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For triangulated categories $T,T'$ I would like to define "weakly exact" functors as those that respect cones, that is, $F(Cone f)\cong Cone(F(f))$ for any $T$-morphism $f$, and I do not require any functoriality for these isomorphisms. Note that one can take $f$ to be a morphism into $0$ to obtain that $F$ ("non-functorially") respects shifts.

Now, do there exist any ("natural"?) examples of functors that are exact in this sense but not exact in the more restrictive sense of https://stacks.math.columbia.edu/tag/014V ? Did anybody consider something similar to "my" definition?

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One way of making examples is as follows. Take your favourite triangulated category $\mathcal{T}$ over a field not of characteristic two. Let $\mathcal T'$ be the triangulated category with the same underlying category, but if the distinguished triangles in $\mathcal T$ are $A\xrightarrow{u}B\xrightarrow{v}C\xrightarrow{w}\Sigma A$ then those in $\mathcal T'$ are $A\xrightarrow{-u}B\xrightarrow{-v}C\xrightarrow{-w}\Sigma A$. The identity functor of categories from $\mathcal T$ to $\mathcal T'$ has your property, but is not a triangulated (i.e., exact) functor.

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  • $\begingroup$ Do you think that this identity is not exact in the sense of stacks.math.columbia.edu/tag/014V ? $\endgroup$
    – Mikhail Bondarko
    Commented Jun 21 at 15:43
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    $\begingroup$ @MikhailBondarko not if you equip it with the identity natural isomorphisms. The thing is that an exact functor must come equipped with such a natural transformation. $\endgroup$
    – Fernando Muro
    Commented Jun 21 at 16:16
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    $\begingroup$ Yes, I understand that you get more exact functors if you vary the transformation in the definition. My question is whether it makes sense not to assume any functoriality for the corresponding isomorphisms at all. $\endgroup$
    – Mikhail Bondarko
    Commented Jun 21 at 19:08

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