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## 6.1 Homology of chain complexes

A graded Abelian group C = {C_i} is a collection of Abelian groups, indexed by the integers. A homomorphism of degree e from {C_i} to {D_i} is a collection f = {f_i} of homomorphisms of Abelian groups, where f_i : C_i -> D_(i+e). A morphism of graded Abelian groups is a homomorphism of degree 0.

A differential on a graded Abelian group is a homomorphism of some degree with square 0. A chain complex (C,d) is a graded Abelian group C with fixed differential d of degree -1. A morphism of chain complexes (a chain map) is a morphism of graded Abelian groups that commutes with the differential.

The homology group H(C) of a chain complex (C,d) is the graded Abelian group H(C) = {H_i(C)} with H_i(C) = ker d_i / im d_(i+1). (We put Z_i(C) = ker d_i, the cycles, and B_i(C) = im d_(i+1), the boundaries.) Given a morphism f : C -> D of chain complexes we find an induced morphism f* : H(C) -> H(D). Thus, H is a functor (with H(f) = f*).

Exercise Check that f* is well-defined and that H commutes with composition.

One has arbitrary sums, products, direct and inverse limits for chain complexes. Taking homology commutes with sums, products and direct limits.

Exercise Show the homology of an inverse limit may differ from the inverse limit of the homology of a sequence of chain complexes.

## 6.2 Homology of exterior algebra

Given a set V, let the exterior algebra E(V) be the Abelian group generated by associative expressions u_1 ^ ... ^ u_m (where the empty product is allowed and denoted by 1), where the u_i are in V, subject to the relations u ^ u = 0 and u ^ v + v ^ u = 0. Let E'(V) be the subgroup of the elements with an expression not involving 1. The group E(V) has a natural grading (say, with u_0 ^ ... ^ u_m having degree m) and a natural differential d of degree -1 defined by d_m(u_0 ^ ... ^ u_m) = Sum_i (-1)^i u_0 ^ ... ^ u_(i-1) ^ u_(i+1) ... ^ u_m. Also E'(V) has a natural differential, defined by the above expression for m > 0 and zero on degree 0 and smaller. Thus, we find homology in both cases.

Exercise Show that E(V) has zero homology. Show that C = E'(V) has all homology groups zero, except for H_0(C), which is isomorphic to the integers when V is nonempty.

[Maybe it feels as if the grading is wrong, shifted by one. I agree. But this is tradition. Similarly, one usually defines homology H using E'(V), notices that there are formal complications in degree 0, and then looks at the augmented chain complex E(V) with reduced homology H-tilde, which equals H everywhere, except in degree 0, in case V is nonempty, where H_0 has an additional summand Z. The theory is simplified by never mentioning E'(V) and only using reduced homology.]

## 6.3 Simplicial homology

A simplicial complex is a set V together with a collection K of nonempty finite subsets called simplices such that any nonempty subset of a simplex (called face of the simplex) is again in K. Each simplicial complex (with underlying set V) gives rise to the subgroup E'(K) of E'(V) generated by the products u_1 ^ ... ^ u_m for each simplex { u_1, ..., u_m } in K, and E'(K) is invariant for the differential d. The corresponding homology is called simplicial homology.

## 6.4 Singular homology

Let X be a topological space. A singular simplex in X is a map from a Euclidean (solid) simplex (with ordered set of vertices) into X, where two singular simplices are considered equal when they differ by the unique affine transformation of the domain simplices that preserves vertex order.

(One does not need a Euclidean space to start with; given an ordered m-set M, one constructs a solid (m-1)-simplex by taking all functions from M to the unit interval with values summing to 1. The coordinates obtained in this way are called barycentric coordinates. Now two singular simplices are considered identical when they only differ by an order preserving bijection of the domains of their domains. Thus, there is no loss in generality to take the set of integers 0..m as domain for all m-simplices.)

We obtain a simplicial complex (called S(X)) by letting the faces of a singular simplex be the restrictions of the map to the faces of the domain simplex. The homology obtained in this way is called the singular homology of the topological space.

