y is path connected, then the number of sheets is independent of p and Uand f : X → Y is called an n-sheeted cover if the number of sheets over each point isn.

Some examples

• The map R→ S1 defined by t 7→ e2πit is a ∞-sheeted cover.

• The winding map wd : S1 → S1 is a d-sheeted cover (for d > 1).

• The quotient map Sn → RP n is a 2-sheeted cover.

• Given a pair of coprime integers p, q ∈ Z, the lense space Lp,q is the quotient spaceS3/ ∼ where we view S3 ⊆ C2 as the unit sphere and impose relation (z, w) =(e2πi/pz, e2πiq/pz). Each equivalence class consists of a exactly p points and thequotient map S3 → Lp,q is a p-sheeted cover.

The basic property of covering spaces used for all applications in topology is thefollowing.

Lemma 3.45. Let f : X → Y be a covering space and let σ : ∆q → Y be a singularsimplex. Let x ∈ X satisfy f(x) = σ(~0). Then there is a unique singular simplex σ :∆q → X such that f ◦ σ = σ.

X

f

��∆q

σ //

σ>>}}}}}}}}Y

Proof. This is proven in the course on homotopy theory. We sketch the argument.If the image of σ lies in an trivializing open set U ⊂ Y , then σ is obtained by composing

σ with the homeomorphism f−1 : U ∼= Vi to the sheet Vi containing x. This lift is clearlyunique.

For the general case, simply decompose ∆q using iterated barycentric subdivision untileach subsimplex lands in a trivializing set U ⊂ Y and then lift σ one simplex at a time.

A deck transformation of a covering space is a homeomorphism φ : X → X such thatf = f ◦ φ.

Xφ //

f AAA

AAAA

X

f~~~~~~

~~~~

Y

54

The deck transformations form a group under composition that acts on X. The coveringspace is called Galois if the deck transformation group acts freely and transitively onfibres. I.e. if for any two points x, x′ ∈ f−1(y) there is a unique deck transformation φsuch that φ(x) = x′. In this case, the preimages of points f−1(y) are equal to the orbitsof a free group action on X.

In the examples above, the deck transformation groups Z acting by translation on R,Zd acting by rotation on S1, Z2 acting by the antipodal map on Sn, and Zp acting on S3.All three examples are Galois.

Theorem 3.46. Suppose that f : X → Y is a n-sheeted covering map for finite n. Thenthere exists a natural homomorphism

τ∗ : Hq(Y ;R) ↪→ Hq(X;R)

such that Hq(f) ◦ τ∗ is multiplication by n on Hq(Y ;R).In particular, if R is a coefficient ring containing 1

nthen τ∗ must be injective. If

furthermore the covering is Galois, then the image of τ∗ is

Hq(Y ;R) ∼= Hq(X;R)G

with the subgroup of Hq(X;R) of elements fixed by the deck transformation group G.

Proof. The transfer map τ∗ is determined by a chain morphism τ : S(Y )→ S(X) definedon simplices σ : ∆q → Y to the sum

τ(σ) =n∑i=1

σi

where σ1, ..., σn are the n lifts of σ (one for each sheet). It is not hard to see that τ∂ = ∂τso τ is a chain morphism. At the level of chains

f∗τ(σ) = f∗(n∑i=1

σi) =n∑i=1

f ◦ σi = nσ.

Thus Hq(f) ◦ (τ) is multiplication by n.Now suppose that n is invertible in R. Then 1

nf∗ ◦ τ∗ is the identity on Hq(Y ;R) so τ∗

must be injective (in fact the inclusion splits!).Finally, suppose additionally that the covering is Galois with deck transformation

group G of order n. Define a homomorphism

π∗ : Hq(X;R)→ Hq(X;R)G

by

π∗(α) =1

n

n∑i=1

g∗(α)

If α ∈ Hq(X;R)G then π∗(α) = α and thus

im(π∗) = Hq(X;R)G.

Note also that nπ∗ = τ∗ ◦ f∗. Since f∗ is surjective and n is a unit it follows that

im(π∗) = im(nπ∗) = im(τ∗ ◦ f∗) = im(τ∗) ∼= Hq(Y ;R).

55

4 Cohomology

Let R be a commutative ring and let M be a R-module. The dual module

M∨ := HomR(M,R)

is the set of R-module homomorphisms from M to R. M∨ is an R module under additionand scalar multiplication of functions. There is a natural isomorphism

(⊕iMi)∨ ∼=

∏i

M∨i ,

defined by the rule (φ1, φ2, ..., )(m1 +m2 + ...) =∑

i φi(mi). In particular, for free moduleswe have

(⊕iR)∨ =∏i

R∨ =∏i

R. (8)

More directly, (8) holds because a homomorphism out of a free module is specified bylisting where the free generators are sent.

Given a R-module homomorphism f : M → N , define the transpose

f∨ : N∨ →M∨

which sends φ ∈ N∨ = HomR(N,R) to f∨(φ) = φ ◦ f .

Mf //

f∨(φ)=φ◦f AAA

AAAA

A N

φ~~~~~~

~~~

R

Dualization is a contravariant functor from R-modules to R-modules. This meansthat Id∨M = IdM∨ and (g ◦ f)∨ = f∨ ◦ g∨. The first is obvious and the second follows fromassociativity of composition:

(g ◦ f)∨(φ) = φ ◦ (g ◦ f) = (φ ◦ g) ◦ f = f∨(g∨(φ)) = (f∨ ◦ g∨)(φ)

