chapter ii: homology · chapter ii: homology in this chapter, groups will usually be abelian....

PMH1: ALGEBRAIC TOPOLOGY

Kevin Coulembier

University of Sydney, Semester 1, 2019. March 20, 2019

Chapter II: HomologyIn this chapter, groups will usually be abelian. Therefore we will follow the convention to denote

the operation as ‘+’, rather than a product. Furthermore, the direct product of two abelian groups(yielding another abelian group) will be seen as a direct sum.

1. Singular homology: definitions and basic results

1.1. Definition. In this subsection, we will associate to an arbitrary topological space X and anyn ∈ N, the singular homology group Hn(X).

Definition 1. An n-simplex is an ordered set [v0, v1, . . . , vn] of points in some Rm such that thedifference vectors v1 − v0, . . . , vn − v0 are linearly independent.

Exercise 2. Show that the vertices of an arbitrary triangle in R2 form a 2-simplex, but the verticesof a quadrilateral in R2 never form a 3-simplex.

Often we will refer to the corresponding convex subspace

{∑i

tivi |∑i

ti = 1, tj ≥ 0} ⊂ Rm

as the n-simplex.

Exercise 3. Every n-simplex is homeomorphic to the standard n-simplex

∆n = {(t0, t1, . . . , tn) ∈ Rn+1 |∑i

ti = 1, tj ≥ 0}.

In particular the standard 0-simplex is 1 ∈ R.

From the point of view of topology, we can (and usually will) just use the standard n-simplex. Foran n-simplex [v0, v1, . . . , vn], omitting some vi yields an n−1-simplex, called a face of [v0, v1, . . . , vn].Observe that for the standard n-simplex, each face is the standard n − 1-simplex ∆n−1, whenconsidered in the appropriate subspace Rn ⊂ Rn+1. The union of all faces gives the boundary ∂∆n.We also write ∆n = ∆n − ∂∆n.

Definition 4. Let X be a topological space.

(i) A (regular) n-simplex in X is a map σ : ∆n → X, with σ|∆n injective.(ii) A singular n-simplex in X is a map σ : ∆n → X.

In this section, we will only use singular simplices.

Remark 5. Since ∆1 ∼= I, singular 1-simplices in X are in natural bijection with paths in X.

We will often define group homomorphisms via the correspondence in the following exercise.1

2 ALGEBRAIC TOPOLOGY, CHAPTER II

Exercise 6. Let S denote a set and ZS (the notation Z⊕S also makes sense) the free abelian groupgenerated by S. In other words, ZS is the additive group of Z-linear combinations of elements ofS. For an arbitrary abelian group G show that we have a bijection

{functions S → G} 1:1→ {group homomorphisms ZS → G},

where the function φ is mapped to the unique group homomorphism which restricts to φ on S ⊂ ZS.

Now fix an arbitrary topological space X.

Definition 7. The group of (singular) n-chains, Cn(X) is the free abelian group of finite formalZ-linear combinations of singular n-simplices in X.

Due to the construction, we denote the operation on the group Cn(X) by ‘+’.

Example 8. The 0-chains are by definition precisely the Z-linear combinations of points in X, soC0(X) = ZX.

The reason for the need of formal sums of singular simplices comes from the idea of restrictinga singular simplex to its boundary. As this boundary is a union of simplices of one dimensionlower, we will see this restriction as a sum of singular simplices. Furthermore, in order to take intoaccount the “orientation” we need to formally invert some singular simplices, such that we arriveat Z-linear combinations.

Definition 9. For n ≥ 1, the boundary homomorphism ∂n : Cn(X) → Cn−1(X) is the uniquegroup morphism satisfying

∂n(σ) =n∑i=0

(−1)iσ|[v0,...,vi,...,vn],

for all singular n-simplices σ : ∆n → X. Furthermore, ∂0 is the zero map C0(X)→ 0.

For example, for a singular 1-simplex σ, we have

∂1(σ) = σ|v1 − σ|v0 = σ(0, 1)− σ(1, 0).

Exercise 10. We have ∂n−1 ◦ ∂n = 0, or im ∂n ⊆ ker ∂n−1.

Example 11. Take the following singular 1-simplices in S1:

σi : ∆1 → S1; σi(t0, t1) =

(cos(

2π(t0 + i)

3), sin(

2π(t0 + i)

3)

), i ∈ {0, 2}

and

σ1 : ∆1 → S1; σi(t0, t1) =

(cos(

2π(2− t0)

3), sin(

2π(2− t0)

3)

).

Make a sketch of this situation. It follows easily that

σ0 − σ1 + σ2 ∈ ker ∂1.

On the other hand, if we were to have σ0 − σ1 + σ2 = ∂2(σ) for some σ : ∆2 → S1, this would“correspond” to constructing a homotopy between the loop which goes around the circle once andthe constant path. Clearly, if we compose the {σi|i ∈ {0, 1, 2}} with the inclusion S1 ↪→ D2, toview them as singular simplices in D2, it would be possible to have σ0 − σ1 + σ2 = ∂2(σ) for someσ : ∆2 → D2. Hence, we see that the question of whether im ∂2 ⊆ ker ∂1 is a strict inclusion ornot is intimately related to the question of whether the fundamental group is trivial. This will bemade very concrete in Section 1.3.

ALGEBRAIC TOPOLOGY, CHAPTER II 3

It is easy to see that, in general, elements of ker ∂1 form “cycles” (when −σ is interpreted asinverting the direction of σ : ∆1 → X). On the other hand, elements in im ∂n are by definitionrestrictions of chains to the boundary of ∆n. This motivates the terminology of boundaries andcycles in the following definition. The observation in Example 11 motivates the introduction of thequotients ker ∂n/im ∂n+1.

Definition 12. The elements of im ∂n+1 are the n-boundaries, the elements of ker ∂n the n-cycles.The singular homology groups are Hn(X) = ker ∂n/im ∂n+1. Elements of Hn(X) are homologyclasses of cycles, denoted by [a] for the corresponding a ∈ ker ∂n. For two n-cycles a and b we thushave [a] = [b] if a− b is in im ∂n+1; we then write a ∼ b and say they are homologous.

This construction is an example of a chain complex of abelian groups. That is a collectionof abelian groups {Ai | i ∈ Z}, with group homomorphisms di : Ai → Ai−1 (the differentials orboundary morphisms) such that di ◦ di+1 = 0, typically denoted as

(1.1) A• : · · ·di+2 // Ai+1

di+1 // Aidi // Ai−1

di−1 // · · · .

The homology groups are generally denoted by Hk(A•) = ker dk/im dk+1 and we say that A• isexact if all homology groups are zero. In our main example of singular homology, correspondingto the chain complex

(1.2) · · ·∂n+2// Cn+1(X)

∂n+1 // Cn(X)∂n // Cn−1(X)

∂n−1 // · · · ∂1 // C0(X)∂0 // 0,

we have thus simplified notation from Hk(C•(X)) to Hk(X).

Remark 13. Note that whereas the operation on the fundamental group had a geometric origin(composition of paths), the operation on the homology groups just comes from the formal algebraicaddition of simplices. Because of this the fundamental group will contain strictly more informationthan the first homology group, see Theorem 30.

Exercise 14. Show that the group Hn(X) is abelian.

For a (possibly infinite) set of abelian groups {Gα}, the direct sum G :=⊕

αGα consists oftuples (gα)α with gα = 0 for all but finitely many α. Then G is again an abelian group for theobvious operation.

Proposition 15. Corresponding to the decomposition into path components X = ∪αXα, we have

Hn(X) ∼=⊕

α∈π0(X)

Hn(Xα).

Proof. As the images of singular n-simplices are path connected, see Exercise I.10, we have anatural direct sum decomposition Cn(X) ∼=

⊕αCn(Xα). Clearly the boundary map preserves this

decomposition. �

Proposition 16. We have a group isomorphism

H0(X) ∼= Z⊕π0(X).

Proof. By Proposition 15 it suffices to show that H0(X) ∼= Z for a path connected space X. Bydefinition, we have

H0(X) = C0(X)/im ∂1,

where C0(X) is the group of Z-linear combinations of points in X. In this interpretation, we have

∂1(σ) = σb − σa,with σa = σ(1, 0) ∈ X and σb = σ(0, 1) ∈ X. From this observation it follows easily that H0(X) ∼=Zv, for an arbitrary v ∈ Z. �


Exercise 17. Show that

Hn(pt) ∼=

{Z if n = 0,

0 if n > 0.