If X and Y are topological spaces, and f is a map from X to Y, then composition with f defines a map from S(X) into S(Y) that commutes with taking faces, and hence defines a homomorphism f* (of degree 0) from H(X) to H(Y).

## 6.5 .Cech homology

[

- unfortunately ISO-8859-1 does not have a hacek.] An entirely different way of finding a simplicial complex in a topological space is by taking an open cover and letting a simplex be a subset of the cover with nonempty intersection. This simplicial complex is called the nerve of the cover. For each cover we obtain a homology group. Given a cover and a refinement, we can map the simplices belonging to the refinement to simplices belonging to the first cover by picking for each open set in the refinement a member of the first cover containing it. This leads to a homomorphism of homology groups. The collection of all open covers is directed for refinement, and we get an inverse system. The inverse limit is the .Cech homology group. (Details later, when we have some more machinery.)

## 6.6 Chain homotopy

Let f,g : C -> D be morphisms of chain complexes. A chain homotopy from f to g is a homomorphism e : C -> D of degree 1 such that f - g = de + ed.

If f,g are homotopic, i.e., if there is a chain homotopy from f to g, then f* = g*, i.e., f and g induce the same map f* : H(C) -> H(D).

[Indeed, if z is a cycle in C, then ed(z) = 0 and hence f(z) - g(z) = de(z) is a boundary.]

Exercise Being homotopic is an equivalence relation. If f,g : C -> D are homotopic, and f',g' : D -> E are homotopic, then f' f and g' g are homotopic.

A chain complex C is called contractible when the identity on C is homotopic to 0. It follows that H(C) = 0, that is, C is acyclic.

For example, a simplex, or, more generally, a simplicial complex all of whose simplices contain some fixed vertex v, is acyclic: the map e defined by e(v_0 ^ ... ^ v_m) = v ^ v_0 ^ ... ^ v_m is a chain homotopy from the identity to 0.

Let us look at .Cech homology again. Given an open cover U and a refinement V, there are in general many ways of assigning to each member of V a member of U containing it. Thus, we find many morphisms of the corresponding chain complexes. However, these are all homotopic, and hence induce the same mapping on the homology groups. [Indeed, if the two maps assign u and u' (respectively) to some v in V, then we can define a chain homotopy e by e(v_0 ^ ... ^ v_m) = Sum_i (-1)^i u_0 ^ ... ^ u_i ^ u'_i ^ ... ^ u'_m, so that (de+ed)(v_0 ^ ... ^ v_m) = u'_0 ^ ... ^ u'_m - u_0 ^ ... ^ u_m.]

Exercise Show that given topological spaces X,Y and a map f:X->Y we find a map f(V)*: H_U(X) -> H_V(Y) on the .Cech homology groups corresponding to open covers U of X and V of Y, where U is the collection of inverse images of V. Show that this induces a map f*: H(X) -> H(Y), on the .Cech homology groups of X and Y (the inverse limit of the groups H_U(X) and H_V(Y)).

Let f : C -> C' be a morphism of chain complexes. The mapping cone of f is the chain complex C'' with C''_i = C_(i-1) + C'_i (direct sum) and d'' (c,c') = (-dc, fc+d'c'). Then f is a chain equivalence (i.e., there exists a morphism of chain complexes f' : C' -> C such that both ff' and f'f are homotopic to 1), if and only if C'' is contractible.

Exercise Verify all details. More precisely: (i) C'' is a chain complex, and (ii) If C'' is contractible, so that d''e''+e''d'' = 1 for some e'', then e''(c,0) = (ec,*) and e''(0,c') = (f'c', -e'c') define a morphism of chain complexes f' and chain homotopies e and e' from f'f and ff' to 1. (iii) If f' : C' -> C is a morphism of chain complexes and e and e' are chain homotopies from f'f and ff' to 1, then e''(c,c') = ((e+f'e'f-f'fe)c+f'c', (e'fe-e'e'f)c-e'c') defines a contraction of C''.

Let us look at singular homology again. Instead of oriented simplices we can use ordered simplices (where the generating elements are sequences of points, possibly with repetitions, that occur in a common simplex).