Mf //

f∨g∨(φ) BBB

BBBB

B N

��

g // P

φ~~~~~~

~~~

R

Given a chain complex of R-modules

...→ Cn+1∂n+1→ Cn

∂n→ Cn−1∂n−1→ ...

we form the dual chain complex (or co-chain complex )

...← Cn+1 δn+1← Cn δn← Cn−1 δn−1← ...

56

where Cn = C∨n and δn = ∂∨n . Note that

δn ◦ δn+1 = ∂∨n ◦ ∂∨n+1 = (∂n+1 ◦ ∂n)∨ = 0∨ = 0

We can now define cochains Zn := ker(δn+1), coboundaries Bn := im(δn) and cohomology

Hn := Zn/Bn.

Definition 19. The singular cohomology of a pair of spaces (X,A), denotedHq(X,A;R)q ≥ 0, is the cohomology of the singular cochain complex

· · · → Sq−1(X,A;R)→ Sq(X,A;R)→ Sq+1(X,A;R)→ . . .

obtained by dualizing the singular chain complex of X.

More geometrically, we can understand a singular co-chain ξ ∈ Sq(X;R) as a functionthat assigns a scalar to every singular simplex σ : ∆q → X. The pairing

Sq(X;R)× Sq(X;R)→ R

is sometimes called integration, because it is an algebraic analogue of integrating adifferential form over parametrized manifold.

For example,

• A 0-cochain is simply a function (of sets) f : X → R, since 0-simplices correspondto points in X.

• A 1-cochain assigns a scalar to every continuous path γ : [0, 1]→ X.

• A 2-cochain assigns a scalar to every map σ : ∆2 → X.

The analogue of Stoke’s Theorem follows just by definition. If α ∈ Sq(X : R) andξ ∈ Sq−1(X;R) then

ξ(∂α) = (δξ)(α).

or in integral notation ∫∂α

ξ =

∫α

δξ.

For example, given a 1 simplex γ : [0, 1]→ X and 0-cochain f : X → R, we have∫γ

δf =

∫∂γ

f = f(γ(1)− γ(0)) = f(γ(1))− f(γ(0)).

This example really illustrate why the boundary of a 1-simple requires signs: to recoverthe Fundamental Theorem of Calculus.

An easy consequence is that this integration pairing descends to homology and coho-mology.

57

Proposition 4.1. There is a natural pairing

Hq(X,A;R)×Hq(X,A;R)→ R

defined by ([ξ], [α]) 7→ ξ(α) for representative (co)cycles ξ and α. This determines anatural homomorphism

Hq(X,A;R)→ Hq(X,A;R)∨. (9)

If Hq(X,A;R) is free in all degrees, then (9) is an isomorphism.

Remark 5. More generally, singular cohomology can always be computed from integralhomology using the Universal Coefficient Theorem for Cohomology. See Hatcher 3.1 fordetails.

Proof. Choosing different representatives of the homology and cohomology classes, we get

(ξ + δν)(α + ∂β) = ξ(α) + ξ(∂β) + (δν)(α) + (δν)(∂β)

= ξ(α) + (δξ)(β) + ν(∂α) + ν(∂2β)

= ξ(α).

establishing that (9) is a well-defined function, which is clearly a homomorphism.We prove the second part only when R = F is a field (the general case is an exercise).

Abbreviate Sq(X,A;R) = Sq. Recall that we have short exact sequences

0→ Zq → Sq → Bq−1 → 0

0→ Bq → Zq → Hq → 0.

These sequences split, giving rise to isomorphisms Sq = Zq⊕B′q−1 and Zq = Bq⊕H ′q. Wehave an isomorphism of chain complexes

. . . // Sq+1

∼=��

∂ // Sq

∼=��

∂ // Sq−1∂ //

∼=��

. . .

. . . // Bq+1 ⊕H ′q+1 ⊕B′q ∂′ // Bq ⊕H ′q ⊕B′q−1∂′ // Bq−1 ⊕H ′q−1 ⊕B′q−2

// . . .

where ∂′ sends summands Bq and H ′q to zero and send B′q−1 isomorphically to Bq−1.Dualizing get

. . . Sq+1δoo Sqδoo Sq−1δoo . . .oo

. . . B∨q+1 ⊕H′∨q+1 ⊕B

′∨q

oo

∼=

OO

B∨q ⊕H′∨q ⊕B

′∨q−1

δ′oo

∼=

OO

B∨q−1 ⊕H′∨q−1 ⊕B

′∨q−2

δ′oo

∼=

OO

. . .oo

where δ′ sends H′∨q and B

′∨q−1 to zero and sends B∨q isomorphically to B′∨q . Taking coho-

mology, we find thatHq = H

′∨q = H∨q .

58

All of the main theorems about homology have versions in cohomology. The maindifference is that arrows have to be reversed. We summarize:

• Hq(−) = Hq(−;R) is a contravariant functor for topological spaces to the categoryof R-modules. Thus for every continuous map f : X → Y , there is a homomorphismof R-modules

Hq(X)Hq(f)=f∗←− Hq(Y )

such that (f ◦ g)∗ = g∗ ◦ f ∗ and (IdX)∗ = IdHq(X;R).

• If f, g : X → Y are homotopic, then f ∗ = g∗.

• A topological pair (X,A) gives rise to a long exact sequence of R-modules

· · · → Hq−1(A)→ Hq(X,A)→ Hq(X)→ Hq(A)→ Hq+1(A)→ ...

where the connecting homomorphism is defined using the same diagram chase asbefore.