For some applications, the fact that the 0-homology group of a point is non-trivial is inconvenient.To salvage this, one introduces the reduced homology groups, by replacing ∂0 with the non-trivial map ε : C0(X)→ Z, which is the group morphism determined by ε(x) = 1, for any x ∈ X:

Hn(X) =

{ker ε/im ∂1 if n = 0,

Hn(X) if n > 0.

The reduced homology groups are thus the homology groups of another chain complex, as in (1.1),namely

· · ·∂n+2// Cn+1(X)

∂n+1 // Cn(X)∂n // Cn−1(X)

∂n−1 // · · · ∂1 // C0(X)ε // Z // 0.

Exercise 18. Show that Hn(pt) = 0, for all n ∈ N, and

Hn(S0) =

{Z if n = 0

0 if n > 0and H0(Sk) =

{Z if k = 0

0 if k > 0.

1.2. Induced homomorphisms and homotopy invariance. In this section we prove that thehomology groups Hn(X) provide indeed algebraic ‘invariants’ in the study of homotopy equiva-lences. First we establish the ‘functorial’ properties of the homology groups, namely we show thatwe have functors

Hn : Top→ Ab and Hn : Top→ Ab.

An arbitrary map f : X → Y induces group homomorphisms f] : Cn(X) → Cn(Y ), determinedby f](σ) = fσ for each singular n-simplex σ : ∆n → X.

Exercise 19. For a subspace ι : A ↪→ X, the induced group homomorphism ι] : Cn(A)→ Cn(X) isa monomorphism.

We will henceforth consider Cn(A) as a (normal) subgroup of Cn(X).

Exercise 20. We have f]∂ = ∂f].

In general, a morphism η between chain complexes A• and B•, usually referred to as a chainmap, is given by a family of group homomorphisms {ηi : Ai → Bi} which lead to a commutativediagram together with the differentials, i.e. dη = ηd.

Exercise 21. For a chain map η : A• → B•, we have group homomorphisms η∗ : Hn(A•)→ Hn(B•),given by η∗(a+ im dn+1) := ηn(a) + im dn+1, for an arbitrary a ∈ ker dn.

Exercise 20 states precisely that f] is a chain map, represented by the commutativity of thefollowing diagram:

· · ·∂n+2// Cn+1(X)

f]��

∂n+1 // Cn(X)

f]��

∂n // Cn−1(X)

f]��

∂n−1 // · · ·

· · ·∂n+2// Cn+1(Y )

∂n+1 // Cn(Y )∂n // Cn−1(Y )

∂n−1 // · · ·

In particular, f] maps boundaries to boundaries and cycles to cycles. Following Exercise 21, therestriction of f] to cycles thus induces a group homomorphism

f∗ : Hn(X) → Hn(Y ), for each n ∈ N,where f∗[α] = [f](α)] for an arbitrary cycle α.


Exercise 22. We have (fg)∗ = f∗g∗ and (1X)∗ is the identity morphism of Hn(X). (compare withExercise I.46)

(SI) Note that we similarly have functors Cn : Top→ Ab and even C• : Top→ Kom(Ab), withKom(Ab) the category of complexes of abelian groups. However, these do not provide invariantsof homotopy equivalence classes.

If we extend f] by also considering it as the identity on Z, if follows identically that we obtain

group homomorphisms f∗ : Hn(X)→ Hn(Y ), for all n ∈ N.

Theorem 23. If f : X → Y is a homotopy equivalence, then f∗ : Hn(X) → Hn(Y ) is a group

isomorphism for each n ∈ N. The same is true for Hn.

First we introduce a general concept in the theory of chain complexes.

Definition 24. For two chain complexes A• and B• of abelian groups and two chain maps η, ξ :A• → B•, a chain homotopy P between η and ξ is a collection of group homomorphisms {Pn :An → Bn+1} (hence not a chain map),

· · ·dn+2 // An+1

η

��

ξ

��

dn+1 // An

P

��

η

��

ξ

��

dn // An−1

P

��

η

��

ξ

��

dn−1 // · · ·

· · ·dn+2 // Bn+1

dn+1 // Bndn // Bn−1

dn−1 // · · · ,

such that Pn−1dn + dn+1Pn = ηn − ξn.

The terminology for this type of map between chain complexes clearly stems from topology andis justified by the following lemma.

Lemma 25. If two maps f, g : X → Y are homotopic, then there exists a chain homotopy betweenf] and g].

Proof. See p112 in Hatcher. �

The usefulness of chain homotopies lies in the following property.

Exercise 26. Using the notation of Definition 24, show that the existence of a chain homotopybetween η and ξ implies that their induced homomorphisms η∗, ξ∗ : Hn(A•)→ Hn(B•) are identical.

The following proposition is an analogue of Proposition I.49.

Proposition 27. If two maps f, g : X → Y are homotopic, then f∗ = g∗.

Proof. This follows immediately from Lemma 25 and Exercise 26. �

Proof of Theorem 23 . Mutatis mutandis the proof of Theorem I.48, using Proposition 27 and Ex-ercise 22. �

Exercise 28. Show that Hk(Dn) = 0 for all k ∈ N and n ∈ Z≥1.

1.3. The first homology group versus the fundamental group. We now have seen twogroups, π1(X,x0) and H1(X), associated to a space which are defined in terms of paths and whichare invariant under homotopy equivalences. In this section we show that they are intimately related.In fact, the fundamental group completely determines the first homology group, see Theorem 30.


To make Remark 5 more precise, we fix a homeomorphism ψ : ∆1 → I as (t0, t1) 7→ t1. Thisinduces a bijection Ψ from the set of paths in X to the set of singular 1-simplices in X, where forf : I→ X we set Ψ(f) = fψ. Now we have

∂1(Ψ(f)) = 0 ⇔ Ψ(f)(0, 1)−Ψ(f)(1, 0) = f(1)− f(0) = 0.

The bijection Ψ thus restricts to a bijection between loops and singular 1-simplices which arecontained in ker ∂1. For a pointed space (X,x0), we can thus define

hx0 : π1(X,x0) → H1(X) ; [f ] 7→ [Ψ(f)].

Note that while [f ] denotes the homotopy class of f , [Ψ(f)] denotes the homology class of Ψ(f).

Lemma 29. The map hx0 is well defined and is a group homomorphism with image contained inthe direct summand H1(Xα) of H1(X), with Xα the path component of X containing x0.

Proof. We leave out the superscript x0 in the proof. To prove that h is well defined, we must showthat f ' g for two loops at x0 implies that Ψ(f) ∼ Ψ(g). This is proved in (ii) on p166 of Hatcher.

To prove that h is a group homomorphism we need to show that h([f ][g]) = h([f ]) + h([g]), orthat Ψ(f • g) ∼ Ψ(f) + Ψ(g). This is proved in (iii) on p166 of Hatcher. Finally we need to showthat h maps the identity element of π1(X,x0) to the identity element of H1(X), or that Ψ(f) ∼ 0for a constant path f . This is proved in (i) on p166 of Hatcher.

That the image is contained in H1(Xα) is obvious. �

By Proposition 15, it is justified to consider only path connected spaces in the following theorem,which states that the first homology group is the abelianisation of the first homotopy group (thefundamental group).

Theorem 30. For any x0 ∈ X in a path connected space X, there is a group epimorphism

h : π1(X,x0) � H1(X)

and the kernel is the commutator subgroup of π1(X,x0).

Proof. See p167 in Hatcher. �

Corollary 31. For two path connected spaces X,Y , we have

H1(X × Y ) ∼= H1(X) ⊕ H1(Y ).

Proof. By Theorem 30, this follows from the corresponding statement for the fundamental groupin Corollary I.28. �

Exercise 32. For X,Y as in Corollary 31, do we have C1(X × Y ) ∼= C1(X) ⊕ C1(Y )?

Remark 33 (SI). Corollary 31 does not extend to higher homology groups. In fact, the Kunneththeorem states that ⊕

i+j=n

Hi(X)⊗Z Hj(Y )

is isomorphic to Hn(X × Y ) only if there is no ‘torsion’ (see Section 1.4) in the homology groups.


H1(S1) ∼= Z and H1(S1 × S1) ∼= Z⊕2 ∼= H1(S1 ∨ S1).


