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Let $f:B\to C$ be a morphism of chain complexes. Exercise 1.5.9 from Weibel's "An introduction to homological algebra", page 24, states that the natural maps $\ker f[-1]\xrightarrow\alpha{\rm cone}(f)\xrightarrow\beta{\rm coker} f$ give rise to a long exact sequence: $$\cdots\xrightarrow\delta H_{n-1}(\ker f)\xrightarrow\alpha H_n({\rm cone}(f))\xrightarrow\beta H_n({\rm coker} f)\xrightarrow\delta H_{n-2}(\ker f)\xrightarrow\alpha\cdots$$ (In the book there is a small mistake. The map $\ker f[-1]\to{\rm cone}(f)$ was denoted by $\delta$ instead of $\alpha$.)

My question is, is there any quotable source for this result? I cannot quote an exercise, especially when it is incomplete. It doesn't state what the map $\delta$ is and in my paper I need that $\delta$.

I figured out that $\delta=\delta'\delta''$, where $\delta'$ is the connecting map $H_{n-1}({\rm Im}f)\to H_{n-2}(\ker f)$, coming from the exact sequence $0\to\ker f\to B\xrightarrow f{\rm Im}f\to 0$, and $\delta''$ is the connecting map $H_n({\rm coker}f)\to H_{n-1}({\rm Im}f)$, coming from the exact sequence $0\to{\rm Im}f\to C\to{\rm coker} f\to 0.$

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    $\begingroup$ This is particular case of the octahedral axiom applied to the distinguished triangles assocated to the exact sequences $0\to ker f\to B \to im f\to 0$ and $0\to Im f\to C \to coker f\to 0$, and the canonical distinguished triangle associated to the cone of $f$. You may thus quote any source claiming to prove the derived category of an abelian category is triangulated. $\endgroup$
    – D.-C. Cisinski
    Jul 8, 2021 at 14:24
  • $\begingroup$ @Constantin-NicolaeBeli Yes, apologies, I was going to say: "If I understand correctly, you are looking for p283 of Gelfand-Manin". $\endgroup$
    – Vidit Nanda
    Jul 8, 2021 at 15:09
  • $\begingroup$ @Denis-CharlesCisinski Thank you for your answer, but I would prefer a more low-tech approach. My paper is aimed at people from number theory, who are not usually experts in cohomology. In fact, I only learned about mapping cones two days ago, when I asked on mathoverflow. mathoverflow.net/questions/396916 $\endgroup$
    – Constantin-Nicolae Beli
    Jul 8, 2021 at 18:13
  • $\begingroup$ Initially, for a morphism of $G$-modules $f:M\to M$ I defined $H^n(G,M,f)=Z^n(G,M,f)/B^n(G,M,f)$, where $Z^n(G,M,f)=\{ (a,b)\in C^n(G,M)\times C^{n+1}(G,M)\mid\, da=f(b),\, db=0\}$ and $B^n(G,M,f)$ is generated by the obvious elements of $Z^n(G,M,f)$ namely, $(dc,0)$, with $c\in C^{n-1}(G,M)$ and $(f(c),c)$, with $c\in C^n(G,M)$. This definition worked fine, until I realized that $H^n(G,M,f)$ is the cohomology of a complex. Now I cannot pretend I don't know about mapping cones and go by the old definition, but I wish I could keep the homological theory to a minimum. $\endgroup$
    – Constantin-Nicolae Beli
    Jul 8, 2021 at 18:16
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    $\begingroup$ If $p:B\to im f$ is the map induced by $f$ and if $i:0\to coker f$ is the inclusion, there is a rather obvious short exact sequence $0\to cone(p)\to cone (f)\to cone(i)\to 0$ inducing a long exact sequence in homology. It remains to check that there is quasi-isomorphism relating $cone(p)$ and $ker(f)[1]$ as well as an identification of $cone(i)$ with $coker(f)$. Note that this kind of reasoning is extremely close to proving the octahedral axiom itself. $\endgroup$
    – D.-C. Cisinski
    Jul 9, 2021 at 11:54

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