Homotopic Chain Maps Induce Equal Maps on Homology

Theorem
Let $A_{\bullet}$, $B_{\bullet}$ be chain complexes of abelian groups. Let $f,g:A_{\bullet} \rightarrow B_{\bullet}$ be chain maps which are homotopic. Then $f$ and $g$ induce equal maps on homology.

Proof
Let $\partial^A_{\bullet},\partial^B_{\bullet}$ be the differentials on $A_{\bullet}$, respectively $B_{\bullet}$. Let $h$ be a homotopy between $f$ and $g$.

Let $a \in H_n(A) \cong \text{Ker}(\partial^A_n)/\text{Im}(\partial^A_{n+1})$. There exists $\tilde{a} \in \text{Ker}(\partial^A_n)$ representing $a$. It is enough to show that $f_n(\tilde{a}) \sim g_n(\tilde{a})$. Equivalently, $f_n(\tilde{a}) - g_n(\tilde{a}) \in \text{Im}(\partial^B_{n+1})$.

We know $\partial h - h \partial = f - g$. This means $\partial^{B}_{n+1}h_n - h_{n-1}\partial^{A}_n = f_n - g_n$. Plugging in $\tilde{a}$:


 * $\displaystyle \partial^B_{n+1}(h_n(\tilde{a})) - h_{n-1}(\partial^A_n(\tilde{a})) = f_n(\tilde{a}) - g_n(\tilde{a})$

Since $\tilde{a} \in \text{Ker}(\partial^A_n)$:


 * $\displaystyle \partial^B_{n+1}(h_n(\tilde{a})) = f_n(\tilde{a}) - g_n(\tilde{a})$

Therefore, $f_n(\tilde{a}) - g_n(\tilde{a}) \in \text{Im}(\partial^B_{n+1})$.