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)=⨁n≥0Hn(X), and to each map f:X→Y a function f∗:Hn(X)→Hn(Y) satisfying (1X)∗=1Hn(X) and (f∘g)∗=f∗∘g∗.
Some properties we would like to have for our construction are:
- Homotopy invariance, if f∼g:X→Y then f∗=g∗.
- 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:
- C=⨁n∈ZCn for R-modules Cn.
- d:C→C where d=⨁dn for R-linear maps dn.
- d∘d=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 x∈kerdn we write [x] for its image in Hn(C).
Example1.3.3
-
C0=C1=Z, Ci=0 otherwise,
0→Z⋅3→Z→0.
Then H1=0, H0=Z/3.
-
Z=⟨e⟩→Z2=⟨f1,f2⟩→Z=⟨g⟩→0
with d(e)=f1−f2, 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,xi≥0∀i}.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:0≤i0≤⋯≤ik≤n}⟩.
To define d it suffices to define d(eI), we let d(eI)=k∑j=0(−1)jei0,…,ij−1,ij+1,…,ik∈Sk−1(Δn).
Example1.3.6
- For Δ1 we have d(e0,1)=e1−e0.
- For Δ2 we have d(e0,1,2)=e12−e02+e01 and so d2(eI=(e2−e1)−(e2−e0)+(e1−e0)=0.
Lemma1.3.7
d2=0Proof
It suffices to show \(d^2(e_I) = 0\) for all \(I\).
\begin{align*}
d^2(e_I) &= d\left(\sum_{j = 0}^{k} (-1)^j e_{i_0,\ldots,\hat{i}_j,\ldots,i_k}\right)\\
&= \sum_{j = 0}^{k} (-1)^j d\left(e_{i_0,\ldots,\hat{i}_j,\ldots,i_k} \right)\\
&= \sum_{j = 0}^{k} (-1)^j\left( \sum_{l \lt j} (-1)^l e_{i_0,\ldots,\hat{i}_l,\ldots,\hat{i}_j,\ldots,i_k} + \sum_{l \gt j} (-1)^{l-1} e_{i_0,\ldots,\hat{i}_j,\ldots,\hat{i}_l,\ldots,i_k}\right)\\
&= \sum_{j = 0}^{k} \left( \sum_{l \lt j} (-1)^{l+j} e_{I - i_l - i_j} - \sum_{l \gt j} (-1)^{l+j} e_{I - i_j - i_l}\right) = 0.
\end{align*}
Example1.3.8Computing H∗(S∗(Δ2))
0→Z=⟨e012⟩→Z3=⟨e01,e02,e12⟩→Z3=⟨e0,e1,e2⟩→0
with d(e012)=e12−e02+e01.
So kerd2=0⟹H2=0.
d(ae01+be02+ce12)=a(e1−e0)+b(e2−e1)+c(e2−e0)
so ae01+be02+ce12∈kerd1⟺−a−b=0,a−c=0,c+b=0⟺a=−b=c hence kerd1=⟨e01−e02+e12=imd2 and so H1=0.
ker(d0)=0, imd1=⟨e1−e0,e2−e1,e2−e0⟩ so H0=Z=⟨[e0]⟩.
Exercise 1.3.9
Show that H∗(S∗(Δn))=0 if k≠0 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:Δn→X any continuous map⟩.
Where
d(eσ)=n∑j=0(−1)jeσ∘Fj∈Cn−1(X)
where Fj:Δn−1→Δn is given by Fj(x0,…,xn1=(x0,…,0,…,xn−1) with the 0 in the jth place.
Subsubsection1.3.3.1Homotopy invariance
Definition1.3.11Chain homotopic maps
Suppose ϕ,ψ:C∗→C′∗ are chain maps, we say that ϕ are chain homotopic if there exists an R-linear map h:C∗→C′∗+1 such that d′∘h+h∘d′=ϕ−ψ.
We denote this relation by ϕ∼ψ.
Lemma1.3.12
If ϕ∼ψ then ϕ∗=ψ∗.
Proof
\begin{align*}
\phi_*([x]) - \psi_*([x]) &= [\phi(x) - \psi(x)]\\
&= [d'hx + hdx] = [d'hx] = 0 \in H_*(C').
\end{align*}
Theorem1.3.13
Suppose f∼g:X→Y via H then f#∼g#⟹f∗=g∗.
Proof
Corollary1.3.14
If X∼Y then H∗(X)≅H∗(Y).
Corollary1.3.15
If X is contractible then H∗(X)≅H∗({p})≅Z if ∗=0, 0 otherwise.