• The excision isomorphism Hq(X \ B,A \ B) ∼= Hq(X,A) hold under the samehypotheses as before.

• H0(X) =∏

π0(X) R is a product of copies of R indexed by the path components ofX.

• H0(pt) = R and Hq(pt) = 0 for q 6= 0. Define Hq(X) to be the cokernel of thenatural map Hq(pt)→ Hq(X).

• Hn(Sn) ∼= R and Hq(Sn) = 0 for q 6= n. If f : Sn → Sn is a degree d map, then f ∗

is simply multiplication by d ∈ R.

• For a cell-complex X, the cohomology of the dual of the cellular chain complex isnaturally isomorphic to the singular cohomology. In this case, where the boundarymaps are given by integer matrices, the co-boundary maps are simply the transposesof those matrices.

• Under the same hypotheses as for homology, there is a long exact Mayer-VietorisSequence

· · · → Hq(X)→ Hq(A)⊕Hq(B)→ Hq(A ∩B)→ Hq+1(X)→ . . . .

Example: RP n Recall that the cellular chain complex has the form

0→ R0→ R

2→ R→ ...→ R2→ R

0→ R

if n is odd and0→ R

2→ R0→ R→ ...→ R

2→ R0→ R

if n is even. The cellular cochain complex is obtained by reversing arrows

0← R0← R

2← R← ...← R2← R

0← R

59

if n is odd and0← R

2← R0← R← ...← R

2← R0← R

If R = Z2 then the coboundary maps are all zero and we get

Hq(RP n; Z2) =

{Z2 for q = 0, ..., n

0 otherwise.

If R = F if a field of odd or zero characteristic, then the coboundary maps alternatebetween zero and isomorphisms and we get

Hq(RP n;F ) =

{F if q = 0 or q = n and n is odd

0 otherwise.

both of which agree with the homology groups (as they must by Proposition 4.1).If R = Z then we obtain

Hq(RP n; Z) =

Z if q = 0 or q = n and n is odd

Z2 if 2 ≤ q ≤ n and is even

0 otherwise

Notice that the integer cohomology groups do not agree with the integer homologygroups in this case (for homology the Z2 groups occur in odd degree). The generalsituation for integer cohomology is as follows (see Hatcher 3.1 for proof).

Proposition 4.2. If the homology groups Hq(X,A; Z) are finitely generated in all degrees,then Hq(X; Z) is isomorphic to a direct sum of the free part of Hq(X; Z) with the torsionpart of Hq−1(X; Z)

4.1 The cup product

A graded R-algebra A∗ is a direct sum of R-modules

A∗ =⊕i∈Z

Ai

equipped with a multiplication Ai × Aj → Ai+j which is distributive with respect toaddition and scalar multiplication.

(r1m1 + r2m2) · n = r1(m1 · n) + r2(m2 · n)

n · (r1m1 + r2m2) = r1(n ·m1) + r2(n ·m2).

The elements of Ai ⊆ A∗ are called homogeneous of degree d. We say A∗ is gradedcommutative if the multiplication satisfies

m · n = (−1)i+jn ·m

for homogeneous elements m ∈ Ai and n ∈ Aj.Examples:

60

• A polynomial ring R[x] is a graded R-algebra if we give x some degree d. It is a freemodule

R[x] =∞⊕i=0

Rxi

where Rxi has degree di and multiplication is xixj = xi+j. If d is even, then R[x] isgraded commutative.

• More generally, the polynomial ring R[x1, ..., xn] with given degrees di = deg(xi)is a graded R-algebra. It is a free R-module with homogeneous generators, themonomials xi11 ...x

inn which have degree d1i1 + ...+ dnin. It is graded commutative if

the degrees di are all even.

• The exterior (or Grassmann) algebra Λ(y1, ..., yn) is the R-algebra generated byy1, ..., yn and identity 1 with multiplication

1. yi ∧ yj = (−1)yj ∧ yi2. yi ∧ yi = 0.

It is a free R-module with 2n generators {yi1 ∧ .... ∧ yik |1 ≤ i1 < ... < ik ≤ n}. It isgraded commutative if the yi all have odd degree.

For example, Λ(y1, y2, y3) has free module generators

1, y1, y2, y3, y1 ∧ y2, y1 ∧ y3, y2 ∧ y3, y1 ∧ y2 ∧ y3.

The goal of this section is to establish the following.

Theorem 4.3. There is a multiplication called the cup product that makes the direct sum

H∗(X;R) :=∞⊕q=0

Hq(X;R)

into a graded commutative, associative R-algebra for which Hq(X;R) has degree q. Thereis a multiplicative identity, denoted 1, which is represented by the 0-cocycle that sendsevery 0-simplex to 1.

Example 30. Recall that for n > 1, the Sn has cohomology Hq(Sn) = R if q = 0, n andzero otherwise. If we select a generator x ∈ Hn(Sn), then x2 ∈ H2n(Sn) = 0 must bezero. Thus there is an isomorphism of graded R-algebras

