Homotopic Chain Maps Induce Equal Maps on Homology

Theorem
Let $A_\bullet$, $B_\bullet$ be chain complexes of abelian groups.

Let $f, g: A_\bullet \to B_\bullet$ 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 $A_\bullet$ and $B_{\bullet}$ respectively.

Let $h$ be a homotopy between $f$ and $g$.

Let:
 * $a \in H_n \left({A}\right) \cong \ker \left({\partial^A_n}\right) / \operatorname{Im} \left({\partial^A_{n + 1} }\right)$

There exists $\tilde a \in \ker \left({\partial^A_n}\right)$ representing $a$.

It is enough to show that:
 * $f_n \left({\tilde a}\right) \sim g_n \left({\tilde a}\right)$

Equivalently:
 * $f_n \left({\tilde a}\right) - g_n \left({\tilde a}\right) \in \operatorname{Im} \left({\partial^B_{n + 1} }\right)$

We know that:
 * $\partial h - h \partial = f - g$

This means:
 * $\partial^B_{n + 1} h_n - h_{n - 1} \partial^A_n = f_n - g_n$

Plugging in $\tilde a$:


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

Since $\tilde a \in \ker \left({\partial^A_n}\right)$:


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

Therefore:
 * $f_n \left({\tilde a}\right) - g_n \left({\tilde a}\right) \in \operatorname{Im} \left({\partial^B_{n + 1} }\right)$