Homomorphism of Chain Complexes induces Homomorphism of Homology

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Theorem

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

Let $f: A_\bullet \to B_\bullet$ be a homomorphism.

Then for every $n$, $f$ induces a morphism $H_n \left({A_\bullet}\right) \to H_n \left({B_\bullet}\right)$ of homology groups.

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Proof

Let $\partial^A_\bullet$, $\partial^B_\bullet$ denote the differential on $A_\bullet$, respectively $B_\bullet$.

First it will be demonstrated that:

$\forall a \in \ker \left({\partial^A_n}\right) \subseteq A_n: f_n \left({a}\right) \in \ker \left({\partial^B_n}\right)$

Thus:

$\partial^B_n f_n \left({a}\right) = f_{n - 1} \left({\partial^A_n a}\right) = f_{n - 1} \left({0}\right) = 0$

Thus there exists a map:

$\bar f_n: \ker \left({\partial^A_n}\right) \to \ker \left({\partial^B_n}\right)$

via the restriction of $f$.

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Next will be shown that:

$\forall a \in \operatorname {Im} \left({\partial^A_{n + 1} }\right): f_n \left({a}\right) \in \operatorname {Im} \left({\partial^B_{n + 1} }\right)$

Let $a = \partial^A_{n + 1}a'$

Then:

$\partial^B_{n + 1} \left({f_{n + 1} \left({a'}\right)}\right) = f_n \left({\partial^A_n a'}\right) = f_n \left({a}\right)$.

Let $\pi: \ker \left({\partial^B_n}\right) \to H_n \left({B_\bullet}\right)$ be the quotient map.

Let $\rho = \pi \circ \bar f_n$.

From above:

$\operatorname{Im} \left({\partial^A_{n + 1} }\right) \subseteq \ker \left({\rho}\right)$

Thus $\rho$ factors through a map:

$\tilde f_n: H_n \left({A_\bullet}\right) \to H_n \left({B_\bullet}\right)$

completing the proof.

$\blacksquare$