$\newcommand{\coker}{\text{coker}}$
$\newcommand{\Ima}{\text{Im}}$
I've started to learn some homological algebra and have been struggling with verifying that if $\mathcal{A}$ is an abelian category then the category of chain-complexes in $\mathcal{A}$ is also an abelian category. In particular, showing if $f_\bullet:A_\bullet \to B_\bullet$ is a chain map then $\coker (f_\bullet) = \dots \to \coker f_{n+1} \to \coker f_n \to \dots$..
The diagram I had in mind was:
Where $\pi_n:B_n \to B_n/\Ima(f_n)$ are natural projection, and I'd like to fill in the maps $\partial_n$ on the bottom row.
There's the obvious guess that you would induce the map $\tilde{d_n^B}: B_m/\Ima f_n \to B_{n-1}$ and then project onto $\coker f_{n-1}$, but to have the induced map I would need that $\Ima f_n \subseteq \ker d_n^B$, and I can't seem to conclude that's the case. I know everything in sight commutes, but $d_n^B \circ f_n = 0$ even seems like perhaps a wrong condition to me, because by commutativity of the square that would mean $f_{n-1} \circ d_n^A = 0$.
I know that if I take the composites $d_n^B \circ f_n \circ d_{n+1}^A = d_n^B \circ d_{n+1}^B \circ f_{n-1} = 0$ since $d^2 = 0$ on a chain complex, but then I'm left with $d_n^B \circ f_n \circ d_{n+1}^A = 0$, and $f_n \circ d_{n+1}^A$ wouldn't correspond to the image of $f_n$.
Any hints would be greatly appreciated, I'm sure the fact is straightforward I'm just not seeing it. Thanks in advance.
Edit: Just in case someone comes across this but would prefer the language of $R$-modules (as I did), the differential in the cokernel complex is just induced by the factor isomorphism theorem, I was just going about it wrong:
The map $\pi \circ d_n^B:B_n \to \coker f_{n-1}$ will factor through $\coker f_n$ if and only if $\Ima f_n \subseteq \ker (\pi \circ d_n^B)$ but note that this kernel is $(d_n^B)^{-1}(\Ima f_{n-1})$ and so this is equivalent to requiring $d_n^B \circ \Ima f_n \subseteq \Ima f_{n-1}$ which is obvious from the commutativity of the top square. Therefore set $\partial_n$ to be the unique map factoring the composite.
Best Answer
This is not only part of a proof that $\mathsf{Ch}(\mathscr{A})$ is Abelian (by in particular showing it has all cokernels) but it is part of a proof that the evaluation functors $e_n:\mathsf{Ch}(\mathscr{A})\to\mathscr{A}$ are exact (by preserving all cokernels).
We are given the zero complex as a clear zero object. It is clear a map of chain complexes is zero (by definition, factors through the zero object) if and only if every single component is zero. Given $f:A_\bullet\to B_\bullet$, $g:B_\bullet\to C_\bullet$, $gf$ is then zero if and only if $g_nf_n$ is zero for all $n$. That is to say, if and only if $g_n$ factors through $\operatorname{coker}f_n$ for every $n$.
There are then two things to check: firstly, that there is a sensible notion of differential making $\operatorname{Cok}_\bullet f$ a chain complex such that $\operatorname{coker}f:B_\bullet\to\operatorname{Cok}_\bullet f$ is a chain map, secondly that the induced map $\operatorname{Cok}_\bullet f\to C_\bullet$ which factors $g$ is a genuine chain map.
Let's sort out the first one. As: $$0=\operatorname{coker}f_{n-1}\circ f_{n-1}\circ\partial_n=\operatorname{coker}f_{n-1}\circ\partial_n\circ f_n$$It follows there exists a unique factorisation $\operatorname{coker}f_{n-1}\circ\partial_n=k\circ\operatorname{coker}f_n$. Let's make the $n$th differential for $\operatorname{Cok}_\bullet f$ the arrow $k$. Then, so long as this satisfies the $\partial^2=0$ property, we'd get a chain complex that by construction makes $\operatorname{coker}f$ a chain map. But this thing squaring to zero is straightforward, since when you precompose with $\operatorname{coker}f_{n+1}$ you see: $$k_n\circ k_{n+1}\circ\operatorname{coker}f_{n+1}=\partial_n\circ\operatorname{coker}f_n\circ\partial_{n+1}=\operatorname{coker}f_{n-1}\circ\partial_n\circ\partial_{n+1}=0$$And as this is an epimorphism we can remove the cokernel from the left hand side and deduce $k_n\circ k_{n+1}=0$ for all $n$.
Now for the second part.
Say $g_n=h_n\circ\operatorname{coker}f_n$ for each $n$. $$\begin{align}h_{n-1}\circ\partial_n\circ\operatorname{coker}f_n&=h_{n-1}\circ\operatorname{coker}f_{n-1}\circ\partial_{n-1}\\&=g_{n-1}\circ\partial_n\\&=\partial_n\circ g_n\\&=\partial_n\circ h_n\circ\operatorname{coker}f_n\end{align}$$Ok. By the universal property of cokernels (well, the uniqueness) we know they are epimorphisms; I can therefore remove $\operatorname{coker}f_n$ from both sides of the above (cancellation doesn't mean it has a right inverse!) and conclude $h_{n-1}\circ\partial_n=\partial_n\circ h_n$. So, $h_\bullet:\operatorname{Cok}_\bullet f\to C_\bullet$ is a chain map and it is unique because $g=hf$ iff. $g_n=h_nf_n$ for all $n$ (by definition) and each $h_n$ is unique in this respect.