§ Singular homology: induced homomorphism

The space of chains $C[i]$ of a topological space $X$ is defined as all functions $\Delta^i \rightarrow X$. The boundary map is defined as:

$$\begin{aligned} &\partial[n]: C[n] \rightarrow C[n-1] \\ &\partial[n](\sigma) \equiv \sum_i (-1)^i \sigma|[v[0], v[1], \dots, \hat{v[i]}, \dots, v[n]] \end{aligned}$$

where $\hat{v[i]}$ means that we exlude this vertex, and $v[0], v[1], \dots$ are the vertices of the domain $\Delta^i$. Now, say we have a function $f: X \rightarrow Y$, and a singular chain complex $D[n]$ for $Y$. In this case, we can induce a chain map $f\sharp: C[n] \rightarrow D[n]$, given by:

$$\begin{aligned} &f\sharp: C[n] \rightarrow D[n] \\ &f\sharp: (\Delta^n \rightarrow X) \rightarrow (\Delta^n \rightarrow Y) \\ &f\sharp(\sigma) = f \circ \sigma \end{aligned}$$

We wish to show that this produces a homomorphism from $H[n](X) \equiv \ker \partial[n]/ im \partial[n+1]$ to $H[n](Y) \equiv \ker \partial[D][n] / im \partial[D][n+1]$. To do this, we already have a map from $C[n]$ to $D[n]$. We need to show that it sends $\ker \partial[n] \mapsto \ker \partial[D][n]$ and. The core idea is that if we have abelian groups $G, H$ with subgroups $M, N$, and a homomorphism $f: G \rightarrow H$, then this descends to a homomorphism $f': G/M \rightarrow H/N$ iff $f(M) \subseteq N$. That is, if whatever is identified in $G$ is identified in $H$, then our morphism will be valid. To prove this, we need to show that if two cosets $g + M$, $h + M$ are equal, then their images under $f'$ will be equal. We compute $f(g+M) = f(g) + f(M) = f(g) + 0$, and $f(h + M) = f(h) + f(M) = f(h) + 0$. Since $g + M = h+M$, we get $f(g) = f(h)$. Thus, the morphism is well-defined.