H∗(Sn;R) ∼= R[x]/(x2)

where we have taken the quotient of the polynomial ring R[x] by the ideal (x2) generatedby x2.

61

We will show later that

H∗(CP n;R) ∼= R[x]/(xn+1)

where deg(x) = 2 and that

H∗(RP n; Z2) ∼= Z2[y]/(yn+1)

where deg(y) = 1.

We begin by defining the cup product at the level of cochains.

Sp(X,A;R)× Sq(X,A;R)→ Sp+q(X,A;R), (φ, ψ) 7→ φ ∪ ψdefined on a (p+ q)-simplex σ by

(φ ∪ ψ)(σ) = φ(σ ◦ [e0, ..., ep])ψ(σ ◦ [ep, ..., ep+q])

where [e0, ..., ep] : ∆p → ∆p+q and [ep, ..., ep+q] : ∆q → ∆p+q are affine simplices.The cup product is both associative and distributive with respect to addition (Home-

work).Drawings ...

Lemma 4.4. The cup product satisfies a “Leibnitz rule”

δ(φ ∪ ψ) = δφ ∪ ψ + (−1)pφ ∪ δψfor φ ∈ Sp(X) and ψ ∈ Sq(X;R).

Proof. For σ : ∆p+q+1 → X we have

(δφ ∪ ψ)(σ) = δφ(σ ◦ [e0, ..., ep+1])ψ(σ ◦ [ep+1, ..., ep+q+1])

=

p+1∑i=0

(−1)iφ(σ ◦ [e0, ..., ei, ..., ep+1])ψ(σ ◦ [ep+1, ..., ep+q+1])

in which the last term is (−1)p+1φ(σ ◦ [e0, ..., ep])ψ(σ ◦ [ep+1, ..., ep+q+1]), and

(φ ∪ δψ)(σ) =

p+q+1∑i=p

(−1)i−pφ(σ ◦ [e0, ..., ep])ψ(σ ◦ [ep, ..., ei, ..., ep+q+1])

in which the first term is φ(σ ◦ [e0, ..., ep])ψ(σ ◦ [ep+1, ..., ep+q+1]).Hence

δ(φ ∪ ψ) =

p+q+1∑i=0

(−1)i(φ ∪ ψ)(σ ◦ [e0, ..., ei, ..., ep+q+1])

=

p∑i=0

(−1)iφ(σ ◦ [e0, ..., ei, ..., ep+1])ψ(σ ◦ [ep+1, ..., ep+q+1])

+

p+q+1∑i=p+1

(−1)iφ(σ ◦ [e0, ..., ep])ψ(σ ◦ [ep, ..., ei, ..., ep+q+1])

= δφ ∪ ψ + (−1)pφ ∪ δψ

62

Observe that (??) implies that the product of two cocycles is a cocycle. Also, if theproduct of a cocycle with a coboundary (in either order) is a coboundary, because

φ ∪ δψ = ±δ(φ ∪ ψ)± δφ ∪ ψ = ±δ(φ ∪ ψ)

if δφ = 0 andδφ ∪ ψ = δ(φ ∪ ψ)± φ ∪ δψ = δ(φ ∪ ψ)

if δψ = 0. It follows that the cup product descends a map

∪ : Hp(X;R)×Hq(X;R)→ Hp+q(X;R)

which is both associative and bilinear with respect to the R-module structure.Thus the cup product makes the direct sum

H∗(X;R) :=∞⊕q=0

Hq(X;R)

into a graded, associative R-algebra. There is a multiplicative identity, denoted 1,which is represented by the 0-cocycle that sends every 0-simplex to 1.

Theorem 4.5. Given α ∈ Hp(X;R) and β ∈ Hq(X;R), the cup product satisfies theidentity

α ∪ β = (−1)pqβ ∪ α. (10)

Sketch. The proof of this result is actually surprisingly difficult, because the identity (10)does not hold for cochains. We will sketch the argument.

The idea of the argument is to construct a morphism of cochain complexes (i.e. a mapρ∗ : Sq(X)→ Sq(X) such that ρ∗ ◦ δ = δ ◦ ρ∗) such that

1. Satisfies ρ∗(φ ∪ ψ) = (−1)pqρ∗(ψ) ∪ ρ∗(φ). for φ ∈ Sp(X) and ψ ∈ Sq(X).

2. Determines the identity map on cohomology H(ρ∗) = IdH∗(X).

Given these two properties, if φ and ψ are cocycles representing cohomology classes[φ] and [ψ] we have

[φ] ∪ [ψ] = [φ ∪ ψ]

= [ρ∗(φ ∪ ψ)]

= [(−1)pqρ∗(ψ) ∪ ρ∗(φ)]

= (−1)pq[ρ∗(ψ)] ∪ [ρ∗(φ)]

= (−1)pq[ψ] ∪ [φ]

thus proving the theorem.

63

Given a singular q-simplex σ : ∆q → X, define

σ = σ ◦ [eq, eq−1, ..., e1, e0]

by composing with the affine simplex that reverses the order of the vertices. Define ahomomorphism

ρq : Sq(X)→ Sq(X),

which sends a simplex σ toρq(σ) = εqσ

where εq = (−1)q(q+1)/2. Clearly ρq is an isomorphism, because (ρq)2 is the identity.

Claim: The ρq combine to form a chain morphism ρ : S∗(X)→ S∗(X).

Proof.

∂ ◦ ρ(σ) = ∂(εqσ ◦ [en, ..., e0]) =

q∑i=0

(−1)iεqσ ◦ [eq, ..., eq−i, ..., e0]

while

ρ ◦ ∂(σ) = ρ(n∑q=0

(−1)iσ ◦ [e0, ..., ei, ..., eq])

=n∑q=0

(−1)iεq−1σ ◦ [eq, ..., ei, ..., e0]).

Thus∂ρ = ±ρ∂

and it is just a matter of checking that the sign

(−1)i(−1)q−iεq−1εq = (−1)q(−1)q(q−1)/2(−1)q(q+1)/2 = 1.

Since ρ is a chain morphism, the transpose ρ∗ = ρ∨ : S∗(X) → S∗(X) is a cochainmorphism, because

δ ◦ ρ∗ = ∂∗ ◦ ρ∗ = (ρ ◦ ∂)∗ = (∂ ◦ ρ)∗ = ρ∗ ◦ ∂∗ = ρ∗ ◦ δ.

Claim: The cochain map ρ∗ induces the identity map on cohomology.

Proof. This is the part we will skip. Topologically this is intuitively plausible, because weare simply reparametrizing simplices, without changing their images. The proof dependson constructing a chain homotopy P : Sq(X)→ Sq+1(X) between ρ and the identity map:

