6 Axiomatic Homology Theory

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1 MATH41071/MATH61071 Algebraic topology 6 Axiomatic Homology Theory Autumn Semester The basic ideas of homology go back to Poincaré in 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. Subsequently, 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. 6.1 Definition. Given a topological space X, a singular n-simplex in X is a continuous map σ : n X where n 0. The singular n-chain group of X, denoted S n (X), is the free abelian group on the set of all singular n-simplices in X. So a singular n-chain is a finite sum i λ iσ i where λ i Z and σ i : n X. We define S n (X) = 0 for n < 0. For 0 i n, define δ i : n 1 n by mapping ε j ε j for 1 j i and ε j ε j+1 for i < j n and extending linearly. (So δ i maps n 1 to the ith face ε 1, ε 2,..., ˆε i+1,..., ε n+1 of n.) We now define, for n > 0, the boundary homomorphism on generators by d n : S n (X) S n 1 (X) d n (σ) = n ( 1) i σ δ i i=0 and extend linearly to the whole of S n (X). Of course, d n = 0 for n 0. The singular homology groups of X are then defined just like the simplicial ones. Thus the singular n-cycle group Z n (X) = Ker ( d n : S n (X) S n 1 (X) ) and the singular n-boundary group B n (X) = Im ( d n+1 : S n+1 (X) S n (X) ). We can prove that d n d n+1 = 0: S n+1 (X) S n 1 (X) so that B n (X) Z n (X) and then the singular n-homology group H n (X) is defined to be the quotient group Z n (X)/B n (X). 6.2 Remark. First of all notice that these groups are obviously topological invariants since a homeomorphism f : X Y induces group isomorphisms S n (X) S n (Y ) given on generators by σ f σ and so isomorphisms 1

2 Z n (X) Z n (Y ), B n (X) B n (Y ), H n (X) H n (Y ). (They are in 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) { H n (P ) Z for n = 0, = 0 for n 0. In the same sort of way we can prove (Exercise) that { H n (S 0 ) Z 2 for n = 0, = 0 for n 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 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 awkward 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. 2

3 6.3 Definition. Suppose that X is a topological space. Let P be a onepoint space and c: X P the constant function. Then the reduced singular homology groups of X, denoted H n (X) are defined by H n (X) = Ker ( c : H n (X) H n (P ) ). 6.4 Proposition. The reduced homology groups of X are related to the usual (unreduced) homology groups by { H0 (X) Z if n = 0, H n (X) = H n (X) if n 0. Proof. For n 0, H n (P ) = 0 and so Ker ( c : H n (X) H n (P ) ) = Ker(c : H n (X) 0) = H n (X). For n = 0, H 0 (P ) = Z and so we have the following short exact sequence. 0 H 0 (X) i H 0 (X) c Z 0 From this it follows (Exercise) that H 0 (X) = H 0 (X) Z. 6.5 Remark. This means in particular from the calculations in Remarks?? that we have the following reduced homology groups. H n (P ) = 0 for all n. H n (S 0 ) = { Z for n = 1, 0 for n 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. 6.6 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. 6.7 Proposition. If (X, A) is a pair of triangulable spaces then the quotient space X/A is a triangulable space. 3

4 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. 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. For a simplicial complex K we set C r (K) = C r (K) for r > 0 and C 0 (K) C 0 (K) the subgroup generated by differences of two points. Starting with (X, A) and K and L as above, after moving to a barycentric subdivision K and subcomplex L of K, such that L and L = L are homotopy equivalent we obtain for all r short exact sequences 0 C r ( L) i C r (K ) q C n (K /L ) 0. Note, that at position r = 0 this statement would be false for the unreduced chain groups C 0 ( ). The maps q and i also commute with the boundary maps. Hence, we obtain a complex of commutative diagrams. Now, we may define a homomorphism H n (K /L ) H n 1 ( L) as follows. Consider an elment z Z r (K /L ), i.e. d(z) = 0. Since the chain map q is surjective, there is a y C r (K ) with q (y) = z. Because of the mentioned commutativity we obtain q (d(y)) = d(q (y)) = d(z) = 0. Hence, d(y) ker q = im i and we find a (unique because of injectivity of i ) element x C r (L ) with i (x) = d(y). Because i (d(x)) = d(i (x)) = d(d(y)) = 0 we have d(x) = 0, since i is injective. Therefore, [x] defines an element of Hr 1 (A) and we set (z) = x. By chasing through the commutative diagrams one checks that is indeed well defined and we obtain a long exact sequence... H n (A) i H n (X) q H n (X/A) H n 1 (A) i H n 1 (X)... This sequence is the statement of the exactness axiom for homology. These considerations lead to the following definition. 6.8 Definition. A reduced homology theory assigns to each non-empty triangulable space X a sequence of abelian groups H n (X) (for n Z) and for each continuous map of triangulable spaces f : X Y a sequence of homomorphisms f : Hn (X) H n (Y ) such that the following axioms hold. (i) [Functorial Axiom 1] Given continuous functions f : X Y g : Y Z, it follows that g f = (g f) : Hn (X) H n (Z) for all i. and 4

