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Let $A$ be any abelian category. Consider the stable $\infty$-category $N_{dg}(Ch(A))$ defined as the differential graded nerve of the category of chain complexes $Ch(A)$ in $A$. We can define an $\infty$-category $$D(A):= N_{dg}(Ch(A))[W^{-1}]$$ by inverting the quasi-isomorphisms.

I want to ask, for general $A$, is $D(A)$ a stable $\infty$-category?

I'm aware of the following facts:

(1) If $A$ is Grothendieck, then the answer is yes (say by the results in Lurie's HA.1.3.5)

(2) If $A$ is small, then the homotopy category $hD(A)$ has a natural triangulated structure (by the classical theory of localization of triangulated categories)

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    $\begingroup$ Perhaps the answer is YES, but it might happen that $D(A)$ lives in another Grothendieck's universe. $\endgroup$
    – Leo Alonso
    Commented Nov 18 at 12:11
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    $\begingroup$ Well, Groth in one of his papers on derivators noted that a derivator/infinity-category is stable iff finite limits commute with finite colimits. I'm not well versed in those matters, but it seems like a property you can check directly. $\endgroup$
    – Denis T
    Commented Nov 18 at 14:54
  • $\begingroup$ The category of unbounded chain complexes on any additive category form a stable model category with chain homotopy equivalences as weak equivalences and degreewise split monomorphisms as cofibrations (exercise!). Thefore, the Dwyer-Kan localization of unbounded chain complexes by chain homotopy equivalences is a stable $\infty$-category. Taking the Verdier quotient by acyclic complexes thus provides a stable $\infty$-category as well, which is the unbounded derived $\infty$-category. $\endgroup$
    – D.-C. Cisinski
    Commented Nov 19 at 11:10
  • $\begingroup$ @D.-C.Cisinski If I understand correctly, given a small additive $\infty$-category $\mathcal A$, its non-abelian derived category $\mathcal P_\Sigma(\mathcal A)$ is pre-stable, thus it embeds into its stabilization. When $\mathcal A$ is an 1-category, I think that, the unbounded chain complexes modulo chain homotopy equivalences form a full subcategory of this stable $\infty$-category. $\endgroup$
    – Z. M
    Commented Nov 19 at 12:23
  • $\begingroup$ @D.-C.Cisinski Thank you! I was confused about how to conduct this Verdier quotient and compare it with the localization, but the theorem cited in David's edited answer explains that. $\endgroup$
    – Lin Chen
    Commented Nov 20 at 2:21

2 Answers 2

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Thanks to Victor Saunier for pointing out a mistake in an earlier edit of this post.

Let $K(A) := N_{dg}(Ch(A))$, with $A$ a small category.

In the Münster video series, around minute 20, it is shown by hand that $K(A)$ is a stable category, this is shown by explicitly writing down what a cofiber square is and seeing that it is both a pushout and a pullback.

The exact same proof applies to show that $K^{\mathrm{coh}=0}(A)$, the subcategory of all complexes with vanishing cohomology groups, is stable. The set $W$ of quasi-isomorphisms is the set of morphisms with cofiber in $K^{\mathrm{coh}=0}(A)$, applying theorem I.3.3 from Nikolaus-Scholze shows that $\mathcal{D}(A)= K(A)[W^{-1}]$ is stable.

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    $\begingroup$ HA.1.1.4.4 is about limits rather than colimits $\endgroup$
    – Lin Chen
    Commented Nov 19 at 7:45
  • $\begingroup$ Sorry, I meant 1.1.4.6, I will correct my post. $\endgroup$
    – David Wiedemann
    Commented Nov 19 at 8:38
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    $\begingroup$ 1.1.4.6 only pertains to filtered colimits, which cofibers are not. Here, I think it's simply not true that the cofiber in $\mathrm{Cat}_{\infty}$ is stable; the localisation on the other hand is indeed stable, and this is proven in Theorem I.3.3 of arxiv.org/pdf/1707.01799 $\endgroup$
    – Victor Saunier
    Commented Nov 19 at 9:05
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    $\begingroup$ Thanks for the correction, I was too hasty in answering. I’ll correct my answer. Is there a way of describing what the cofiber in $Cat_\infty$ is? $\endgroup$
    – David Wiedemann
    Commented Nov 19 at 17:01
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If $A$ is small, it might be better to consider $D^b(A)$, the derived bounded category, which is also small. In this case, more is known: Corollary 7.4.12 of https://arxiv.org/pdf/1911.02338 proves that for an exact category $E$ (in the Quillen sense, so this in particular applies to abelian categories), the map $E\to D^b(E)$ is the initial exact functor from E to a stable category.

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    $\begingroup$ I agree. But in the classical theory, people also considered unbounded derived category of small abelian categories. For instance, arxiv.org/pdf/2003.11261. $\endgroup$
    – Lin Chen
    Commented Nov 19 at 7:48
  • $\begingroup$ As David's answer below illustrates, the large derived bounded category is often constructed as a localisation, and this means one should probably be careful with the set-theory. For example, if your abelian category is locally presentable, then you should be able to reduce to the small case. $\endgroup$
    – Victor Saunier
    Commented Nov 19 at 8:32
  • $\begingroup$ I am confused. What is the relation between the bounded derived category of a small abelian category with its stable envelope. $\endgroup$
    – Z. M
    Commented Nov 19 at 12:26
  • $\begingroup$ @Z.M unless I misunderstand what you ask, the above-cited theorem precisely says the two agree. $\endgroup$
    – Victor Saunier
    Commented Nov 19 at 14:07
  • $\begingroup$ Oh, what I had in my mind was incorrect. I thought of stable envelope of additive categories, but here, we need stable envelope of exact categories. $\endgroup$
    – Z. M
    Commented Nov 19 at 17:01

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