Exercise The map that assigns to an ordered simplex (u_0,...,u_m) the oriented simplex u_0 ^ ... ^ u_m is a chain equivalence. Answer

Exercise A simplicial complex K and its barycentric subdivision (that has as vertices the simplices of K, and as m-simplices the chains consisting of i-simplices of K for i=0,...,m, totally ordered by inclusion) are chain equivalent.

## 6.7 Singular Homology and Homotopy

Homotopy is a finer invariant than homology. We give two results in this direction.

Theorem Let X,Y be two topological spaces, and f,g : X -> Y two homotopic maps. Then we find for singular homology that the maps f*,g* : H(X) -> H(Y) coincide.

Let us set up a little bit of machinery, so that the proof will be obvious. Given a solid simplex A, let i,j : A -> A × I be the two injections sending a to (a,0) and (a,1), respectively. Let us define a canonical subdivision (triangulation) T(A) of A × I as follows. Let c be the center of A. Then T(A) consists of the simplices <c,iA> and <c,jA> (and <c,iF> and <c,jF> for all faces F of A) and all simplices <c,S> with S in T(F) for some face F of A. (Given a point x distinct from c in A × I, the line cx hits the boundary of A × I either in iA or in jA or in F × I for some face F of A, so we do indeed cover A × I.)

Given a simplicial complex K, let T(K) be the simplicial complex that is the union of T(A) for all simplices A of K. The maps i and j extend linearly to chain maps (denoted by the same symbols) from the chain complex defined by K to that defined by T(K).

Lemma The chain maps i,j : K -> T(K) are chain homotopic.

Proof Consider the chain map k : K -> T(K) that assigns to a simplex A the element <c,iA> - <c,jA> - <c,kdA> (defined by induction of dim A; <c,*> extended linearly). We see by induction that dk+kd = i-j. Indeed, (dk+kd)(A) = iA - jA - <c,idA> + <c,jdA> + <c,dkdA> = iA - jA - <c,kddA> = iA - jA. QED

Proof of the theorem Let i,j : X -> X × I be the two injections sending x to (x,0) and (x,1), respectively. Since f and g are homotopic, there is a map F : X × I -> Y such that f = Fi and g = Fj. Thus, it suffices to show that i* = j*. We do this by exhibiting a chain homotopy h from i to j considered as maps from S(X) to S(X × I). Thus, h will be a map from S(X) to S(X × I) of degree 1, such that dh+hd = i-j. Elements of S(X) are (linear combinations of) maps s from a (solid) simplex A into X. Define h(s) = (s × 1) k(A) where k is the map defined in the proof of the lemma, viewed as a linear combination of maps into A × I and 1 is the identity on I. We find (dh+hd)(s) = (s × 1)(dk+kd)(A) = (s × 1)(i-j)(A) = (i-j)(s). QED

The second result here is that H_1 equals pi_1 made Abelian, that is, the first homology group of a space (for singular homology) equals the quotient of the fundamental group by its commutator subgroup.

Theorem H_1(X) is the quotient of P(X,x) by its commutator subgroup.

Proof Let P(X;x) be the fundamental group of X and let H_1(X) be the first homology group. We find a homomorphism h : P(X;x) -> H_1(X) as follows. A map f : I -> X with f(0) = f(1) = x is a singular simplex of dimension 1, so we can let h map the homotopy class of f to the homology class of f. This is well-defined: we can view f as a map from the circle S(1) into X, and then find a map f* : H(S(1)) -> H(X), and as we just saw, f* only depends on the homotopy class of f. Let e : I -> S(1) be the map sending t to exp(2 pi i t). It is a singular simplex and hence represents an element of H(S(1)). The image f*(e) is an element of H(X), that has representative f. Thus h is well-defined.

Moreover, h is a homomorphism. Indeed, we have to check that h(f#g) = h(f)+h(g), that is, that f + g - f#g = 0 in H_1(X), that is, that f + g - f#g is a boundary, and clearly it is.

