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This question seems trivial, so hopefully it will be resolved quickly.

As pointed out in this question on quotient categories and localization, the two constructions are sometimes related, but in general quite different.

Let $C$ be a triangulated subcategory of a triangulated category $D$, and let $$Q: D \to D/C$$ be the Verdier quotient. As I understand it, to show the existence of such a quotient, Verdier constructed it as a localization (see the proof in these notes, for example). But it is still referred to as the Verdier quotient.

My question is Is the Verdier quotient category $D/C$ actually a quotient category of $D$, say in the sense of the Wikipedia article?

The motivation of this question is that I am looking at a particular Verdier quotient and I want to understand certain properties of the hom-sets, and it would be easier (for me) to think about a quotient, rather than a localization for this.

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    $\begingroup$ My understanding is that the Verdier quotient is an example of cofiber (see en.wikipedia.org/wiki/Mapping_cone_%28topology%29 ), which is a sort of "homotopy quotient". I am skeptical that the notion of "quotient category" is related in general (although it might well coincide in some special cases) $\endgroup$
    – Denis Nardin
    Commented Aug 31, 2016 at 17:19
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    $\begingroup$ No, it's not a quotient in Wikipedia's sense. Consider the derived category as the Verdier quotient of the homotopy category by acyclic complexes. $\endgroup$
    – Fernando Muro
    Commented Aug 31, 2016 at 17:26
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    $\begingroup$ The sense in which Verdier quotients are quotients is that you are trying to identify some objects (not morphisms), namely you are trying to identify a bunch of objects with $0$. Wikipedia's notion of quotient category is only about identifying morphisms. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 31, 2016 at 18:09
  • $\begingroup$ I see. Thanks for the quick responses! $\endgroup$
    – Spencer Leslie
    Commented Aug 31, 2016 at 19:56
  • $\begingroup$ @DenisNardin this can actually be made precise: arxiv.org/abs/1511.08287 $\endgroup$
    – fosco
    Commented Sep 10, 2016 at 16:10

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