1.4. Betti numbers and Euler characteristic. First we recall some basis structure theory forabelian groups.

Exercise 35. (i) Show that any proper subgroup of Z is of the form mZ for some m ∈ N.(ii) We set Zm := Z/mZ. Show that Zpq ∼= Zp ⊕ Zq, for two distinct prime numbers p, q.

The fundamental theorem of finitely generated abelian groups states that every finitely generatedabelian group G is isomorphic to a group of the form

Z⊕n ⊕ Zq1 ⊕ · · · ⊕ Zqk ,for some n ∈ N and q1, · · · , qk powers of prime numbers. The above is known as the primarydecomposition of G. The part Z⊕n is a free group, the remainder is the torsion part.

Definition 36. For a finitely generated abelian group G, the rank rkG ∈ N is the number ofsummands in the primary decomposition of G which are isomorphic to Z.

Exercise 37. For a short exact sequence

0→ A→ B → C → 0

of finitely generated abelian groups, we have

rkB = rkC + rkA.

Definition 38. Let X be a topological space, such that each group Hn(X) is finitely generated.

• The n-th Betti number of X is βn(X) := rkHn(X).• If Hn(X) = 0 for n large enough, the Euler characteristic of X is defined as

χ(X) :=∞∑n=0

(−1)nrkHn(X).

We will encounter in Section 4 large classes of spaces for which the conditions in Definition 38are satisfied. The Betti numbers (and hence also the Euler characteristic) are homotopy invariants,by Theorem 23.

Exercise 39. Show that for a contractible space X, we have χ(X) = 1.

Remark 40. The Betti numbers do not contain all information about the homology groups, sinceoften there appears torsion in the homology groups (see e.g. Theorem 113), meaning that there arefinite cyclic groups in the primary decomposition of Hn(X).

1.5. Recommended exercises: 11, 12, 13, 14, 15, 30 on p132.


2. Relative homology groups

Before going deeper into the theory of the singular homology groups, we define a ‘relative’analogue. This generalisation will actually prove to be very useful in the theory of the originalsingular homology groups in following sections.

2.1. Definition and long exact sequence.

Definition 41. For a pair (X,A), the abelian group Cn(X,A) is given by the quotient groupCn(X)/Cn(A) (see Exercise 19). The ordinary boundary map induces the relative boundarymap ∂rn : Cn(X,A)→ Cn−1(X,A), as

∂rn(a+ Cn(A)) := ∂n(a) + Cn−1(A), for a ∈ Cn(X).

The homology groupsHn(C•(X,A)) of the corresponding chain are the relative homology groupsHn(X,A).

We thus have

(2.1) Hn(X,A) = {[a+ Cn(A)] | a ∈ Cn(X), ∂a ∈ Cn−1(A)}.

The definition Cn(X,A) := Cn(X)/Cn(A) implies a quotient map j : Cn(X) → Cn(X,A), withkernel ι](Cn(A)) for ι : A ↪→ X. The notion of j allows to rewrite the above definition of ∂r as∂rn(j(a)) := j(∂n(a)). We this have a commutative diagram with exact columns

(2.2) 0

��

0

��

0

��· · · ∂ // Cn+1(A)

∂ //

ι]

��

Cn(A)∂ //

ι]

��

Cn−1(A)∂ //

ι]

��

· · ·

· · · ∂ // Cn+1(X)∂ //

j

��

Cn(X)∂ //

j

��

Cn−1(X)∂ //

j

��

· · ·

· · · ∂r// Cn+1(X,A)∂r //

��

Cn(X,A)∂r //

��

Cn−1(X,A)∂r //

��

· · ·

0 0 0

or, in other words, a short exact sequence of chain complexes. By Exercise 21, the above diagramleads to group homomorphisms j∗ : Hn(X)→ Hn(X,A), with j∗([α]) = [j(α)].

Now we investigate the group homomorphisms ι∗ and j∗

(2.3) Hn(A)ι∗ // Hn(X)

j∗ // Hn(X,A).

Exercise 42. Consider the group homomorphisms in (2.3).

(i) Show that im ι∗ = ker j∗.(ii) Show that im j∗ = {[a+ Cn(A)] | a ∈ Cn(X), ∂a = 0}.

(iii) Show that ker ι∗ = {[∂b] | b ∈ Cn+1(X) with ∂b ∈ Cn(A)}.

We thus find that (2.3) is exact, but does not always lead to a short exact sequence, sinceim j∗ ( Hn(X,A), by equation (2.1), and ker ι∗ 6= 0.

The reasons that im j∗ ( Hn(X,A) and ker ι∗ 6= 0 boil down to the same principle. Namely, thepossible existence of a ∈ Cn(X) with 0 6= ∂a ∈ Cn−1(A). We will therefore be able to remedy both‘issues’ with one solution.


Clearly, for a ∈ Cn(X) with ∂a ∈ Cn−1(A), we find that ∂a is a cycle in Cn−1(A), it is howevernot necessarily a boundary since we need not have ∂a = ∂b for some b ∈ Cn(A). We define a grouphomomorphism

(2.4) δ : Hn(X,A)→ Hn−1(A), [a+ Cn(A)] 7→ [∂a],

for an arbitrary cycle a+ Cn(A) ∈ Cn(X,A).

Exercise 43. Show that δ is well defined, meaning the homology class of ∂a does not depend onthe explicit choice of representative a in Cn(X), only on [a+ Cn(A)].

All constructions we have made so far in this section also make sense for the reduced homologygroups. For this we add two copies of Z at the right end of the first two rows in diagram (2.2).

Theorem 44. We have long exact sequences of group homomorphisms

· · · δ // Hn(A)ι∗ // Hn(X)

j∗ // Hn(X,A)δ // Hn−1(A)

ι∗ // Hn−1(X)j∗ // · · ·

and

· · · δ // Hn(A)ι∗ // Hn(X)

j∗ // Hn(X,A)δ // Hn−1(A)

ι∗ // Hn−1(X)j∗ // · · ·

Proof. We clearly have im δ = ker ι∗, using the expression in Exercise 42(iii). It is also easilychecked that ker δ = im j∗. �

Exercise 45. Show that for a space V with deformation retract A we have Hn(V,A) = 0 for alln ∈ N.

Exercise 46. For any point x ∈ X viewed as a subspace {x} ⊂ X, we have group isomorphisms

Hn(X) → Hn(X, {x}), [α] 7→ [α+ Cn({x})] for any n-cycle α, for all n ∈ N.

Remark 47. Since our proof of Theorem 44 was entirely algebraic and did not use any specificsabout the groups Hn(X), we actually proved the following general principle of homological algebra.For a short exact sequence A• ↪→ B• � C• of chain complexes, there exist group homomorphismsδ : Hn(C•)→ Hn−1(A•), leading to an exact sequence

· · · → Hn+1(C•)→ Hn(A•)→ Hn(B•)→ Hn(C•)→ Hn−1(A•)→ · · · .

2.2. Induced morphisms and naturality. We expand the idea of maps of pointed spaces topairs.

Definition 48. For two pairs (X,A), (Y,B), the notation f : (X,A) → (Y,B) stands for a mapf : X → Y with f(A) ⊆ B.

Just as in Subsection 1.2, a map f : (X,A) → (Y,B) induces group homomorphisms f r] :

Cn(X,A)→ Cn(Y,B), asf r] (a+ Cn(A)) = f](a) + Cn(B).

This is easily seen to be a chain map, so that we also have group homomorphisms f r∗ : Hn(X,A)→Hn(Y,B).

When we consider a homotopy {ft : (X,A) → (Y,B) | t ∈ I}, it means an ordinary homotopy,with additional condition ft(A) ⊆ B for each t ∈ I.

Example 49. With the above notation, {ft : (I, ∂I) → (X, {x0})} thus stands for a homotopy ofloops (see Section I.1.1).

Definition 50. A homotopy equivalence of pairs (X,A) and (Y,B) is a map f : (X,A)→ (Y,B)for which there exists g : (Y,B) → (X,A) and homotopies {ht : (X,A) → (X,A) | t ∈ I} and {h′t :(Y,B)→ (Y,B) | t ∈ I} with h0 = 1X , h1 = gf , h′0 = 1Y and h′1 = fg. We write (X,A) ' (Y,B) ifthere exists such a homotopy equivalence.


Proposition 51. For a homotopy {ft : (X,A) → (Y,B) | t ∈ I}, we have f r0∗ = f r1∗ : Hn(X,A) →Hn(Y,B).

