Homomorphism of Chain Complexes induces Homomorphism of Homology

Theorem
Let $A_{\bullet}$ and $B_{\bullet}$ be chain complexes of abelian groups, and $f:A_{\bullet} \rightarrow B_{\bullet}$ a homomorphism. Then for every $n$, $f$ induces a morphism $H_n(A_{\bullet}) \rightarrow H_n(B_{\bullet})$ of homology groups.

Proof
Let $\partial^{A}_{\bullet}$, $\partial^{B}_{\bullet}$ denote the differential on $A_{\bullet}$, respectively $B_{\bullet}$.

First, for $a \in \text{Ker}(\partial^A_{n}) \subseteq A_n$, $f_n(a) \in \text{Ker}(\partial^B_n)$. To prove this, compute $\partial^B_nf_n(a) = f_{n-1}(\partial^A_na) = f_{n-1}(0) = 0$. That means we have a map $\bar{f}_n:\text{Ker}(\partial^A_n) \rightarrow \text{Ker}(\partial^B_n)$ via the restriction of $f$.

Second, for $a \in \text{Im}(\partial^A_{n+1})$, $f_n(a) \in \text{Im}(\partial^B_{n+1})$. To prove this, write $a = \partial^A_{n+1}a'$. Then $\partial^B_{n+1}(f_{n+1}(a')) = f_n(\partial^A_na') = f_n(a)$.

If $\pi:\text{Ker}(\partial^B_n) \rightarrow H_n(B_{\bullet})$ is the quotient map, let $\rho = \pi \circ \bar{f}_n$. From the previous paragraph we have $\text{Im}(\partial^A_{n+1}) \subseteq \text{Ker}(\rho)$, and so $\rho$ factors through a map $\tilde{f}_n:H_n(A_{\bullet}) \rightarrow H_n(B_{\bullet})$, completing the proof.

Remark
This result holds with 'Abelian Groups' replaced with an arbitrary abelian category.