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Let $R$ be a ring and $f:A\to A'$ and $g:B\to B'$ be morphisms of $R$-modules. Let $h:C_{\bullet}\to C_{\bullet}'$ be a morphism of $R$-module complexes fitting in a morphism of distinguished triangles:

$$\require{AMScd} \begin{CD} C_{\bullet} @>>> A[0] @>>> B[0] @>>> C_{\bullet}[1] \\ @VVhV @VVfV @VVgV @VVh[1]V \\ C_{\bullet}' @>>> A'[0] @>>> B'[0] @>>> C_{\bullet}'[1] \end{CD}$$

From a paper I am reading, the following is used:

If $f$ and $g$ are injective morphisms, we have a quasi-isomorphism $\operatorname{cone}(h)\cong[\operatorname{coker}(f)\to \operatorname{coker}(g)]$.

It seems that the equivalent general statement is wrong, i.e. if $f$ and $g$ are not assumed to be injective and $A[0]$, $A'[0]$, $B[0]$ and $B'[0]$ are not assumed to be concentrated in degree $0$ anymore (see for instance this question).

Does anyone have a proof - if true - of this statement?

Many thanks!

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    $\begingroup$ This is easy to show for the functorial cone (when $f,g$ are injective). Are you specifically worried about choosing $h$ in a way that disagrees with the functorial cone? $\endgroup$
    – Phil Tosteson
    Commented Aug 24, 2020 at 21:14
  • $\begingroup$ A priori $h$ is arbitrary here (but still such that it forms a morphism of distinguished triangles). Is this an issue regarding the truth of the statement? $\endgroup$
    – Stabilo
    Commented Aug 24, 2020 at 22:00
  • $\begingroup$ Unfortunately, I'm not sure. It seems likely that by modifying $h$ you can end up with a non isomorphic cone, but I don't know how to construct a pathological example right now. $\endgroup$
    – Phil Tosteson
    Commented Aug 24, 2020 at 22:22
  • $\begingroup$ After a small computation, it seems that $\operatorname{cone} h$ do depend on the differential of $C_{\bullet}$ and $C_{\bullet}'$ while $[\operatorname{coker}f\to \operatorname{coker} g]$ do not. Maybe this explains why $h$ cannot be arbitrary? $\endgroup$
    – Stabilo
    Commented Aug 25, 2020 at 13:15

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