§C. Axiomatic Homology Theory

Spring Semester 2015–2016 MATH41072/MATH61072 Algebraic Topology Supplementary read- ing §C. Axiomatic Homology Theory

C.1 Introduction. The basic ideas of homology go back to Poincaréin 1895 when he defined the Betti numbers and torsion numbers of a complex. The idea of defining homology groups was developed from about 1925 by Heinz Hopf in Göttingen under the influence of Emmy Noether. Subse- quently, various different way of defining homology groups were developed in order to extend their application to non-triangulable spaces. One can in fact define the homology groups of any topological space using what is called singular homology. This is defined as follows.

C.2 Definition. Given a topological space X, a singular n-simplex in X n is a continuous map σ : ∆ → X where n > 0. The singular n-chain group of X, denoted Sn(X), is the free abelian group on the set of all singular P n-simplices in X. So a singular n-chain is a finite sum i λiσi where λi ∈ Z n and σi : ∆ → X. We define Sn(X) = 0 for n < 0. i n−1 n For 0 6 i 6 n, define δ : ∆ → ∆ by mapping ej 7→ ej for 1 6 j 6 i i n−1 and ej 7→ ej+1 for i < j 6 n and extending linearly. (So δ maps ∆ to n the ith face he1, e2,..., eî+1,..., en+1i of ∆ .) We now define, for n > 0, the boundary homomorphism

dn : Sn(X) → Sn−1(X) on generators by n X i i dn(σ) = (−1) σ ◦ δ i=0 and extend linearly to the whole of Sn(X). Of course, dn = 0 for n 6 0. The singular homology groups of X are then deﬁned just like the simplicial ones. Thus the singular n-cycle group Zn(X) = Ker dn : Sn(X) → Sn−1(X) and the singular n-boundary group Bn(X) = Im dn+1 : Sn+1(X) → Sn(X) . We can prove that dn ◦ dn+1 = 0: Sn+1(X) → Sn−1(X) so that Bn(X) ⊂ Zn(X) and then the singular n-homology group Hn(X) is deﬁned to be the quotient group Zn(X)/Bn(X).

C.3 Remarks. First of all notice that these groups are obviously topological invariants since a homeomorphism f : X → Y induces group isomorphisms Sn(X) → Sn(Y ) given on generators by σ 7→ f ◦ σ and so isomorphisms Zn(X) → Zn(Y ), Bn(X) → Bn(Y ), Hn(X) → Hn(Y ). (They are in

1 fact homotopy invariants but this is harder to prove.) This excellent result is counterbalanced by the fact that it is not all clear how to compute any of these homology groups since for most spaces X the singular n-chain group is uncountably generated. It is possible to calculate the homology groups of very simple spaces from the deﬁnition. For example, if P is a one-point space then (Exercise)

for n = 0, H (P ) =∼ Z n 0 for n 6= 0.

In the same sort of way we can prove (Exercise) that

2 for n = 0, H (S0) =∼ Z n 0 for n 6= 0.

Calculations for other spaces are then carried out by deriving various properties of these groups and then using these properties to compute the homology groups from those of the above simple spaces. There are in fact several ways of deﬁning homology groups but for some spaces they do not necessarily give the same groups. However, it turns out that for the underlying spaces of a simplicial complex (these spaces are usually called polyhedra) they do always give the same answer. This was ﬁrst proved by Samuel Eilenberg and Norman Steenrod in their fundamental book Foundations of Algebraic Topology published in 1952. They showed that for polyhedra (and in fact for spaces homotopy equivalent to polyhedra) homology groups are characterized by certain axioms. They then showed that simplicial homology groups and singular homology groups satisfy these axioms. The axioms they state are for homology groups of a general topological space and require some very technical details. To avoid these technical details these notes do two things.

• The material below restricts attention to the homology groups of triangulable spaces.

• The notes below are couched in terms of reduced homology groups which allow for a slightly simpler statement of the axioms.

C.4 Definition. Suppose that X is a topological space. Let P be a one- point space and c: X → P the constant function. Then the reduced singular homology groups of X, denoted Hñ(X) are defined by Hñ(X) = Ker c∗ : Hn(X) → Hn(P ) .

2 C.5 Proposition. The reduced homology groups of X are related to the usual (unreduced) homology groups by

˜ H0(X) × Z if n = 0, Hn(X) = H˜n(X) if n 6= 0.

