Let $\sf{A}$ be an abelian category and $\sf{C}(\sf{A})$ be its category of complexes. I want to prove that the $i$-th cohomology is an additive functor $\sf{C}(\sf{A})\to\sf{A}$. For that, let $\varphi^\bullet:M^\bullet\to N^\bullet$ be a morphism of complexes. By the universal property of kernels, we have an induced morphism
In order for the universal property of cokernels to induce a morphism $H^i(\varphi^\bullet):H^i(M^\bullet)\to H^i(N^\bullet)$ making the diagram
commute, we have to show that the morphism $I^{i-1}_{M^\bullet}\to K^i_{M^\bullet}\to K^i_{N^\bullet} \to H^i(N^\bullet)$ is zero. That would follow from the existence of an induced map $I^{i-1}_{M^\bullet}\to I^{i-1}_{N^\bullet}$, but I can't seem to figure out why this exists.
Also, I would love to know if there's a simpler way to construct the induced morphism $H^i(\varphi^\bullet):H^i(M^\bullet)\to H^i(N^\bullet)$ since my construction seems a little tough to work with.
Best Answer
You have a commutative square $$\require{AMScd}\begin{CD}X @>f>> Y \\ @VVV @VVV \\ Z@>g>> W\end{CD}$$ From this you can extract a commutative diagram $$\begin{CD}X @>>> \mathrm{im}(f) @>>> Y \\ @VVV @VVV @VVV \\ Z@>>> \mathrm{im}(g) @>>> W\end{CD}$$
by using the definition of the image either as the kernel of the cokernel, or as the cokernel of the kernel.
Let me do the first one : you get an induced diagram $$\begin{CD}X @>f>> Y @>>> \mathrm{coker}(f) \\ @VVV @VVV @VVV\\ Z@>g>> W @>>> \mathrm{coker}(g)\end{CD}$$ by universal property of the cokernel, and now it's clear that you also get a diagram $$\begin{CD}\mathrm{im}(f) @>>> Y @>>> \mathrm{coker}(f)\\ @VVV @VVV @VVV \\ \mathrm{im}(g) @>>> W @>>> \mathrm{coker}(g)\end{CD}$$ by universal property of the kernel.
Moreover, by uniqueness in the universal property of the kernel, you also get a commutative diagram $$\begin{CD}X @>>> \mathrm{im}(f) \\ @VVV @VVV \\ Z@>>> \mathrm{im}(g)\end{CD}$$ and pasting them gives the one I claimed.
Since everything here is defined in terms of universal properties this obviously gets you all the functoriality properties you might want