Singular Homology

2021年4月21日

Given a space, we want to study it “step by step”. We want to sort out the points, the edges, the faces etc. All those things are connected: edges connecting points, faces gluing to edges etc, and those connecting can be viewed as maps. In algebra, such a long chain is called a chain complex, and our first goal is to assign a chain complex to a given space. Then we can use homology to understand the maps.

Now we have a topological space $X$.

Topological n simplex

The “simplest” objects is triangular, and the analog here is topological $n$-simplex $\Delta^n_{top}$, they are defined as:

$$ \{ (x0,\dots,xn)\in \mathbb{R}^n \mid xi\geq 0, \sumi x_i =1 \} $$

As for the map, all maps between those $n$-simplexs are boil down to face maps:

$$d^i(x0,\dots,xn) = (x0,\dots,x{i-1},0, xi,\dots,xn)$$

Singular simplicial set

We look at the homomorphisms from topological $n$-simplex to sets, which are called singular simplicial set. They are defined by:

$$ Singn(X):=\hom{Top}(\Delta^{n}_{top},X) : \Delta^{op} \rightarrow Set $$

Together with the face maps between topological $n$-simplex, we now have a chain of sets:

$$ \dots \xleftarrow{di}\hom{Top}(\Delta^n{top},X) \xleftarrow{di} \hom{Top}(\Delta^{n+1}{top},X) \xleftarrow{d_i}\dots $$

For example, $Sing0(X):=\hom{Top}(\Delta^{0}{top},X)$ is set of points of $X$, and $Sing1(X):=\hom{Top}(\Delta^{1}{top},X)$ are paths in $X$. $f: \Delta^{1}{top}\rightarrow X$ is sent to $f(1)\in X$ by $d0$.

Singular chain complex

Recall that for each set $S$, we can find a free abelian group $\mathbb{Z}(S)$ such that every map from $S$ to any abelian group $G$ factors through $\mathbb{Z}(S)$. We abelianize the singular simplicial set to get a chain of abelian groups:

$$ Cn(X;\mathbb{Z}):=\mathbb{Z}Singn(X) $$

This thing is not a chain complex yet, we still need a differential map. It is of course built upon the face maps:

$$ d:= \sum{i=0}^n (-1)^i di:An \rightarrow A{n-1} $$

where $d_i$ is the $i$-th face map.

Check: pick $\alpha \in A{n+1}$, WTS $d^2(\alpha) = 0 \in A{n-1}$. Indeed,

$$ d^2(\alpha)= \sumi \sumj (-1)^{i+j} di dj \alpha $$

which can be broken into two pieces:

$$ \sum{0\leq i < j \leq n+1} + \sum{0\leq j \leq i <n+1} $$

It is easy to check the following simplicial identities:

$$ didj = d{j-1}di,\quad i<j $$

as well as

$$ dj dj = dj di, \quad j \leq i $$

And eventually we can prove that $d^2=0$.

Singular homology

Then we define $H_n(X;\mathbb{Z})$ as usual.

Property

H0

We can immediately prove $H0$ is a free abelian group. By definition, $H0=Z0/B0$. Here $Z0=C0$, which is the set of points in $X$. Since $d=d0-d1$, $d(\sigma) = \sigma(1)-\sigma(0)$. Hence

$$ H_0 = \mathbb{Z}\{\text{pts of X}\}/\sim $$

where $x0\sim x1$ if there's a path connecting them.

Disjoint spaces

Claim: $\mathbb{Z}\{\amalg{\lambda\in \Lambda} S\lambda\}\cong \bigoplus{\in \Lambda} \mathbb{Z}\{S\lambda\}$, where $\amalg$ is the disjoint union.

The claim follows from the following facts: 1. free abelian group functor preserves coproducts; 2. $Hn(\bigoplus{\lambda\in \Lambda} A^\lambda\bullet ) \cong \bigoplus Hn(A^\lambda\bullet)$ where $A\bullet= \bigoplus{\lambda \in \Lambda} A\bullet^\lambda$ of chain complexes $A\bullet^\lambda$, where $An = \oplus An^\lambda$ and $d= \oplus d^\lambda$; 3. $\hom{Top}(\Delta^n{top},\amalg{\lambda \in \Lambda}X\lambda) \cong \amalg \hom{Top}(\Delta^n{top},X\lambda)$ ( because each topological n-simplex is connected); 4. $C\bullet (\amalg{\lambda\in\Lambda}) X\lambda; \mathbb{Z}) \cong \oplus{\lambda \in \Lambda} C\bullet(X\lambda;\mathbb{Z})$

Preview

The next question is: What spaces for which we can compute $H_n(X;\mathbb{Z})$ for all $n$?