Proof. Mutatis mutandis the proof of Proposition 27. �

Corollary 52. For a homotopy equivalence f : (X,A) → (Y,B) of pairs, f r∗ : Hn(X,A) →Hn(Y,B) is a group isomorphism, for each n ∈ N.

(SI) Now we demonstrate that the group homomorphisms δ : Hn(X,A) → Hn−1(A) actuallyconstitute a ‘natural transformation’ between functors Pair → Ab, where Hn−1(−) has to beinterpreted as such a functor appropriately.

Exercise 53 (naturality). Show that for a map f : (X,A) → (Y,B), with restriction f0 : A → B,we have a commutative diagram

· · · δ // Hn(A)ι∗ //

f0∗��

Hn(X)j∗ //

f∗��

Hn(X,A)δ //

fr∗��

Hn−1(A)ι∗ //

f0∗��

Hn−1(X)j∗ //

f∗��

· · ·

· · · δ // Hn(B)ι∗ // Hn(Y )

j∗ // Hn(Y,B)δ // Hn−1(B)

ι∗ // Hn−1(Y )j∗ // · · ·

This naturality has very useful applications, as the following exercise demonstrates.

Exercise 54. Let f : X → Y be a homotopy equivalence, such that the restriction f0 : A → B(with A ⊂ X and B ⊂ Y ) is also a homotopy equivalence. Use the five lemma (see e.g. p129 inHatcher) to show that the homomorphisms

f r∗ : Hn(X,A)→ Hn(Y,B)

are isomorphisms.

When we have a space X with subspaces A1 ⊂ A2 ⊂ X, we can also construct a short exactsequence using ι : A2 ↪→ X as follows

0 // Cn(A2, A1)ιr] // Cn(X,A1)

(1X)r] // Cn(X,A2) // 0,

where (1X)r] , by definition, is the map α+ Cn(A1) 7→ α+ Cn(A2). Since these also lead to a shortexact sequence of chain complexes, by Remark 47 we get another long exact sequence

(2.5) · · · → Hn+1(X,A2)→ Hn(A2, A1)→ Hn(X,A1)→ Hn(X,A2)→ Hn−1(A2, A1)→ · · · .The case A1 = ∅ yields the first sequence in Theorem 44, the case A1 = pt, the second sequence.

2.3. Barycentric subdivision. In this section we will informally describe (a complete treatmentis given on p120-122 of Hatcher) a procedure for singular homology which a priori is not related torelative homology groups. However, we will rely on these ideas to prove the Excision Theorem inthe next section. The main idea is to divide the standard n-simplex into (n+1)! smaller n-simplices,which will lead to a chain map which is chain homotopic to the identity. The barycentre of ∆n is

bn := (1

n+ 1, . . . ,

1

n+ 1) ∈ ∆n.

We define the barycentric subdivision of the standard n-simplex by induction on n. Forn = 0, there is only one 0-simplex in the subdivision of ∆0. For n = 1, the standard 1-symplexis the union of the 1-simplices [(1, 0), b1] and [b1, (0, 1)]. In general, taking bn together with one ofthe n− 1-simplices in the barycentric subdivision in one of the faces of ∆n yields (n+ 1)! choices ofn-simplices, which together build ∆n. In particular, we have a natural bijection between the n− 1-simplices in the barycentric subdivision of ∂∆n and the n-simplices in the barycentric subdivisionof ∆n.

To each simplex in the subdivision of ∆n we associate its ‘inclusion map’ α : ∆n → ∆n.


Example 55. We have αi : ∆1 → ∆1, for i ∈ {0, 1}, given by α0(t0, t1) = (t0/2 + t1, t0/2) andα1(t0, t1) = (t0/2, t0/2 + t1).

We fix a topological space X. The group homomorphisms S : Cn(X)→ Cn(X) are defined by

S(σ) =∑α

(−1)sασα

for a singular n-simplex σ : ∆n → X. Here, the summation over α corresponds to all (n+ 1)! mapsin the barycentric subdivision, where the signs (−1)sα are chosen such that S is a chain map, i.e.

(2.6) S∂ = ∂S.

Example 56. For σ : ∆1 → X, we have S(σ) = σα0 − σα1. On C0(X), S is the identity.

Exercise 57. Using the general description on p120 of Hatcher, determine the explicit form of S onC2(X) and show that ∂2S = S∂2 on C2(X).

We have a map

γ : ∆n+1 → ∆n, (t0, t1, . . . , tn+1) 7→ t0bn + (t1, t2, . . . , tn+1).

There exist group homomorphisms T : Cn(X) → Cn+1(X) which are given by σ 7→ σγ, for σ :∆n → X, with additional ‘boundary terms’. We refer to p121 in Hatcher for a precise definition.

Example 58. For σ : ∆0 → X, we have T (σ) = σγ. For σ : ∆1 → X, we have

T (σ) = σγ − σ(0, 1)γγ + σ(1, 0)γγ.

In general, it can be proved that

(2.7) ∂T + T∂ = 1− S, on Cn(X).

In particular, this means that the induced homomorphisms S∗ : Hn(X)→ Hn(X) are the identity,by Exercise 26.

Exercise 59. Show that equation (2.7) is satisfied for n = 1.

Corollary 60. For any m ∈ N, a chain homotopy between Sm and 1 is given by

Tm =m−1∑i=0

TSi.

In other words, we have ∂Tm + Tm∂ = 1− Sm.

Proof. This follows from a direct computation using equations (2.7) and (2.6). �

For m = 0, we interpret the above expressions as S0 = 1Cn(X) and T0 = 0.Now we will use the principle of barycentric subdivision to prove that, when X = A∪B for two

open subspaces A and B, we can calculate Hn(X) by only looking at singular simplices with imagein A or B. This is an analogue of van Kampen’s theorem for the fundamental group.

Definition 61. For a collection U = {Uα} of subspaces Uα ⊂ X, let CUn (X) =∑

αCn(Uα) be thesubgroup of Cn(X) generated by the subgroups {Cn(Uα) < Cn(X)}. The boundary map ∂ restrictsto ∂ : CUn (X)→ CUn−1(X) and the resulting homology groups are denoted by HUn (X). The inclusion

iU : CUn (X) ↪→ Cn(X) (which restricts to the identity map on each Cn(Uα)) is a chain map andhence induces group homomorphisms iU∗ : HUn (X)→ Hn(X), by Exercise 21.

Proposition 62. Consider X and U = {Uα} a collection of subspaces with X = ∪αint(Uα).

(i) The map iU∗ : HUn (X)→ Hn(X) is a group isomorphism for each n ∈ N.


(ii) There is a chain map ρ : Cn(X) → CUn (X), with ρiU = 1CUn (X), and group homomorphisms

D : Cn(X)→ Cn+1(X) such that

∂D +D∂ = 1Cn(X) − iUρ.

Furthermore, D is identically zero on CU• (X).

Proof. Part (i) follows from Exercise 26 and part (ii), so we only prove part (ii).We will use, without proving it, the Lebesgue covering lemma, which implies that for each singular

n-simplex σ : ∆n → X, there exists m ∈ N, such that Sm(σ) ∈ CUn (X). We define m(σ) ∈ N to bethe minimal such number.

We start by defining group homomorphisms D : Cn(X) → Cn+1(X), where we set D(σ) =Tm(σ)(σ), with Tm as defined in Corollary 60. Note that T0 = 0, so D is identically zero on CUn (X).Now we define group homomorphisms ρ′ : Cn(X)→ Cn(X) as

ρ′ := 1− ∂D −D∂.By construction ρ′ constitutes a chain map. Furthermore, we claim that im ρ′ ⊂ CUn (X). To provethis, observe that Corollary 60 shows that for any singular n-simplex σ

ρ′(σ) = σ − ∂Tm(σ)σ −D∂σ = Sm(σ)σ + (Tm(σ) −D)∂σ.

By definition, Sm(σ)σ ∈ CUn (X). Furthermore, the restriction to each face of σ, is a singular n− 1-simplex η with m(η) ≤ m(σ). The second term above is a linear combination of (Tm(σ)−Tm(η))(η)for such η. By definition of Tm, we have

(Tm(σ) − Tm(η))(η) =

m(σ)−1∑i=m(η)

TSi(η).

Since each Si(η), with i ≥ m(η) is in CUn−1(X) and T maps CUn−1(X) to CUn (X), the claim follows.

