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Section1.3Homology

Our goal is to construct a functor H from the category of topological spaces and continuous maps to the category of Z-modules and Z-linear maps. This means to each space X we associate an abelian group H(X)=n0Hn(X), and to each map f:XY a function f:Hn(X)Hn(Y) satisfying (1X)=1Hn(X) and (fg)=fg.

Some properties we would like to have for our construction are:

  1. Homotopy invariance, if fg:XY then f=g.
  2. The dimension axiom, Hn(X)=0 for any n>dimX.

Subsection1.3.1Chain complexes

Definition1.3.1Chain complex
If R is a commutative ring then a chain complex over R is a pair (C,d) satisfying:
  1. C=nZCn for R-modules Cn.
  2. d:CC where d=dn for R-linear maps dn.
  3. dd=0.
The indexing by n is called a grading. Usually we take Cn=0 for n<0. An element of kerdn is called closed or a cycle. An element of imdn is called a boundary. d is the boundary map or differential.
Definition1.3.2Homology groups
If (C,d) is a chain complex, its nth homology group is Hn(C,d)=kerdn/imdn+1. If xkerdn we write [x] for its image in Hn(C).
Example1.3.3
  1. C0=C1=Z, Ci=0 otherwise, 0Z3Z0. Then H1=0, H0=Z/3.
  2. Z=eZ2=f1,f2Z=g0 with d(e)=f1f2, d(f1)=d(f2)=g, then H(C)=0(exercise).

Subsection1.3.2The chain complex of a simplex

Definition1.3.4n-simplex
The n-dimensional simplexΔn is Δn={(x0,,xn)Rn:ixi=1,xi0i}.Δn has verticesv0,,vn which are the intersections with the coordinate axes. The k-dimensional faces are in bijection with the k+1element subsets of {0,,n}.
Definition1.3.5Simplicial chain complex
S(Δn) is the chain complex with Sk(Δn) the free Z-module generated by the k-dimensional faces of Δn. So Sk(Δn)=eI:I={i0,,ik:0i0ikn}. To define d it suffices to define d(eI), we let d(eI)=kj=0(1)jei0,,ij1,ij+1,,ikSk1(Δn).
Example1.3.6
  1. For Δ1 we have d(e0,1)=e1e0.
  2. For Δ2 we have d(e0,1,2)=e12e02+e01 and so d2(eI=(e2e1)(e2e0)+(e1e0)=0.
Example1.3.8Computing H(S(Δ2))

0Z=e012Z3=e01,e02,e12Z3=e0,e1,e20 with d(e012)=e12e02+e01. So kerd2=0H2=0. d(ae01+be02+ce12)=a(e1e0)+b(e2e1)+c(e2e0) so ae01+be02+ce12kerd1ab=0,ac=0,c+b=0a=b=c hence kerd1=e01e02+e12=imd2 and so H1=0. ker(d0)=0, imd1=e1e0,e2e1,e2e0 so H0=Z=[e0].

Exercise 1.3.9
Show that H(S(Δn))=0 if k0 and Z if k=0.

Subsection1.3.3The singular chain complex

Definition1.3.10Singular chain complex
If X is a space, the singular chain complex of X, C(X) is defined by Cn(x)=eσ:σ:ΔnX any continuous map. Where d(eσ)=nj=0(1)jeσFjCn1(X) where Fj:Δn1Δn is given by Fj(x0,,xn1=(x0,,0,,xn1) with the 0 in the jth place.

Subsubsection1.3.3.1Homotopy invariance

Definition1.3.11Chain homotopic maps
Suppose ϕ,ψ:CC are chain maps, we say that ϕ are chain homotopic if there exists an R-linear map h:CC+1 such that dh+hd=ϕψ. We denote this relation by ϕψ.