∂P + P∂ = ρ− IdS∗(X).

64

The construction is quite similar to the one used to prove homotopy invariance, but alteredto deal with the reordering of vertices. The transpose of P is a cochain homotopy betweenρ∗ and the identity, because

ρ∗ − IdS∗(X) = (ρ− IdS∗(X))∗ = (∂P + P∂)∗ = P ∗δ + δP∨

which thus induces identity on cohomology.

Claim: For φ ∈ Sp(X) and ψ ∈ Sq(X), we have

ρ∗(φ ∪ ψ) = (−1)pqρ∗(ψ) ∪ ρ∗(φ).

Proof of claim. Let σ be a p+ q simplex. Then

ρ∗(φ ∪ ψ)(σ) = (φ ∪ ψ)(ρ(σ))

= (φ ∪ ψ)(εp+qσ ◦ [ep+q, ..., e0])

= εp+qφ(σ ◦ [ep+q, ..., ep])ψ(σ ◦ [ep, ..., e0])

while

(ρ∗(ψ) ∪ ρ∗(φ))(σ) = ρ∗(ψ)(σ ◦ [e0, ..., ep])ρ∗(φ)(σ ◦ [ep, ..., ep+q])

= εpεqψ(σ ◦ [ep, ..., e0])φ(σ ◦ [ep+q, ..., ep])

= εpεqφ(σ ◦ [ep+q, ..., ep])ψ(σ ◦ [ep, ..., e0]).

It remains to show the signs work out : εp+q = εpεq(−1)pq.

4.2 The Kunneth formula

Given R-modules M and N , the tensor product M ⊗R N = M ⊗ N is aa R-modulepossessing the following universal property. There is a canonical bilinear map t : M×N →M ⊗N such that for any R-module P and any bilinear map b : M ×N → P , there existsa unique homomorphism φb such that b = φb ◦ t:

M ×N t //

b

&&MMMMMMMMMMMM M ⊗R Nφb��P

Of course, this works backwards as well. Given a homomorphism φ : M ⊗R N → Pthe composition φ ◦ t is a bilinear map from M × N to P . Thus we have a one-to-onecorrespondence

{ bilinear maps b : M ×N → P } 1:1←→ { homomorphisms φ : M ⊗R N → P }

The tensor product of two modules always exists and is unique (up to canonical isomor-phism). We use notation t(m,n) = m⊗n. It is a little complicated to describe in general,but for free modules it is very easy to understand.

65

Proposition 4.6. If M =⊕

i∈I Rxi and N =⊕

j∈J Ryj are free modules with generatorsxi and yj respectively. Then

M ⊗R N =⊕

(i,j)∈I×J

R(xi ⊗ yj)

is a free module with generators xi ⊗ yj.

Proof. Define the map

M ×N →⊕

(i,j)∈I×J

R(xi ⊗ yj)

by

t(∑i

rixi,∑j

rjyj) =∑i,j

rirjxi ⊗ yj.

The map t is clearly bilinear. Given another bilinear map b : M × N → P , define thehomomorphism

φb :⊕

(i,j)∈I×J

R(xi ⊗ yj)→ P

which sends generatorsφb(xi ⊗ yj) = b(xi, yj).

It is easy to see that b = φb ◦ t and that φb is the unique homomorphism satisfying thisproperty.

Now we connect this idea with topology. Given two spaces, we may form the productspace X × Y which comes equipped with projection maps

X × YπY

##GGGGGGGGG

πX

{{vvvv

vvvv

v

X Y

These can be used to define a bilinear map

Hp(X)×Hq(Y )→ Hp+q(X × Y ), (α, β) 7→ π∗X(α) ∪ π∗Y (β).

By the universal property, this determines a homomorphism

κ : Hp(X)⊗Hq(Y )→ Hp+q(X × Y ). (11)

Theorem 4.7. If both Hp(X;R) and Hq(Y ;R) are free R-modules for all degrees p andq, then there is an isomorphism

Hn(X × Y ;R) ∼=⊕p+q=n

Hp(X;R)⊗R Hq(Y ;R)

defined by adding up the natural homomorphisms 11 .

66

Proof. A proof in the case when X and Y are cell complexes is found in Hatcher 3.2.

Example 31. Consider the 3-manifold S1 × Σg where Σg is a genus g surface. Then

H0(S1 × Σg) = H0(S1)⊗H0(Σg) ∼= R.

H1(S1 × Σg) = H0(S1)⊗H1(Σg) +H1(S1)⊗H0(Σg) ∼= R2g ⊕R = R2g+1

H2(S1 × Σg) = H0(S1)⊗H2(Σg) +H1(S1)⊗H1(Σg) ∼= R⊕R2g = R2g+1

H3(S1 × Σg) = H1(S1)⊗H2(Σg) ∼= R.

In fact, the Kunneth Theorem also determines the multiplicative structure of H∗(X×Y ).

Given two graded associative R-algebras, A∗ and B∗, the tensor algebra is (A⊗RB)∗ =A∗ ⊗R B∗ is a the graded, associative algebra with graded modules

(A⊗R B)n := ⊕p+q=nAp ⊗R Bq

and multiplication

(a1 ⊗ b1) · (a2 ⊗ b2) = (−1)deg(a2) deg(b1)(a1 · a2)⊗ (b1 · b2).

Observe that if A∗ and B∗ are graded commutative, then so is A∗⊗B∗, because if ai ∈ Apiand bi ∈ Bqi then

(a1 ⊗ b1) · (a2 ⊗ b2) = (−1)p2q1(a1a2)⊗ (b1b2)

= (−1)p2q1((−1)p1p2a2a1)⊗ ((−1)q1q2b2b1)

= (−1)p2q1+p1p2+q1q2(a2a1)⊗ (b2b1)

= (−1)p2q1+p1p2+q1q2+p1q2(a1 ⊗ b1) · (a2 ⊗ b2)

= (−1)(p1+q1)(p2+q2)(a1 ⊗ b1) · (a2 ⊗ b2)

Example 32. If x1, ..., xn have even degrees, then

