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Assume that $T$ is a symmetric monoidal triangulated category, and $X$ is an object in it. Then the functors $X\otimes -$ and $-\otimes X: T\to T$ are not necessarily exact since they send distinguished triangles into triangles that are not necessarily "quite distinguished" (see https://ncatlab.org/nlab/show/tensor+triangulated+category), yes? Is there any nice way to "fix this problem" (which seems to be rather technical)? In particular, are all functors of this sort (canonically?) isomorphic to truly exact ones? Did anybody treat this question earlier; maybe, some terminology (for "nice" $X$ or "almost exact" functors) was introduced?

Moreover, I am also interested in external tensor products (actions) as well (cf. Does the homotopy category of finite spectra act on stable homotopy categories?). In particular, note that for a stable infinity category $T$ one can describe the shift functor $[1]_T$ as the external tensor product $S^1\otimes -$.

Upd. This was so stupid of me! It appears that one does not usually demand an exact functor to commute with shifts and respect distinguished triangles "strictly"; see Definition 2.1.1 in Neeman's "Triangulated categories". However, any comments on this matter would still be very welcome! Is $[1]$ always exact?

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    $\begingroup$ Perhaps I'm missing something obvious here but could you spell out what do you mean by "quite distinguished" ? What is the exact difference between this and one might call a triangulated functor? $\endgroup$
    – AT0
    Commented Jul 13, 2023 at 21:03
  • $\begingroup$ Exact functor is the same thing as a triangulated one. But the shift functor [1] is not ("quite") exact since one should change the signs of arrows in the shift of a distinguished triangle to make this shifted triangle distinguished. On the other hand, the functor [2] is exact. $\endgroup$
    – Mikhail Bondarko
    Commented Jul 14, 2023 at 8:33
  • $\begingroup$ I see, thanks for the clarification. I guess my question is then if you're asking or expecting your functors X\otimes to commute with taking cones? Otherwise already on the third term of the candidate triangle you'll need to make an adjustement, well before any shift issue arises, no? $\endgroup$
    – AT0
    Commented Jul 14, 2023 at 9:15
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    $\begingroup$ Also, suspension is exact, you just didn’t specify the natural transformation ;) $\endgroup$
    – Fernando Muro
    Commented Jul 17, 2023 at 7:54
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    $\begingroup$ Yes; I have realized that the standard definition involves a natural transformation.:) $\endgroup$
    – Mikhail Bondarko
    Commented Jul 17, 2023 at 19:46

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