Proof of the exact sequence $0\to H^i(M)\to \operatorname{coker} d^{i-1}\to\ker d^{i+1}\to H^{i+1}(M)\to 0$.

Question

Let $M^\bullet$ be a complex in an abelian category. By the universal property of cokernels, there is a unique induced morphism making the diagram

commute. I want to prove that its kernel is $H^i(M^\bullet)$ and that its cokernel is $H^{i+1}(M^\bullet)$.

Since $H^i(M^\bullet)$ is, by definition, the cokernel of $I^{i-1}_{M^\bullet}\to K^i_{M^\bullet}$, the universal property of cokernels induces a morphism $H^i(M^\bullet)\to C^{i-1}_{M^\bullet}$ making the diagram

commute. I see that the composition $H^i(M^\bullet)\to C^{i-1}_{M^\bullet}\to K^{i+1}_{M^\bullet}$ is zero but I don't understand why $H^i(M^\bullet)\to C^{i-1}_{M^\bullet}$ satisfies the universal property of the kernel. After all, we understand the morphisms getting out of $H^i(M^\bullet)$, not those going in.

I have the same problem showing that $H^{i+1}(M^\bullet)$ is the cokernel of $C^{i-1}_{M^\bullet}\to K^{i+1}_{M^\bullet}$.

Answer 1

After a lot of thought, I think I solved this problem by finding a more natural candidate for the morphism on the middle. (What follows are my own notes.)

Then our desired result becomes way simpler.