Proof. For n 6= 0, Hn(P ) = 0 and so Ker c∗ : Hn(X) → Hn(P ) = Ker(c∗ : Hn(X) → 0) = Hn(X). For n = 0, H0(P ) = Z and so we have the following short exact sequence.

˜ i c∗ 0 → H0(X) → H0(X) → Z → 0 ˜ From this it follows (Exercise) that H0(X) = H0(X) × Z.

C.6 Remarks. This means in particular from the calculations in Re- marks C.3 that we have the following reduced homology groups. ∼ H˜n(P ) = 0 for all n.

for n = 1, H˜ (S0) =∼ Z n 0 for n 6= 0.

The second of these results is one of the axioms. Basically you have to know the homology groups of one space in order to get started with calculations. The other axioms are the functorial properties of homology (see Theorem 5.25), the homotopy property (see Theorem 5.26) and one further property known as the exactness property which is the key to relating the homology groups of diﬀerent spaces. In order to state the exactness property we need to introduce the idea of a triangulable pair of spaces.

C.7 Deﬁnition. A triangulable pair of spaces (X,A) is a topological space X with a subspace A such that there is a homeomorphism h: X → |K|, the underlying space of a simplicial complex K, with h(A) = |L| the underlying space of a subcomplex L of K.

C.8 Proposition. If (X,A) is a pair of triangulable spaces then the quotient space X/A is a triangulable space.

Proof. The proof of this result is omitted. In general you cannot form a quotient simplicial complex K/L when L is a subcomplex of K but you can construct a triangulation of X/A by using an appropriate barycentric subdivision of K and collapsing an appropriate subcomplex.

3 C.9 Remarks. These remarks are intended to motivate the statement of the exactness axiom. They present an outline of some ideas which it would take quite a bit of work to develop in detail. Given a pair of triangulable spaces (X,A), we may define the homology groups of the pair, Hn(X,A), by defining the n-chain group of the pair to be Sn(X,A) = Sn(X)/Sn(A). The boundary homomorphisms dn : Sn(X) → Sn−1(X) and Sn(A) → Sn−1(A) then induce a boundary homomorphism dn : Sn(X,A) → Sn−1(X,A) and so we may define Zn(X,A), Bn(X,A) and finally Hn(X,A) = Zn(X,A)/Bn(X,A). The homology groups of the pair (X,A) provide a measure of the difference between the homology groups of X and the homology groups of A. To see this, notice that if z ∈ Sn(X) represents an n-cycle in Zn(X,A) = Ker dn : Sn(X)/Sn(A) → Sn−1(X)/Sn−1(A) then dn(z) ∈ Sn−1(A). So we can define a homomorphism ∂ : Zn(X,A) → Zn−1(A) by ∂([z]) = dn(z) This induces a homomorphism ∂ : Hn(X,A) → Hn−1(A). We can then prove that these groups and homomorphisms fit into an exact sequence as follows.

i∗ j∗ ∂ i∗ ... → Hn(A) → Hn(X) → Hn(X,A) → Hn−1(A) → Hn−1(X) → ... ∼ Finally. we can prove that Hn(X,A) = H˜n(X/A) (it is this step which uses the fact that the pair (X,A) is triangulable) so that this exact sequence can be written as follows.

i∗ q∗ ∂ i∗ ... → Hñ(A) → Hñ(X) → Hñ(X/A) → Hñ−1(A) → Hñ−1(X) → ...

Here the groups Hn(X) and Hn(A) have been replaced by the reduced groups. For n 6= 0 this doesn’t make any difference and for n = 0 where the groups are different it is not difficult to check that the sequence is still exact. This is done so that that the whole sequence is written in terms of the reduced groups. This sequence is the statement of the exactness axiom for homology. These considerations lead to the following definition.

C.10 Definition A reduced homology theory assigns to each non-empty ˜ triangulable space X a sequence of abelian groups Hn(X) (for i ∈ Z) and for each continuous map of triangulable spaces f : X → Y a sequence of homomorphisms f∗ : Hñ(X) → Hñ(Y ) such that the following axioms hold. (i) [Functorial Axiom 1] Given continuous functions f : X → Y and g : Y → Z, it follows that

g∗ ◦ f∗ = (g ◦ f)∗ : H˜n(X) → H˜n(Z) for all i.

(ii) [Functorial Axiom 2] For the identity map I : X → X,

I∗ = I : H˜n(X) → H˜n(X) (the identity map) for all i.

