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

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.