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

Let $MC(\varphi^\bullet)^\bullet$ be the mapping cone of $\varphi^\bullet$. Recall that the mapping cone fits into an exact sequence

and that there's a natural quasi-isomorphism $\rho^\bullet:MC(\varphi^\bullet)^\bullet\to N^\bullet$, given by composing the projection with $\psi^\bullet$. Most books in homological algebra (Proposition IX.4.1 in Aluffi's Algebra: Chapter 0, for example) affirm that the connecting morphism $H^i(L^\bullet)\to H^i(M^\bullet)$ coincides with $H^i(\varphi^\bullet)$. I wonder what this even means. As in Uniqueness of the connecting morphism in the snake lemma, the connecting morphisms are not really unique. Since we only know that there exists connecting morphisms, does it even make sense to say that, paraphrasing Aluffi, "the connecting morphism is the morphism induced by $\varphi^\bullet$ in cohomology"?

If this indeed makes sense, is it also true that the diagram

commutes?

Gabriel
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    Connecting morphisms are unique : it's not because there are other morphisms that would do the job that we can't give a definition that uniquely specifies them – Maxime Ramzi Feb 27 '21 at 20:21
  • (e.g. there are many automorphisms $\mathbb Z/n\mathbb Z$, but I can nonetheless specify the identity of this group) – Maxime Ramzi Feb 27 '21 at 20:21
  • @MaximeRamzi If we use the "standard" definition given by using elements (i.e. Freyd-Mitchell), then my diagram indeed commutes? (I've seen some places saying that it commutes and some places saying that it anticommutes, which made me confused.) – Gabriel Feb 27 '21 at 20:47
  • I think it might anticommute, but I'm not entirely sure. Ultimately, one would have to go through all the proofs to figure that out, but one rarely needs this because for most purposes it doesn't matter. If at some point, it does (I doubt it), then it might be good to check it out. – Maxime Ramzi Feb 27 '21 at 21:01

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