R[x1]⊗ ...⊗R[xn] = R[x1, ..., xn].

Example 33. If y1, ..., yn have odd degrees, then

Λ(y1)⊗ ....⊗ Λ(yn) = Λ(y1, ..., yn).

.

Theorem 4.8 (Kunneth formula). The Kunneth isomorphisms described in ... combineto form an isomorphism of graded R-algebras

H∗(X × Y ;R) ∼= H∗(X;R)⊗R H∗(Y ;R).

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Proof. Recall that if αi ∈ Hpi(X;R) and βj ∈ Hpj(Y ;R) then the Kunneth homomor-phism κ : Hpi(X)⊗Hqi(Y )→ Hpi+qi(X × Y ) sends αi ⊗ βj to π∗1(αi)⊗ π2(βj). Thus

κ(α1 ⊗ β1) ∪ κ(α2 ⊗ β2) = πa(α1) ∪ π∗b (β1) ∪ π∗a(α2) ∪ π∗b (β2)

= πa(α1) ∪ π∗a(α2) ∪ π∗b (β1) ∪ π∗b (β2)

= πa(α1 ∪ α2) ∪ π∗b (β1 ∪ β2)

= κ((α1 ∪ α2)⊗ (β1 ∪ β2))

= κ((α1 ⊗ β1) · (α2 ⊗ β2))

This extends by linearity to an isomorphism of graded R-algebras.

Example 34. for integers k1, ..., kn ∈ Z, the product of spheres Sk1 × ... × Skn hascohomology ring

H∗(Sk1 × ...× Skn) ∼= Λ(x1, ..., xk)

where the generator xi has degree ki.

4.3 Manifolds and orientations

An n-manifold (a.k.a. manifold of dimension n) is a Hausdorff topological space Mwhich is locally Euclidean, meaning that for each point p ∈ M there exists an openneighbourhood p ∈ U ⊂M such that U is homeomorphic to Rn. We say that M is closedif it is compact. We assume for simplicity that M is path connected.

Examples:

• Euclidean space Rn is an n-manifold that is not closed.

• The sphere Sn is a closed n-manifold.

• The orientable and non-orientable surfaces Σg and Ng are closed 2-manifolds.

• Real projective space RP n is a closed n-manifold (it looks locally like Rn).

• Complex projective space CP n is a closed 2n-manifold (it looks locally like Cn ∼=R2n).

• The product on a m-manifold with an n-manifold is an (m+n)-manifold, by takingproducts of euclidean neighbourhoods.

Using excision, we have isomorphisms

Hn(M,M \ p) = Hn(U,U \ p) ∼= Hn(Rn,Rn \ p) ∼= Z.

A local orientation of M is a choice of generator µp ∈ Hn(M,M \ p).

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Given a Euclidean neighbourhood Rn ∼= U ⊆ M , then for any two points p, q ∈ U wecan find a compact set B ⊂ U corresponding to a closed ball in Rn and containing p andq. We have isomomorphisms

Hn(U,U \ p)∼=← Hn(U,U \B)

∼=→ Hn(U,U \ q) (12)

The local orientations at p and q are called consistent if the isomorphism (12) sendsone to the other. A (global) orientation of M is a set {µp | p ∈M} of local orientationsfor every point in M , which is consistent with respect to every Euclidean neighbourhood.If an orientation exists, we say M is orientable. Otherwise M is non-orientable.

The concept of a global orientation can be understood more geometrically as follows.Consider the set M consisting of ordered pairs (p, µp) where p ∈ M and µp is one of thetwo generators of Hn(M,M \ p). We have a forgetful function (of sets)

π : M →M

which sends (p, µp) to p and the pre-image of any point p ∈ M is the two possibleorientations at that point. We topologies M by partitioning the pre-image of a Euclideanneighbourhood U ⊂M according to consistent orientations, an having each partition maphomeomorphically onto U . This makes π : M → M into a 2-fold covering space, calledthe orientation cover. Notice that M is a manifold which is orientable because it comesequipped with a global orientation by construction.

A global orientation for M is equivalent to a continuous map s : M → M such thatπ ◦ s = IdM . We call s section of the covering map π.

If an orientation s exists, then the map

−s : M → M,

which takes the opposite value at each point is also an orientation. In this case coveringspace decomposes into two copies of M

M ∼= M qM

each mapping homeomorphically to M via π.If no orientation of M exists, then some loop in M lifts (via the simplex lifting prop-

erty) to a non-closed path in M from which it follows that M is path connected.

Example 35. A loop that winds once around RP 2 = N0 lifts to a path connectingantipodal points on S2 = Σ0.