It thus follows that we can write ρ′ = iUρ, for some chain map ρ. By construction, D yields achain homotopy between 1 and iUρ.

Finally, we remark that DiU = 0 implies that ρ′iU = iU , or ρiU = 1CU (X). This concludes the

proof of part (ii). �

2.4. Excision.

Theorem 63 (Excision).

(i) For a pair (X,A) and Z ⊂ X such that clZ ⊂ intA, the inclusion ι : (X−Z,A−Z) ↪→ (X,A)induces isomorphisms ιr∗ : Hn(X − Z,A− Z)→ Hn(X,A) for all n ∈ N.

(ii) For subspaces A,B of X such that X = intA ∪ intB, the inclusion ι : (B,A ∩ B) ↪→ (X,A)induces isomorphisms ιr∗ : Hn(B,A ∩B)→ Hn(X,A) for all n ∈ N.

Exercise 64. Show that the two parts of the theorem are equivalent (using Exercise 0.6).

Proof of Theorem 63. We can restrict to proving part (ii), by Exercise 64. We set U = {A,B}. Westudy the chain map ιr] : Cn(B,A ∩B)→ Cn(X,A), which allows a commutative diagram

(2.8) Cn(B)/Cn(A ∩B)ιr] //

∼

))

Cn(X)/Cn(A)

CUn (X)/Cn(A)

iU66

Here, the isomorphism Cn(B)/Cn(A∩B)→ CUn (X)/Cn(A) is given by a+Cn(A∩B) 7→ a+Cn(A),

and iU is the relative version of iU , meaning

iU (a+ Cn(A)) = iU (a) + Cn(A) = a+ Cn(A).


We can interpret (2.8) as a commutative diagram of chain maps, for the logical choice of boundarymorphism on CUn (X)/Cn(A).

Consider ρ and D from (the proof of) Proposition 62. As ρ(Cn(A)) ⊂ Cn(A) and D(Cn(A)) =0 ⊂ Cn+1(A), we can also define corresponding relative versions

ρ : Cn(X)/Cn(A)→ CUn (X)/Cn(A) and D : Cn(X)/Cn(A)→ Cn+1(X)/Cn+1(A).

By construction, ρ is a chain map and by Proposition 62(ii) we have

∂rD +D∂r = 1Cn(X,A) − iUρ and ρiU = 1CUn (X)/Cn(A).

By Exercise 26, iU∗ yields isomorphisms of homology groups. Diagram (2.8) thus implies that ιr∗yields isomorphisms as well. �

2.5. Recommended exercises: 16, 18, 20, 24, 26 and 27 on p132-133.


3. Calculation of singular homology groups

3.1. The long exact sequence for good pairs. For a pair (X,A) the canonical inclusion ι :A ↪→ X and the quotient map q : X → X/A (see Section 0.1) form a ‘short exact sequence oftopological spaces’

∅ // Aι // X

q // X/A // ∅.In general, the induced group homomorphisms ι∗ and q∗ will not yield a short exact sequence ofhomology groups. However, for certain pairs, we will be able to deform the long exact sequence ofrelative homology (in Theorem 44) into a long exact sequence generated by the above short exactsequence.

Definition 65. A pair (X,A) is a good pair if A is a nonempty closed subspace that is a deformationretract of some neighbourhood in X.

Exercise 66. Show that (Dn,Sn−1) is a good pair for all n ∈ Z≥1.

Theorem 67. For a good pair (X,A), there exist group homomorphisms ς : Hn(X/A)→ Hn−1(A),which lead to a long exact sequence

// Hn(A)ι∗ // Hn(X)

q∗ // Hn(X/A)ς // Hn−1(A)

ι∗ // · · ·q∗ // H0(X/A) // 0.

We start the preparations for the proof of Theorem 67.

Exercise 68. Consider a pair (X,A), with quotient map q : X → X/A, which we also interpretas q : (X,A) → (X/A,A/A). Recall the induced group homomorphisms q∗ and qr∗, the grouphomomorphisms j∗ introduced in Section 2.1, and the group isomorphism of Exercise 46:

Hn(X,A)qr∗

''Hn(X)

j∗99

q∗

%%

Hn(X/A,A/A)

Hn(X/A)

∼77

Show that this diagram is commutative.

Proposition 69. For a good pair (X,A), the quotient map q induces an isomorphism qr∗ : Hn(X,A)→Hn(X/A,A/A), such that the corresponding isomorphism Hn(X,A)

∼→ Hn(X/A) (using Exer-cise 46) admits a commutative diagram

Hn(X)j∗ //

q∗

%%

Hn(X,A)

∼��

Hn(X/A)

Proof. That the diagram is commutative follows immediately from Exercise 68. It thus remains tobe proved that

qr∗ : Hn(X,A)→ Hn(X/A,A/A)

is an isomorphism for a good pair (X,A).Let V be a neighbourhood of A such that A is a deformation retract of V . We reserve ι

for the inclusion ι : X − A ↪→ X (and hence also for (X − A, V − A) ↪→ (X,V )). The latter


map, 1X : (X,A) → (X,V ) and 1X/A : (X/A,A/A) → (X/A, V/A) induce the following grouphomomorphisms:

Hn(X,A)(1X)r∗ //

qr∗��

Hn(X,V )

qr∗��

Hn(X −A, V −A)ιr∗

oo

θ��

Hn(X/A,A/A)(1X/A)r∗ // Hn(X/A, V/A) Hn(X/A−A/A, V/A−A/A)τ

oo

The homomorphism θ is induced from the canonical homeomorphism X −A→ X/A−A/A, whichclearly leads to a homeomorphism of pairs. In particular, θ is an isomorphism. The homomorphismτ is induced from the inclusion X/A − A/A ↪→ X/A. Commutativity of the diagram follows fromthe fact that the maps from which the morphisms are induced already yield commutative diagrams.

By assumption, we have A = clA ⊂ intV , so ιr∗ is an isomorphism by Theorem 63(i). Similarly,A/A is closed in X/A by definition of the quotient topology and V/A is a neighbourhood of A/A.Theorem 63(i) thus also implies that τ is an isomorphism.

By Exercise 45, we have Hk(V,A) = 0 and Hk(V/A,A/A) = 0, for all k ∈ N. It then followsfrom equation (2.5) that the left horizontal arrows are also isomorphisms. Another way to provethe latter uses Exercise 54.

Commutativity of the diagram then implies that also the left vertical arrow is an isomorphism,concluding the proof. �

Proof of Theorem 67. Let φ : Hn(X,A)→ Hn(X/A) be the isomorphism in Proposition 69. Recallδ : Hn(X,A)→ Hn−1(A) from equation (2.4). We define

ς := δφ−1 : Hn(X/A)→ Hn−1(A).

The long exact sequence thus follows from Theorem 44. �

Exercise 70. Assume (X,A) is a good pair where both X and A satisfy the properties in Defini-tion 38, show that X/A also satisfies these properties and

χ(X) = χ(A) + χ(X/A).

We conclude with another application of Proposition 69. We call a pointed space (X,x0) goodif the corresponding pair (X, {x0}) is good in the sense of Definition 65.

Corollary 71. For a family of good pointed spaces {(Xα, xα)}, the inclusions ια : Xα ↪→ ∨αXα

induce an isomorphism ⊕α

Hn(Xα) → Hn(∨αXα).

Proof. We consider the good pair (X,A), with X = qαXα and A = qα{xα}. We then have adiagram of group homomorphisms⊕

α Hn(Xα)⊕ια∗ // Hn(∨αXα)

⊕αHn(Xα, xα)

∼

OO

∼ // Hn(X,A)

∼

OO

where the isomorphisms in the vertical arrows are given by Proposition 69 and Exercise 46. It canbe checked that the diagram is commutative. �

Exercise 72. Show that the analogue of Corollary 71 does not hold true for (non-reduced) singularhomology groups.


3.2. The homology of the sphere. The following simple calculation of the homology groups ofthe spheres demonstrate the power of the developed techniques.

Theorem 73. For any k ∈ N, we have

Hn(Sk) =

{0 if n 6= k

Z if n = k.

Proof. The good pair (Dk+1,Sk) in Exercise 66 implies a short exact sequence

0→ Sk → Dk+1 → Sk+1 → 0,

see Example 0.52. By Theorem 67 and Exercise 28, we thus have Hn+1(Sk+1) ∼= Hn(Sk) for alln, k ∈ N. The result then follows from Exercise 18. �

Exercise 74. If Sn ' Sm, show that m = n.

