Category of cochain complexes is abelian

Question

It is well known that if $\mathbf{A}$ abelian, then so is $\mathcal{Cochain}(\mathbf{A})$, the category of cochain complexes in $\mathbf{A}$.

Clearly $\mathcal{Cochain}(\mathbf{A})$ has all finite (co)products and (co)kernels, but it is unclear how every monomorphism $f:K^{\bullet}\to X^{\bullet}$ is a kernel of some morphism $g:X^{\bullet}\to Y^{\bullet}$. More precisely, how do you construct the differentials $\delta_i:Y_i\to Y_{i+1}$ for each $i\in\mathbb{Z}$?

I suppose you can show that for each morphism $g:X^{\bullet}\to Y^{\bullet}$, the image of $g$ coincides the coimage of $g$, by computing the kernels and cokernels using universal properties.

Edit: It sounds silly but we can set each $\delta_i:Y_i\to Y_{i+1}$ zero.

Answer 1

In an abelian category every monomorphism is the kernel of its cokernel. If you have already constructed the cokernels in the category of cochain complexes, then you just have to check that your monomorphism is exactly the kernel of this cokernel. Note that the differentials in kernels and cokernels are in general non-zero.