Recall that when we studied spheres, a global orientation was defined to be a generatorof Hn(Sn), which determines local orientations at each point p ∈ Sn via the homomor-phism

Hn(Sn)∼=→ Hn(Sn, Sn \ p)

It turns out that for closed manifolds, the same idea works.

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Theorem 4.9. Let M be a n-manifold. Then Hq(M ;R) = 0 for q > n. If M is compact,then

Hn(M ; Z) =

{Z if M is orientable

0 if M is non-orientable

If M is orientable, we call a generator of Hn(M ; Z) an orientation class.

Proof. Sketch: The proof uses a version of Mayer-Vietoris that I haven’t covered yet. LetA and B be closed subsets of M , then we have a short exact sequence

Hq+1(M,M\(A∩B))→ Hq(M,M\(A∪B))→ Hq(M,M\A)⊕Hq(M,M\B)→ Hq(M,M\A∩B)→ ...

Given a closed ball B in M contained in a Euclidean neighbourhood U , then by excisionwe have

Hq(M,M \B; Z) ∼= Hq(B, ∂B; Z) =

{Z if q = n

0 if q > n

Now choose a collection of balls Bi such that their interiors cover M . Since M is compact,we may choose this covering to be finite {B1, ..., BN}. We use the Mayer-Vietoris sequenceand induction to show that

Hq(M,M \ (∪Ni=1Bi)) = Hq(M)

{Z if q = n

0 if q > n

This gets a bit tricky, because we have to keep track of the intersections between balls.One must prove that the balls can be chosen to be “good” in the sense that any intersectionof Bi1 ∩ ... ∩ Bin is also homeomorphic to a ball. In Rn this holds by convexity, but forgeneral manifolds it takes some care.

Examples:

• The spheres Sn, the orientable surfaces Σg and CP n are all orientable.

• The non-orientable surface Ng have orientation cover Σg.

• RP n is orientable when n is odd and non-orientable when n is even.

If f : M → N is a continuous map between two oriented, closed n-manifolds, itdetermines a group homomorphism

f∗ : Hn(M ; Z)→ Hn(N ; Z)

which by ... is a homomorphism from Z to Z and thus corresponds to multiplication by aninteger d ∈ Z. We call d the degree of f . By a similar argument as worked for spheres,one can calculate the degree of d by adding up local degrees of a finite set of points f−1(p)for “regular value” p ∈ N .

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More generally, one can define local/global orientations of a manifold M with respectto a coefficient ring R; that is, a choice of generator µp ∈ Hn(M,M \p;R) = R. The mostinteresting case is when R = Z2, because then there is only one choice of generator ateach point and choices have to be consistent. It follows that every manifold is orientablewith respect to Z2 and we get the following result.

Theorem 4.10. If M is closed n-manifold, then Hn(M ; Z2) ∼= Z2.

For instance, we see this with Hn(RP n; Z2) = Z2 and H2(Ng; Z2) = Z2.

4.4 The cap product and Poincare Duality

The cap product is a bilinear map that takes as input a singular n-chain and a singulark-cochain, and outputs a n− k-chain (we assume n ≥ k):

∩ : Sn(X;R)× Sk(X;R)→ Sn−k(X;R).

Given a n-simplex σ : ∆n → X and a cochain ψ ∈ Sk(X;R), the cup product isdefined

σ ∩ ψ = ψ(σ ◦ [e0, ..., ek]) σ ◦ [ek, ..., en]

which extends by linearity to any n-chain.The cap product is dual to the cup product in the sense that if φ ∈ Sn−k(X;R) is a

cochain, then

φ(σ ∩ ψ) = (ψ ∪ φ)(σ),

so that the homomorphism

ψ ∪ ( ) : Sn−k(X;R)→ Sn(X;R)

is the transpose of the linear map

( ) ∩ ψ : Sn(X;R)→ Sn−k(X;R).

Theorem 4.11. The cap product determines a bilinear map (also called the cap product)

∩ : Hn(X;R)×Hk(X;R)→ Hn−k(X;R)

by the rule[α] ∩ [ψ] = [α ∩ ψ]. (13)

Proof. We begin by proving Leibnitz rule:

Lemma 4.12. If α ∈ Sn(X) and ψ ∈ Sk(X) then

∂(α ∩ ψ) = (−1)k(∂α ∩ ψ − α ∩ δψ)

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Proof. It is enough to consider case that α = σ is a singular n-simplex. The result isverified by direct calculation.

∂(σ ∩ ψ) = ∂(ψ(σ ◦ [e0, ..., ek]) σ ◦ [ek, ..., en])

=n∑i=k

(−1)i−k(ψ(σ ◦ [e0, ..., ek]) σ ◦ [ek, ..., ei, ..., en])

σ ∩ δ(ψ) = (δψ)(σ ◦ [e0, ..., ek+1])σ ◦ [ek+1, ..., en]

=k+1∑i=0

(−1)iψ(σ ◦ [e0, ..., ei, ..., ek+1])σ ◦ [ek+1, ..., en]