Exercise 75. Show that the Euler characteristics of the spheres are χ(Sn) = 1 + (−1)n.

Exercise 76. Study Theorem 2.26 on p126 in Hatcher.

Now we introduce the notion of degree for any map from Sn to itself.

Exercise 77. Show that any group homomorphism Z→ Z is of the form a 7→ da for some d ∈ Z.

Definition 78. For any map f : Sn → Sn, with n > 0, its degree deg(f) ∈ Z is defined as f∗(1),with f∗ : Z→ Z, the induced homomorphism of the n-th singular homology group of Sn.

Exercise 79. Consider maps f, g : Sn → Sn. Show that

(i) deg(1Sn) = 1;(ii) deg(fg) = deg(f) deg(g);

(iii) deg(f) = deg(g) if f ' g;(iv) if f is a homotopy equivalence, then deg(f) ∈ {1,−1}.

Exercise 80. Find a homeomorphism f : S1 → S1 with deg f = −1.

3.3. Attaching cells. In this subsection, we fix an arbitrary space A and a map ϕ : Sk−1 → A,for some k ∈ Z≥1. We set

X := A tϕ Dk and ι : A ↪→ X.

Exercise 81. Show that (X,A) forms a good pair, using Exercises 0.44 and 0.45.

Proposition 82. Consider the pair (X,A) above.

(i) Hn(X,A) ∼=

{Z if n = k

0 otherwise.

(ii) We have an isomorphism ι∗ : Hn(A) → Hn(X) if n 6∈ {k, k − 1}.(iii) There exists an exact sequence of abelian groups

0 // Hk(A)ι∗ // Hk(X) // Z // Hk−1(A)

ι∗ // Hk−1(X) // 0.

(iv) Assuming A satisfies the restrictions in Definition 38, then

χ(X) := χ(A) + (−1)k.

Proof. We start by proving part (i). By Exercise 81 and Proposition 69, we have

Hn(X,A) ∼= Hn(X/A).

Moreover, we clearly have X/A ∼= Dk/Sk−1 ∼= Sk. Theorem 73 then concludes the proof of part (i).Parts (ii) and (iii) are then immediate consequences of Theorem 44.


Part (ii) implies that βn(X) = βn(A) if n 6∈ {k − 1, k}, while part (iii) and Exercise 37 implythat

βk(X)− βk−1(X) = βk(A)− βk−1(A) + 1,

which implies part (iv). �

Remark 83. Proposition 82(iii) implies in particular a monomorphism ι∗ : Hk(A) ↪→ Hk(X) andan epimorphism ι∗ : Hk−1(A) � Hk−1(X).

Observe the resemblance between, on the one hand, the latter epimorphism and Proposition 82(ii),and on the other hand, Propositions I.72 and I.74(i).

We can iteratively apply the results in Proposition 82 to obtain the following corollary. Inparticular, this states that the Euler characteristic (as in Definition 38) of a finite cell complexcoincides with the notion of Euler characteristic already introduced in Definition 0.50.

Corollary 84. Let X = Xm be a (finite) m-dimensional cell complex.

(i) We have Hn(X) = 0 for n > m.(ii) For k < m, the map Hn(Xk) → Hk(X) induced from the inclusion Xk ↪→ X is an isomor-

phism for n < k and an epimorphism for k = n.(iii) The Euler characteristic χ(X) is the number of even cells minus the number of odd cells.(iv) Hn(Xk, Xk−1) = 0 if k 6= n.(v) Hn(Xn, Xn−1) ∼= Z⊕d, with d the number of the n-cells of X.

Proof. Parts (i)-(iii) follow from Proposition 82. Parts (iv) and (v) follow from the followingobservations. Since (Xk, Xk−1) is a good pair, Proposition 69 implies that Hn(Xk, Xk−1) ∼=Hn(Xk/Xk−1). Furthermore, we have

Xk/Xk−1 ∼= ∨α(Dkα/∂Dkα) ∼= ∨αSkα.

The results thus follow from Corollary 71 and Theorem 73. �

Remark 85. We can formulate Corollary 84(v) slightly more precisely. The characteristic maps

Φkα : Dkα → Xk of Definition 0.54 lead to maps of pairs Φk

α : (Dkα, ∂Dkα) → (Xk, Xk−1) which yield

an isomorphism

⊕α(Φkα)r∗ :

⊕α

Hk(Dkα, ∂D

kα) → Hk(X

k, Xk−1).

This can be proved rigorously from commutative diagram

Hk(qαDkα,qα∂Dkα)(qαΦkα)r∗ //

∼��

Hk(Xk, Xk−1)

∼��

Hk(∨α(Dkα/∂Dkα))

∼ // Hk(Xk/Xk−1)

Remark 86. For k = 0, we interpret Xk−1 = X−1 = ∅ in formulas as in Corollary 84.

3.4. Recommended exercises: 17, 22 on p132.


4. Simplicial and cellular homology

We end the chapter on homology by discussing two methods to compute singular homologygroups in practical situations. Both methods correspond to homology theories in their own right,which are only concerned with specific types of topological spaces, but recover the singular homologygroups for those spaces.

The first theory, simplicial homology corresponds to calculating the homology groups of aspecific sub-complex of (1.2), which happens to yield the same homology groups (we have alreadyencountered another example of this general principle in Proposition 62(i)). It should be pointed outthat historically the development of simplicial homology actually preceded singular homology. Theconstructions of simplicial homology are somewhat reminiscent of the definition of cell complexes.Correspondingly, the second theory, cellular homology, will be concerned with cell complexesand it replaces simplices by cells. Contrary to simplicial homology, in order to define it we willalready need notions and results from singular homology. However, the end result is similar, inthe sense that we obtain a method to compute the singular homology groups for specific spaces,namely cell complexes.

4.1. ∆-complexes and simplicial homology. In this subsection we consider topological spaceswith a fixed ‘decomposition’ into regular n-simplices, called ∆-complexes. To such a ∆-complex X,we will associate simplicial homology groups H∆

n (X), by restricting the boundary homomorphismsin (1.2) to subgroups.

Definition 87. A ∆-complex structure on a space X is a collection of (regular) simplices σα :∆nα → X (where nα ∈ N varies) in X, such that

(i) as sets, we have X = qασα(∆nα);(ii) each restriction of σα to a face of ∆nα is one of the nα − 1-simplices σβ in X;

(iii) each set U ⊂ X is open if and only if σ−1α (U) is open for all α.

A ∆-complex structure is called finite, if there are only finitely many simplices involved.

We will only work with finite ∆ complexes.

Exercise 88. By parts (i) and (iii) of the definition, X is a quotient of the space qα∆nα . Show that

the restriction of the quotient map q : qα∆nα → X to qα∆nα is not necessarily a homeomorphism,despite being a bijective map.

(SI) In general a quotient space of a Hausdorff space need not be Hausdorff. However, due tothe nature of (regular) simplices and the above definition, spaces with ∆-complex structure mustbe Hausdorff. In particular, not every space admits a ∆-structure.

Example 89. Let [x1,x2,x3,x4] be four points in R2 forming the vertices of a convex quadrilateralQ. We can define two 2-simplices in A σa : ∆2 → Q with σa(t0, t1, t2) = t0x1 + t1x2 + t2x3 andσa : ∆2 → Q with σb(t0, t1, t2) = t0x1 + t1x4 + t2x3. This defines precisely five distinct 1-simplices,from the six faces. We also have four 0-simplices, each one mapping one point to one of the vertices.These 11 (2 + 5 + 4) simplices together form a ∆-complex on Q.

Example 90. Consider the torus in the form T = Q/∼, with Q a square, where we identify oppositeedges. We have two 2-simplices given by the composition of σa and σb in Example 89 with q : Q→ S.Taking the faces of the two 2-simplices yields three distinct 1-simplices, and taking their faces yieldsone 0-simplex. These 6 (2 + 3 + 1) simplices form a ∆-complex for the torus.

A space X with a fixed ∆-complex structure will simply be called a ∆-complex X. Similarlyto cell complexes, we will refer to the enα := σα(∆nα) in Definition 87(i) as the open n-simplicesof X. In fact, one sees that the ∆-complex structure on X makes X into a cell complex withcharacteristic maps σα. We will not prove this explicitly, but it essentially follows from the trivialobservation that that ∆n ∼= Dn.