5 (ii) [Functorial Axiom 2] For the identity map id X : X X, (id X ) = id: Hn (X) H n (X) (the identity map) for all i. (iii) [Homotopy Axiom] For homotopic maps f g : X Y, f = g : Hn (X) H n (Y ) for all i. (iv) [Exactness Axiom] For any triangulable pair (X, A) there are boundary homomorphisms : Hn (X/A) H n 1 (A) for all i which fit into a long exact sequence as follows.... H n (A) i H n (X) q H n (X/A) H n 1 (A) i H n 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) Hn 1 (A) f H n (Y/B) f Hn 1 (B) (v) [Dimension Axiom] H0 (S 0 ) = Z and H n (S 0 ) = 0 for all n Theorem. The properties in Definition 6.8 are satisfied by the reduced singular homology groups. Moreover, it can be shown that the properties in Definition 6.8 uniquely determine the homology groups of a triangulable space and the homomorphisms induced by continuous functions between triangulable spaces. In particular, for a triangulable spaces simplicial and singular homology coincide Theorem. A homotopy equivalence of triangulable spaces f : X Y induces isomorphisms f : Hn (X) H n (Y ) of their reduced homology groups. Proof. Exercise. 5

6 6.11 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 id: P P be the identity map. Then by (ii) id = id: Hn (P ) H n (P ) is an isomorphism for all n. Now consider the homology exact sequence coming from the pair (P, P ). This gives the following exact sequence.... H n (P ) id H n (P ) q H n (P/P ) H n 1 (P ) id H n 1 (P )... Since the homomorphisms id 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 Theorem. The reduced homology groups of the n-sphere S n are given by H i (S n ) = { Z for i = n, 0 for i 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 pair (D k+1, S k ) (Exercise) Theorem. Given a finite wedge sum of triangulable spaces X = n i=1 X i. Then one has H k (X) = n i=1 H k (X i ). Proof. We prove the claim by induction. Consider X = n i=1 X i and the inclusions ι : X n X and ῑ : n 1 i=1 X i X and the two projections p : X X/X n ( n 1 = i=1 X n 1 i and p : X X/ i=1 i) X = Xn. It is easy to see that p ι = id Xn and p ῑ = id n 1 i=1 X. It follows by the axioms i that p ῑ and p ι both give the identity on the corresponding homology groups. In particular p and p must be surjective and ι and ῑ must both be injective. Hence, we obtain exactness on the left and one the right of the following sequence ( 0 H k (X n+1 ) ι H k (X) p H n k i=1 X i ) 0 But exactness in the middle follows from the long exact sequence for the pair (X, X n ). Now, Exercise 3 implies the claim. (Why exactly?) 6

7 6.14 Remark. Note, that it also follows from the proof that the inclusions ι l : X l X on the level of homology induces the natural inclusions H k (X l ) n i=1 H k (X i ) and the projections p l : X X/ ( i l X i) = Xl the natural projections n i=1 H k (X i ) H k (X l ) Remark. 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 = S n /(x ±x), we can observe that the inclusion map S k S k+1, given by x (x, 0), induces an inclusion map P k P k+1 such that the quotient space P k+1 /P k = S k+1. So the Exactness Axiom gives an exact sequence as follows.... H n (P k ) i H n (P k+1 ) q H n (S k+1 ) H n 1 (P k ) i H n 1 (P k+1 )... 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 = S 1 and using the above sequence and our knowledge of the homology of S k+1 and for the inductive step. The only difficulty in the calculation is identifying the homomorphism : Hn (S k+1 ) H n 1 (P k ) 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. 7