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Let $(X,\mathcal{O}_X)$ be a topological ringed space and $(U,V)$ be an open covering of $X$ (i.e. $U$ and $V$ are two open subsets of $X$ such that $U\cup V=X$). Let $\mathcal{F}$ and $\mathcal{F}'$ be sheaves of $\mathcal{O}_X$-modules.

By the Mayer-Vietoris triangle (see 73.10.5), one has a distinguished triangle in the derived category of $\mathcal{O}_X(X)$-modules: $$R\Gamma(X,\mathcal{F})\to R\Gamma(U,\mathcal{F})\oplus R\Gamma(V,\mathcal{F})\stackrel{(1,-1)}{\longrightarrow}R\Gamma(U\cap V,\mathcal{F})\to [1].$$ Let $f:\mathcal{F}\to \mathcal{F}'$ be a morphism of sheaves of $\mathcal{O}_X$-modules. It induces maps of complexes in the derived category: $f_X:R\Gamma(X,\mathcal{F})\to R\Gamma(X,\mathcal{F}')$, $f_U:R\Gamma(U,\mathcal{F})\to R\Gamma(U,\mathcal{F}')$, $f_V:R\Gamma(V,\mathcal{F})\to R\Gamma(V,\mathcal{F}')$ and $f_{U\cap V}:R\Gamma(U\cap V,\mathcal{F})\to R\Gamma(U\cap V,\mathcal{F}')$. It allows us to complete the diagram into a morphism of distinguished triangles: $$\require{AMScd} \begin{CD} R\Gamma(X,\mathcal{F}) @>>> R\Gamma(U,\mathcal{F})\oplus R\Gamma(V,\mathcal{F}) @>>> R\Gamma(U\cap V,\mathcal{F}) @>>> [1] \\ @VVf_XV @VVf_U\oplus f_{V}V @VVf_{U\cap V}V @VVV \\ R\Gamma(X,\mathcal{F}') @>>> R\Gamma(U,\mathcal{F}')\oplus R\Gamma(V,\mathcal{F}') @>>> R\Gamma(U\cap V,\mathcal{F}') @>>> [1] \end{CD}$$ Not every morphism of triangles can be completed into a $3\times 3$ square (see this question, or Neeman's paper). Here is my question:

Is the morphism of triangle $(f_X, f_{U}\oplus f_{V}, f_{U\cap V})$ middling good in the sense of Neeman? In other terms, can the above diagram be completed into $$\require{AMScd} \begin{CD} R\Gamma(X,\mathcal{F}) @>>> R\Gamma(U,\mathcal{F})\oplus R\Gamma(V,\mathcal{F}) @>>> R\Gamma(U\cap V,\mathcal{F}) @>>> [1] \\ @VVf_XV @VVf_U\oplus f_{V}V @VVf_{U\cap V}V @VVV \\ R\Gamma(X,\mathcal{F}') @>>> R\Gamma(U,\mathcal{F}')\oplus R\Gamma(V,\mathcal{F}') @>>> R\Gamma(U\cap V,\mathcal{F}') @>>> [1] \\ @VVV @VVV @VVV @VVV \\ \operatorname{cone}(f_X) @>>> \operatorname{cone}(f_{U}\oplus f_{V}) @>>> \operatorname{cone}(f_{U\cap V}) @>>> [1] \\ @VVV @VVV @VVV @VVV \\ [1] @>>> [1] @>>> [1] @>>> [1] \end{CD}$$ where all columns and all rows are distinguished, and all squares commute, except for the bottom right square, which anti-commutes?

In particular, we would be able to deduce a Mayer-Vietoris triangle not only for sheaves, but also morphisms of sheaves:

$$R\Gamma\left(X,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)\to R\Gamma\left(U,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)\oplus R\Gamma\left(V,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)\longrightarrow R\Gamma\left(U\cap V,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)\to [1]$$ where the notation $R\Gamma\left(X,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)$ stems for $\operatorname{Tot}(f_X)=\operatorname{cone}[-1](f_X)$.

Aside from an answer, any comment, reference, example or discussion would be surely very inspiring to me. Many thanks in advance!

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    $\begingroup$ Yes it can. It seems that your last couple questions have to do with the pathologies that arise in triangulated categories due to non-functorial cones. These pathologies almost never matter in practice, because the triangulated categories that mathematicians actually use are in fact the homotopy categories of dg-enhanced/stable infinity categories-- these have functorial cones "built in." From a lowbrow perspective, you can just construct your diagram as a strictly commuting diagram of honest chain complexes and take the cones of this diagram to obtain the desired square. $\endgroup$ Aug 25, 2020 at 15:20
  • $\begingroup$ @PhilTosteson Thank you very much. I agree with your comment. However, the question is rather then: how does one show that $f_X$ indeed comes from the functorial cone construction? (maybe I am missing something...) $\endgroup$
    – Stabilo
    Aug 25, 2020 at 19:11

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