Remark 91. (SI) Conversely, although harder to prove, it is also true that any cell complex has a∆-complex structure, even though the number of simplices might be higher than the number ofcells.

Definition 92. A subspace A ⊂ X of a ∆-complex X is a subcomplex if it is the union of acollection of σα(∆nα).

Exercise 93. Show that the n-simplex ∆n itself is a ∆-complex where, for each 0 ≤ k ≤ n, thenumber of k-simplices is

(n+1k+1

). Show that ∂∆n = ∆n − int∆n constitutes a subcomplex.

Definition 94. For a ∆-complex X, the group of n-chains, ∆n(X) is the subgroup of Cn(X)generated by all n-simplices in the ∆-complex structure of X. For a subcomplex A, we have thegroup of relative chains ∆n(X,A) = ∆n(X)/∆n(A).

By definition, the boundary homomorphism ∂n in Definition 9 restricts to a homomorphism∆n(X) → ∆n−1(X). Since ∂∆n(A) ⊂ ∆n(A) for a subcomplex A, we can define similarly therelative boundary homomorphism ∆n(X,A)→ ∆n−1(X,A).

Definition 95. The quotient H∆n (X) = ker ∂n/im ∂n+1 (for the above restriction of the boundary

homomorphisms) is the n-th simplicial homology group of X. Similarly, we have the relativesimplicial homology group H∆

n (X,A), for any subcomplex A.

Exercise 96. Compute H∆n (Q) for Q the quadrilateral as a ∆-complex as in Example 89.

Exercise 97. Compute H∆n (T ) for T the torus as a ∆-complex as in Example 90.

Exercise 98. Compute H∆n (∆2), for ∆2 as a ∆-complex as in Exercise 93.

4.2. Equivalence of simplicial and singular homology. In this section we show that for any∆-complex X, we actually have group isomorphisms H∆

n (X) ∼= Hn(X). There is an obvious chainmap ∆•(X,A)→ C•(X,A), which leads to canonical homomorphisms H∆

n (X,A)→ Hn(X,A).

Theorem 99. For a ∆-complex X with subcomplex A (including the case A = ∅), the homomor-phisms H∆

n (X,A)→ Hn(X,A) are isomorphisms for all n ∈ N.

We start the proof with the following exercise.

Exercise 100. Since we have a homeomorphism (∆n, ∂∆n) ∼= (Dn, ∂Dn), Corollary 84 implies that

Hn(∆n, ∂∆n) ∼= Z.Observe that 1∆n : ∆n → ∆n is an element of Cn(∆n) and that 1∆n + Cn(∂∆n) is a cycle inCn(∆n, ∂∆n). Show that [1∆n + Cn(∂∆n)] actually generates Hn(∆n, ∂∆n). The outline of theproof is given in Example 2.23 on p125 in Hatcher.

Proof of Theorem 99. Similarly to cell complexes, we set Xk = ∪nα≤kim (σα) for a ∆-complex X.We can also compute Hn(Xk, Xk−1) similarly as for cell complexes, for the good pair (Xk, Xk−1),where moreover, Xk−1 is a subcomplex of the ∆-complex Xk. We will first prove that the homo-morphism is indeed always an isomorphism for the pair (Xk, Xk−1).

We construct a commutative diagram

Hn(Xk, Xk−1)

qr∗��

⊕αHn(∆k

α, ∂∆kα)

⊕α(σα)r∗

oo

qr∗��

Hn(Xk/Xk−1) Hn(qα∆kα/qα ∂∆k

α)oo

The lower horizontal arrow is an isomorphism, since it is induced from a homeomorphism. Thevertical arrows are isomorphisms, since they involve good pairs, by Proposition 69. Hence the top


horizontal arrow is an isomorphism and we find by Exercise 100 that Hn(Xk, Xk−1) = 0 when n 6= kand Hn(Xn, Xn−1) is a free abelian group with generators [σα + Cn(Xn−1)], since (σα)](1∆n) =

σα. It is clear that H∆n (Xk, Xk−1) = 0 when n 6= k and H∆

n (Xn, Xn−1) ∼= ∆n(Xn, Xn−1) isthe free abelian group with generators [σα + ∆n(Xn−1)]. This means that the homomorphismH∆n (Xk, Xk−1)→ Hn(Xk, Xk−1) is indeed always an isomorphism.Now we use this observation to prove the claim for pairs (X,∅), by induction on the dimension

of the largest simplex in the complex. If X = X0 = qαpt, the claim is easily checked. Assumethe claim holds true up to dimension k − 1 for k > 0. For any X = Xk, we have a commutativediagram

· · · // H∆n+1(Xk, Xk−1) //

��

H∆n (Xk−1) //

��

H∆n (Xk) //

��

H∆n (Xk, Xk−1) //

��

· · ·

· · · // Hn+1(Xk, Xk−1) // Hn(Xk−1) // Hn(Xk) // Hn(Xk, Xk−1) // · · ·

The lower horizontal row is Theorem 44, the upper horizontal row is similarly obtained from theshort exact sequence ∆•(X

k−1) ↪→ ∆•(Xk) � ∆•(X

k, Xk−1), see Remark 47. By the inductionhypothesis and the previous paragraph, all vertical arrows except H∆

n (Xk)→ Hn(Xk) are isomor-phisms. That H∆

n (Xk)→ Hn(Xk) is an isomorphism then follows from the five-lemma.For the case (X,A) with A 6= ∅ follows from the previous paragraph and the 5-lemma, using the

long exact sequence in Theorem 44. �

A nice consequence is that we can compute the Euler characteristic of a finite ∆-complex withoutactually computing any homology, only by looking at the ranks of the groups ∆n(X).

Corollary 101. For a finite ∆-complex X, we have

χ(X) =∞∑n=0

(−1)nrk∆n(X).

Proof. By Theorem 99 and Definition 38, we have

χ(X) =

∞∑n=0

(−1)nrkH∆n (X).

We have short exact sequences

0→ im ∂n+1 → ker ∂n → H∆n (X)→ 0

and

0→ ker ∂n → ∆n(X)→ im ∂n → 0.

By Exercise 37, we thus have

rkH∆n (X) = rk ker ∂n − rk im ∂n+1 = rk∆n(X)− rk im ∂n+1 − rk im ∂n.

Since the sum in the definition of the Euler characteristic is alternating, the claim now follows. �

Remark 102. Note that rk∆n(X) is just the number of n-simplices in the ∆-complex structure.


χ(S1 × S1) = 0, χ(�2) = 1 and χ(∆n) = 1 for all n ∈ N.


4.3. Cellular homology. Let X = Xm be an m-dimensional cell complex and recall the nota-tion ekα for the k-cells of X and ϕα : Sk−1 → Xk−1 for the corresponding attaching map.

Definition 104. Let CCWn (X) denote the free abelian group of formal Z-linear combinations of then-cells of X. For n > 1, the cellular boundary map ∂CWn : CCWn (X)→ CCWn−1(X) is defined by

∂CWn (enα) =∑β

dαβen−1β ,

where dαβ = deg(ϕαβ). The map ϕαβ : Sn−1 → Sn−1 defined as the composition

Sn−1 ϕα // Xn−1 q // Xn−1/(Xn−1 − en−1β )

∼ // Dn−1β /∂Dn−1

β∼ // Sn−1 ,

with q is the canonical quotient map. Furthermore, we set ∂CW1 (e1α) = ϕα(1)−ϕα(−1) and ∂CW0 = 0.

Remark 105. Since the last homeomorphism in the sequence of maps in Definition 104 is notcanonical, the resulting map is only defined up to composition with a homeomorphism. The degreedαβ is therefore only defined up to sign, see Exercises 79 and 80. However, if we fix the chosen

homeomorphism (meaning, we take one and the same Dn−1β /∂Dn−1

β∼→ Sn−1 for each α), the resulting

homology groups will be isomorphic. We will ignore this subtlety from now on.

We will not prove directly that (CCW• (X), ∂CW• ) is a chain complex, but take a detour as follows.By Corollary 84(v), we have Hn(Xn, Xn−1) ∼= CCWn (X). For any n ≤ m, we construct a

group homomorphism Hn(Xn, Xn−1)→ Hn−1(Xn−1, Xn−2), by composing the homomorphism δ :Hn(Xn, Xn−1)→ Hn−1(Xn−1) introduced in equation (2.4) and j∗ : Hn−1(Xn−1)→ Hn−1(Xn−1, Xn−2).By construction, we thus have a commutative diagram

Hn(Xn, Xn−1)dn //

δ

((

Hn−1(Xn−1, Xn−2).

