Direct proof that a short exact sequence induces a quasi-isomorphism from the mapping cone

224 Views Asked by At

Consider the following exact sequence of complexes in an abelian category:

I want to prove that there exists a quasi-isomorphism $\rho^\bullet:\operatorname{MC}(\varphi^\bullet)^\bullet\to N^\bullet$, where $\operatorname{MC}(\varphi^\bullet)^\bullet$ is the mapping cone of $\varphi^\bullet$.

It is clear that $\rho^\bullet$ should be the composition of the projection $\operatorname{MC}(\varphi^\bullet)^\bullet \to M^\bullet$ with the given morphism $\psi^\bullet:M^\bullet\to N^\bullet$. The usual way to continue seems to be noticing that

is an exact sequence, that $H^\bullet(K^\bullet)=0$ and using the long exact sequence in cohomology to conclude.

However, I'm trying to use this statement to prove the existence of the long exact sequence in cohomology, so I'm trying to prove that $\rho^\bullet$ is a quasi-isomorphism directly. It seems that it shouldn't be really difficult, since everything is somewhat explicit here, but I can't seem to do it. (I know how to do it using Freyd-Mitchel but I really would prefer an arrow-theoretic argument.)