Remains to show that the commutator subgroup of P(X,x) is the full kernel of h. To this end we construct a left inverse h' to h from H_1(X) into AP(X,x), the quotient of P(X,x) by its commutator subgroup. The elements of H_1(X) are equivalence classes of 1-cycles in S(X), and these equivalence classes of 1-cycles have representatives that have boundary contained in {x}. Given such a 1-cycle z, let h' map it to z, regarded as a path in X. We have to show that this is well-defined, that is, that z is null-homotopic (0 in AP(X,x)) when it is a boundary. But that is clear. QED

## 6.8 Exact sequences

Let A -> B -> C be a chain of Abelian groups, connected by mappings f : A -> B and g : B -> C. The chain is called exact at B when the image of f equals the kernel of g. A longer chain is called exact when it is exact at the middle of each 3-term subsequence. An exact sequence of chain complexes is a sequence of morphisms of chain complexes giving an exact sequence of Abelian groups for each index.

Theorem Given a short exact sequence 0 -> C -> D -> E -> 0 of chain complexes, we find a long exact sequence ... -> H_i(C) -> H_i(D) -> H_i(E) -> H_(i-1)(C) -> .... The correspondence is functorial: a morphism between two short exact sequences is mapped to a morphism between two long exact sequences, where a morphism of exact sequences is a collection of mappings making all squares commute.

Proof We already saw that a morphism C -> D gives a morphism H(C) -> H(D). Our first task is to define the map H_i(E) -> H_(i-1)(C).

Given z in Z_i(E), we find y in C_i(D) of which it is the image. Since dz = 0 the element dy has zero image in E_(i-1) and hence is the image of a unique element x in C_(i-1). Moreover, dx has image ddy = 0, so dx = 0, i.e., x is a cycle. Map the homology class of z to that of x. This is well-defined: if we replace y by y' with the same image z, then y'-y is the image of a x' in C_i and x becomes x+dx', which lies in the same homology class. And if we change z by a boundary: z' = z + dz'' with z'' in E_(i+1), then y is changed by an element dy'', where z'' is the image of y'', so that dy is not changed at all.

Thus we find a function from H_i(E) to H_(i-1)(C), and one quickly verifies that it is a homomorphism.

Remains to verify exactness at each point, and functoriality. Let us leave that as an exercise.

Exercise Prove that the above long sequence is exact at each point. Prove that the construction behaves well under morphisms.

## 6.9 The Maier-Vietoris sequence

As a special example of the situation of the previous section we have the case of two simplicial complexes B,C of which the union again is a simplicial complex D. Let their intersection be A. Then we have a short exact sequence 0 -> A -> B+C -> D -> 0 (where + denotes direct sum) if we define the arrows by mapping a to (a,-a) and (b,c) to b+c.

Consequently, we find a long exact sequence, called the Maier-Vietoris sequence, ... -> H_i(A) -> H_i(B)+H_i(C) -> H_i(D) -> H_(i-1)(A) -> ....

This enables one to compute the homology H_i(D) of a union of two complexes, given the homology of these complexes and that of their intersection.

For example, look at the reduced homology of the n-sphere S = S(n). Let x,y be two points of S. Then S\{x} and S\{y} are homeomorphic to Euclidean n-space E(n) (by `stereographic projection') and hence are contractible and have zero homology. And S\{x,y} is homeomorphic to E(n) minus a point, which is homotopic to S(n-1). Thus, H_i(S(n)) = H_(i-1)(S(n-1)). At the bottom we have the empty set S(-1) with reduced homology vanishing everywhere except that H_(-1) is isomorphic to the integers. It follows that H_i(S(n)) = 0 for all i except that H_n(S(n)) is isomorphic to the integers.

Exercise Prove that E(m) is not homeomorphic to E(n) when m < n.

Exercise A map f from S(n) to itself has a (mapping) degree, namely the number d such that f* is multiplication by d on H_n(S(n)). Viewing the complex plane together with a point infinity as S(2), show that the mapping degree of a polynomial of degree d equals d. Conclude that a nonconstant polynomial has a root.