4 (iii) [Homotopy Axiom] For homotopic maps f ' g : X → Y ,

f∗ = g∗ : H˜n(X) → H˜n(Y ) for all i.

(iv) [Exactness Axiom] For any triangulable pair (X,A) there are boundary homomorphisms ∂ : Hñ(X/A) → Hñ−1(A) for all i which fit into a long exact sequence as follows.

i∗ q∗ ∂ i∗ ... → Hñ(A) → Hñ(X) → Hñ(X/A) → Hñ−1(A) → Hñ−1(X) → ...

Furthermore, given any continuous function of triangulable pairs f :(X,A) → (Y,B) (i.e. f : X → Y such that f(A) ⊂ B) this induces a continuous function of quotient spaces f¯: X/A → Y/B. Then the following diagram commutes for all n.

∂- H˜n(X/A) H˜n−1(A)

f¯∗ f∗ ? ? ∂- H˜n(Y/B) H˜n−1(B)

˜ 0 ∼ ˜ 0 (v) [Dimension Axiom] H0(S ) = Z and Hn(S ) = 0 for all n 6= 0.

C.11 Remark. It can be shown that the properties in Definition C.10 are satisfied by the singular homology groups of Definition C.2 and furthermore that these properties uniquely determine the homology groups of a triangulable space and the homomorphisms induced by continuous functions between triangulable spaces.

C.12 Theorem. A homotopy equivalence of triangulable spaces f : X → Y induces isomorphisms f∗ : H˜n(X) → H˜n(Y ) of their reduced homology groups.

Proof. Exercise.

C.13 Proposition. The reduced homology groups of a contractible triangulable space are all trivial.

Proof. First of all we calculate the homology groups of a one point space P . Let I : P → P be the identity map. Then by (ii) I∗ = I : H˜n(P ) → H˜n(P ) is an isomorphism for all n.

5 Now consider the homology exact sequence coming from the pair (P,P ). This gives the following exact sequence.

I∗ q∗ ∂ I∗ ... → Hñ(P ) → Hñ(P ) → Hñ(P/P ) → Hñ−1(P ) → Hñ−1(P ) → ...

Since the homomorphisms I∗ are all isomorphisms it follows that the homomorphisms q∗ and ∂ are trivial and so H˜i(P/P ) = H˜i(P ) = 0 for all i. The result now follows from Theorem C.12.

C.14 Theorem. The reduced homology groups of the n-sphere Sn are given by for i = n, H˜ (Sn) =∼ Z i 0 for i 6= n.

Proof. The proof is by induction on n. The result for n = 0 is the Dimension Axiom. For the inductive step, consider the exact sequence coming from the k+1 k pair (D ,S ) (Exercise).

C.15 Remarks. Calculating the homology groups of spheres from the axioms is of course just the beginning. But it turns out that the spheres are the key spaces in building up most standard spaces. So, for example, to compute the homology groups of projective n-space P n = Sn/(x ∼ ±x), we can observe that the inclusion map Sk → Sk+1, given by x 7→ (x, 0), induces an inclusion map P k → P k+1 such that the quotient space P k+1/P k =∼ Sk+1. So the Exactness Axiom gives an exact sequence as follows.

k i∗ k+1 q∗ k+1 ∂ k i∗ k+1 ... → Hñ(P ) → Hñ(P ) → Hñ(S ) → Hñ−1(P ) → Hñ−1(P ) → ... This leads to a calculation of the homology groups of P n by induction on n with the base case given by the fact that P 1 =∼ S1 and using the above sequence and our knowledge of the homology of Sk+1 and for the inductive step. The only difficulty in the calculation is identifying the homomorphism k+1 k ∂ : Hñ(S ) → Hñ−1(P ) in the case n = k + 1 (the only non-trivial case) and this does require further analysis of the topology of P n which would take us too long.

Exercises

1. Prove from Deﬁnition C.2 that the singular homology groups of a one- point space P are given by if n = 0, H (P ) = Z n 0 if n 6= 0.

6 2. Prove from Deﬁnition C.2 that the singular homology groups of the two-point space S0 (with the usual discrete topology) are given by

2 if n = 0, H (S0) = Z n 0 if n 6= 0.

3. Prove that if there is a short exact sequence of abelian groups

i q 0 → H → G → Z → 0 ∼ then G = H × Z. [Hint: Choose an element g1 ∈ G such that q(g1) = 1 and use this to deﬁne an isomorpism H × Z → G.]

4. Prove Theorem C.12.

5. Complete the proof of Theorem C.14.