How to show a morphism $f^•: A^•\to B^•$ of complexes induce the morphism of cohomology objects $H^i(A^•)\to H^i(B^•)$?

Question

Let $\mathscr C$ be a Abelian category, how to show a morphism $\varphi^•: A^•\to B^•$ of complexes induce the morphism of cohomology objects $H^i(A^•)\to H^i(B^•)$?

$A^•:\qquad \cdots\to A^{i-1}\stackrel{f^{i-1}}{\to}A^i\stackrel{f^i}{\to}A^{i+1}\stackrel{f^{i+1}}\to\cdots$

$B^•:\qquad \cdots\to B^{i-1}\stackrel{g^{i-1}}{\to}B^i\stackrel{g^i}{\to}B^{i+1}\stackrel{g^{i+1}}\to\cdots$

However, I can only give the morphism $\operatorname{ker}f^i\to \operatorname{ker}g^i$.

Answer 1

Let $\varphi^\bullet : (A^\bullet, d_A^\bullet)\to (B^\bullet,d_B^\bullet)$ be a morphism of cochain complexes. Concretely, recall that this means that for any $i,$ we have $d_B^{i}\circ\varphi^i = \varphi^{i+1}\circ d_A^{i}$ as maps $A^i\to B^{i+1}.$

1. First, observe that $\varphi^i$ induces a morphism $\varphi^i : \ker d_A^i\to\ker d_B^i$ by restriction. Given $\alpha\in\ker d_A^i\subseteq A^i,$ we have $d_B^i\circ\varphi^i(\alpha) = \varphi^{i+1}\circ d_A^i(\alpha) = \varphi^{i+1}(0) = 0.$ This means that $\varphi^i(\alpha)\in\ker d_B^i,$ which is what we wanted.
2. In order to show that we obtain a well-defined map at the level of cohomology, we must show that the composition $$ \ker d_A^i\xrightarrow{\varphi^i}\ker d_B^i\xrightarrow{\pi}\ker d_B^i/\operatorname{im}d_B^{i-1} = \mathrm{H}^i(B^\bullet) $$ factors through $H^i(A^\bullet);$ i.e., that any $\alpha\in\operatorname{im}d_A^{i-1}$ is sent to $0$ in the quotient. Precisely, this means that we must check that such an $\alpha$ is mapped into $\operatorname{im}d_B^{i-1}$ under $\varphi^i.$ To that end, let $\alpha\in\operatorname{im}d_A^{i-1},$ and write $\alpha = d_A^{i-1}(\beta)$ for some $\beta\in A^{i-1}.$ Then it follows that $$ \varphi^i(\alpha) = \varphi^i(d_A^{i-1}(\beta)) = d_B^{i-1}(\varphi^{i-1}(\beta))\in\operatorname{im}d_B^{i-1}. $$ This shows that the map $\pi\circ\varphi^i : \ker d_A^i\to \mathrm{H}^i(B^\bullet)$ factors through a unique map $(\varphi^i)^\ast : \mathrm{H}^i(A^\bullet)\to\mathrm{H}^i(B^\bullet).$ This is the desired map.

As a remark addressing a comment: though diagram chases do not work in an arbitrary Abelian category a priori, in situations like this they are fine. In general, the Freyd-Mitchell embedding theorem tells you that a small Abelian category is equivalent to a full subcategory of $R$-$\mathsf{Mod}$ for some commutative ring $R.$ While your original Abelian category might not be small, this allows you to diagram chase using elements nonetheless: you can consider the smallest Abelian subcategory of your original category generated by all the objects and morphisms in the relevant diagram, which will again be small. Then the embedding theorem will apply, and diagram chasing may commence!