where the last term equals, (up to sign) the first term of the previous sum. Finally,

∂(σ) ∩ ψ = (n∑i=0

(−1)iσ ◦ [e0, ..., ei, ..., en]) ∩ ψ

=k∑i=0

(−1)iψ(σ ◦ [e0, ..., ei, ..., ek+1])σ ◦ [ek+1, ..., en]

+n∑k+1

(−1)iψ(σ ◦ [e0, ..., ek])σ ◦ [ek, ..., ei, ..., en]

From this it follows that if ∂α = 0 and δψ = 0, then

∂(α ∩ ψ) = ±∂α ∩ ψ ± α ∩ δψ = 0,

so the cap product α∩ψ appearing in (13) is a cycle that represents a homology class. Tosee that it is well-defined, we use different representatives [α+∂β] = [α] and [ψ+δφ] = [ψ]and check

[(α + ∂β) ∩ (ψ + δφ)] = [α ∩ ψ + ∂β ∩ ψ + β ∩ δφ+ (∂β) ∩ (δφ)]

= [α ∩ ψ + ∂(β ∩ ψ)± ∂(β ∩ φ)± ∂(∂β ∩ φ)]

= [α ∩ ψ + ∂(β ∩ ψ ± β ∩ φ± ∂β ∩ φ

)]

= [α ∩ ψ].

We are now able to state the Poincare Duality Theorem.

Theorem 4.13. Let M be a closed, R-oriented n-manifold M with orientation class[M ] ∈ Hn(M ;R). Cap product with respect to [M ] defines an isomorphism

[M ] ∩ ( ) : Hk(M ;R)∼=→ Hn−k(M ;R).

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Proof. We unfortunately don’t have time to prove this. The argument is similar to theone constructing the orientation class in that it begins with a local result, that leads toa global result via Mayer-Vietoris and induction with respect to a finite cover of M byballs. To do the proof properly, would require introducing the notion of “cohomologywith compact supports”.

Example 36. Recall that if Hq(X;R) is a free R-module in all degrees, then there is anatural isomorphism

Hq(X;R)∨ = Hq(X;R).

Combined with Poincare duality, this implies for R-oriented closed n-manifold M , thatHq(M ;R) and Hn−q(M ;R) are dual modules and that Hq(M ;R) and Hn−q(M ;R) aredual modules, which in particular means that they have the same rank. This explains theword duality in Poincare duality.

Example 37. In case the homology groups are not all free, Theorem 4.13 still works. Forodd n, the projective space RP n is orientable and we have

Hk(RP n; Z) ∼= Hn−k(RP n; Z) ∼=

{Z2 if 0 < k < n is odd (thus n− k is even)

0 0 < k < n is even (thus n− k is odd)

The duality between Hk(M ;R) and Hn−k(M ;R) is better understood using the cupproduct.

Corollary 4.14. If M is closed and R-oriented, and Hq(X;R) is a free R-module in alldegrees, then the bilinear pairing

Hk(M ;R)×Hn−k(M ;R)→ R

sending (α, β) to α∪ β([M ]) is the duality pairing and thus determines a natural isomor-phism

Hk(M ;R) = (Hn−k(M ;R))∨

and vice-versa.

Proof. This follows immediately from Theorem ... and Theorem 4.13 from the identity

(α ∪ β)([M ]) = β([M ] ∩ α).

Example 38. Suppose X = (S1)n is a product of n circles. The cohomology ring is

H∗(X;R) = Λ(x1, ..., xn).

In this case, the generator in top degree is x1 ∧ ...∧ xn and the duality pairing is betweenΛk(x1, ..., xn) and Λn−k(x1, ..., xn), each of which has rank

(nk

)=(

nn−k

).

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Example 39. Recall Hq(RP n) = Z2 in degrees 0 ≤ q ≤ n. Denote by y the generator ofH1(RP ; Z2). The sequence of inclusions

RP 1 ↪→ RP 2 ↪→ RP 3 ↪→ ...

determines a sequence of surjections of cohomology rings

H∗(RP 1; Z2) � H∗(RP 2; Z2) � H∗(RP 3; Z2) � ...

and is an isomorphism in degrees q less than the dimension. For n = 2, Poincare dualityimplies that the generator y ∈ H1(RP 2; Z2) must satisfy

(y ∪ y)([RP 2]) = 1 6= 0

so y2 = y ∪ y is the generator of H2(RP 2; Z2). Since ... are isomorphisms in degree 1, y2

must be the generator of H2(RP n; Z2) for all n. Similarly, when n = 3 Poincare dualityimplies that y3 := y ∪ y2 must be a generator of H3(RP n; Z2) and thus this holds for alln. Proceeding inductively in this manner, we prove that

H∗(RP n; Z2) ∼= Z2[y]/(yn+1).

A similar argument works for to prove that

H∗(CP n;R) ∼= R[x]/(xn+1)

for deg(x) = 2.

Example 40. For the genus g orientable surface Σg, the cup product boils down to thepairing

H1(Σg;R)×H1(Σg;R)→ H2(Σg;R) ∼= R

Because the cup product is graded commutative, the pairing above is anti-symmetric. S

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