Hn−1(Xn−1)

j∗55

Exercise 106. Show that dndn+1 = 0.

We thus get a chain complex of abelian groups

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

dn // Hn−1(Xn−1, Xn−2) // · · ·

We will show below that the isomorphisms Hn(Xn, Xn−1) ∼= CCWn (X) actually allow to identify∂CW and d. This will show that (CCW• (X), ∂CW• ) is indeed a chain complex and that the homologygroups can be defined as follows.

Definition 107. The cellular homology groups of the finite cell complexX are given byHCWn (X) :=

ker dn/im dn+1, for all n ∈ N.

Theorem 108. For any finite cell complex X, we have HCWn (X) ∼= Hn(X), for all n ∈ N.

Proof. By Corollary 84(ii), we have Hn(X) ∼= Hn(Xn+1). Furthermore, by Theorem 67, we havean exact sequence

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

By Corollary 84(iv), the last term is zero, so in conclusion we find

(4.1) Hn(X) ∼= Hn(Xn)/im δn+1.


By Theorem 44, we have an exact sequence

Hn−1(Xn−2) // Hn−1(Xn−1)j∗ // Hn−1(Xn−1, Xn−2).

By Corollary 84(i), Hn−1(Xn−2) = 0, meaning that j∗ is injective. We thus find ker dn = ker δnand therefore

(4.2) HCWn (X) = ker δn/im dn+1.

Using again the injectivity of j∗, we find an isomorphism

j∗ : im δn+1 → im (j∗δn+1) = im dn+1.

Moreover, since the long sequence in Theorem 67 is exact, we also have an isomorphism

j∗ : Hn(Xn) → ker δn.

It thus follows that j∗ induces an isomorphism between the groups in equations (4.1) and (4.2),concluding the proof. �

We use this result to improve the statement in Corollary 84(i).

Corollary 109. If X is a finite cell complex without k-cells, then Hk(X) = 0. In general, Hk(X)is finitely generated with number of generators bounded by the number of k-cells.

Proof. By Corollary 84(iv), the group Hk(Xk, Xk−1) is zero. In particular, ker dk = 0, which

implies HCWk (X) = 0. The first conclusion then follows from Theorem 108. The more general

statement similarly follows from Corollary 84(v) and Exercise 35(i). �

Now we show that the boundary homomorphisms dn ‘coincide’ with the homomorphisms ∂CWn .

Proposition 110. We can choose isomorphisms CCWk (X)∼→ Hk(X

k, Xk−1) for all k, such thatwe get commutative diagrams

CCWn (X)∂CWn //

∼��

CCWn−1(X)

∼��

Hn(Xn, Xn−1)dn // Hn−1(Xn−1, Xn−2).

Proof. First we make the vertical isomorphisms explicit. We take CCWn (X) → Hn(Xn, Xn−1)such that enα, as an abstract element in CCWn (X), gets mapped to the image of a generator ofHn(Dnα, ∂D

nα) ∼= Z, under

(Φα)∗ : Hn(Dnα, ∂Dnα) ↪→ Hn(Xn, Xn−1).

Note that this construction indeed yields an isomorphism, by Remark 85. That the diagram in theproposition then commutes for n = 1 is then immediate by definition. Now we focus on n > 1.Therefore, we can use reduced homology groups in the arguments below, which is slightly moreconvenient.

Next we observe that we have a commutative diagram (where isomorphisms are written as ⇒)

Hn(Dnα, ∂Dnα)

δn +3

(Φα)r∗��

Hn−1(∂Dnα)

(ϕα)∗��

Hn(Xn, Xn−1)δn // Hn−1(Xn−1)

j∗ //

q∗��

q∗

**

Hn−1(Xn−1, Xn−2)

��

Hn−1(Xn−1/(Xn−1 − en−1β )) Hn−1(Xn−1/Xn−2)q∗

oo


for arbitrary α, β. The right-most commutative triangle is Proposition 69. Following the path of 1 ∈Z ∼= Hn−1(∂Dnα) down the middle column, leads by Definition 104 to dαβ ∈ Z ∼= Hn−1(Xn−1/(Xn−1−en−1β )).

Now we consider enα ∈ CCWn (X) as an element in Hn(Xn, Xn−1) by the isomorphism introducedin the first paragraph. By definition of the latter isomorphism, and commutativity of the diagram,we find q∗(δ(e

nα)) = dαβ. On the other hand, starting from an arbitrary en−1

γ ∈ Hn−1(Xn−1, Xn−2),

we find that the quotient map sends it to 0 ∈ Hn−1(Xn−1/(Xn−1 − en−1β )) if γ 6= β and to 1 if

γ = β. In other words, we must have j∗(δ(enα)) =

∑β dαβe

n−1β , which concludes the proof. �

4.4. Projective space. We will calculate the singular homology groups of real projective space RPkand complex projective space CPk, by establishing cell complex structures on them.

For a field k, k-dimensional projective space kPk is the space of all lines through the originin kk+1. In other words, it is the quotient space (kk+1 − {0})/∼, with ∼ the equivalence relationidentifying points on the same line through 0, i.e. v ∼ λv for λ ∈ k× and v ∈ kn+1 − {0}.When k = R, this equips RPk with the quotient topology coming from the euclidean topology onRk+1. For k = C, this similarly equips CPk with a topology as a quotient of the metric (and thustopological) space Ck+1 − {0}.

Lemma 111. The space RPk is a k-dimensional cell complex, with exactly 1 j-cell, for j ∈ [0, k].

Proof. We have RPk ∼= Sk/∼, with ∼ the relation identifying the vectors v and −v. We fix thenotation qk : Sk → RPk for the corresponding quotient map. We consider the hemisphere

hSk = {v ∈ Sk | v1 ≥ 0} with boundary ∂hSk = Sk−1.

We have RPk ∼= hSk/∼ with ∼ the relation on Sk−1 = ∂hSk which again identifies antipodal vectors.Since moreover, hSk ∼= Dk, this implies that we actually have

RPk ∼= RPk−1 tqk−1Dk.

Since RP0 ∼= pt, the statement follows by iteration. �

Lemma 112. The space CPk is a 2k-dimensional cell complex, with exactly 1 2j-cell, for j ∈ [0, k].

Proof. See p6-7 in Hatcher. �

Theorem 113. For any n, k ∈ N, we have

(i) We have

Hn(RPk) =

Z if n = 0

Z2 if n ∈ {1, 3, . . . , 2bk/2c − 1}Z if n = k is odd

0 otherwise.

(ii) We have

Hn(CPk) =

{Z if n ∈ {0, 2, 4, . . . , 2k}0 otherwise.

Proof. Theorem 108 and Lemma 111 imply that Hn(RPk) = 0 for n > k and that the remainingsingular homology groups are the homology groups of the chain complex

0 // Zdk // Z

dk−1 // · · · d2 // Z d1 // Z // 0.

That d1 = 0 follows easily from the definition of d1, or that of ∂CW1 . The degree of the composition

S1 q1 // RP1 ∼ // S1


is clearly 2. By Proposition 110, we thus have d2(a) = 2a, for any a ∈ Z. In particular, this meanswe can already compute the homology groups for k ≤ 2.

Now assume that we already know the result for all k = 2l, with l ∈ N. What is missing tocalculate H•(RP2l+1) is only knowledge of d2l+1. That is obtained from the degree of the map

S2l q2l // RP2l // RP2l/RP2l−1∼ // S2l .

However, since H2l(RP2l) = 0, this degree must be zero, so d2l+1 = 0. This allows to obtain theresult for k = 2l + 1.

To conclude the proof of part (i) it suffices to show that d2l acts by multiplying with 2 for alll ∈ Z≥1. This is proved in Examples 2.23 and 2.42 in Hatcher.

Part (ii) follows immediately from Theorem 108. Concretely, by Lemma 112 those results implythat Hn(CPk) can be computed as the homology of a chain complex of the form

· · · → 0→ 0→ 0→ Z→ 0 · · · → Z→ 0→ Z→ 0→ Z→ 0,

where we have k + 1 times the combination Z→ 0 and all groups to the left of that are 0. �

4.5. Recommended exercises: 1, 5 on p131.

chapter ii: homology · chapter ii: homology in this chapter, groups will usually be abelian....

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