Homotopic Chain Maps Induce Equal Maps on Homology

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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:

$a \in \map {H_n} A \cong \map \ker {\partial^A_n} / \Img {\partial^A_{n + 1} }$

There exists $\tilde a \in \map \ker {\partial^A_n}$ representing $a$.

It is enough to show that:

$\map {f_n} {\tilde a} \sim \map {g_n} {\tilde a}$

Equivalently:

$\map {f_n} {\tilde a} - \map {g_n} {\tilde a} \in \Img {\partial^B_{n + 1} }$

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 $\tilde a$:

$\map {\partial^B_{n + 1} } {\map {h_n} {\tilde a} } - \map {h_{n - 1}} {\map {\partial^A_n} {\tilde a} } = \map {f_n} {\tilde a} - \map {g_n} {\tilde a}$

Since $\tilde a \in \map \ker {\partial^A_n}$:

$\map {\partial^B_{n + 1} } {\map {h_n} {\tilde a} } = \map {f_n} {\tilde a} - \map {g_n} {\tilde a}$

Therefore:

$\map {f_n} {\tilde a} - \map {g_n} {\tilde a} \in \Img {\partial^B_{n + 1} }$

$\blacksquare$