## 6.10 Relative homology

A topological pair is a pair (X,A) where X is a topological space, and A a subspace. A morphism f : (X,A) -> (Y,B) is a map f : X -> Y such that f(A) is contained in B. The singular chain complex S(X,A) is defined as S(X)/S(A) (singular simplices in X modulo those entirely contained within A). The homology is called H(X,A). All concepts generalize in an obvious way. Regard X as a pair by identifying it with (X,empty).

Exercise Prove that if A is a retract of X then H(X) = H(A) + H(X,A).

Exercise Prove that S(n-1) is not a retract of E(n). Conclude that the two theorems from the Introduction hold.

## 6.11 Betti numbers, Euler characteristic, Lefschetz number, Hopf trace formula, Lefschetz fixed point theorem

Let A be an R-module. An element a of A is called a torsion element when ra = 0 for some nonzero r in R.

Exercise Give an example to show that the set of torsion elements need not be a submodule. Show that it is a submodule when R is an integral domain (i.e., is commutative, without zero-divisors, with nonzero 1).

Assume that R is an integral domain. Let Tor(A) be the torsion submodule of A, that is, the submodule consisting of all torsion elements. The module A is called torsion-free when Tor(A) = 0. The module A' = A / Tor(A) is torsion free. If A is finitely generated, then A' is free, with a finite basis. The cardinality of such a basis is called the rank of A. Let f : A -> A be a morphism. The trace Tr(f) if the trace of f induced on A' (e.g., the sum of the diagonal elements when written on a basis).

The above applies (with R the ring of integers) to Abelian groups. The Betti numbers of a simplicial complex are the ranks of the homology groups H_i.

Let C be a finitely generated chain complex, and let f : C -> C be a homomorphism of degree 0. The Lefschetz number L(f) is defined as L(f) = sum (-1)^i Tr(f|C_i).

The Euler characteristic chi(C) is the Lefschetz number of the identity map.

Theorem (Hopf trace formula) Let C be a finitely generated chain complex, and let f : C -> C be a homomorphism of degree 0. Then L(f) = L(f*). (In particular, the Euler characteristic is the alternating sum of the Betti numbers.)

Exercise Prove this. (Hint: do the case of the Euler characteristic first.)

Theorem (Lefschetz fixed point theorem) Let X be a nice compact topological space and let f : X -> X be a map. If L(f) is nonzero then f has a fixed point.

(Proof: "nice" here may mean carrier of a finite simplicial complex, or connected orientable n-dimensional manifold, or so. Use a suitable simplicial approximation. If f has no fixed points then all simplices will be moved far away by f, so that f will have trace 0.)

## 6.12 Axiomatic homology

Thee are many homology theories (we have seen singular homology and .Cech homology), and it is possible to develop the theory axiomatically. See S. Eilenberg & N.E. Steenrod, Foundations of Algebraic Topology, Princeton, 1952.

A homology theory is a pair (H,d), where H is a covariant functor from the category of topological pairs to that of graded Abelian groups (i.e., H commutes with composition of maps and sends the identity to the identity), and d assigns to each pair (X,A) a homomorphism of degree -1 from H(X,A) to H(A), where for f : (X,A) -> (Y,B) we have d(Y,B) H(f) = H(f|A) d(X,A), satisfying the following four axioms:
Homotopy Axiom If f,g : (X,A) -> (Y,B) are homotopic, then H(f) = H(g).
Exactness Axiom Consider the pair (X,A) and the two inclusion maps A -> X -> (X,A). We find an exact sequence ... -> H_i(A) -> H_i(X) -> H_i(X,A) -> H_(i-1)(A) -> ... (with arrows given by ...,H,H,d,...).
Excision Axiom If U is an open subset of X with closure contained in the interior of A, then the excision map i : (X\U,A\U) -> (X,A) induces an isomorphism H(i) between H(X\U,A\U) and H(X,A).
Dimension Axiom If X is a single point, then H_i(X) = 0 for nonzero i and H_0 (X) is isomorphic to the additive group of integers.

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