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Suppose $\mathcal{C}$ is a stable $\infty$-category whose homotopy category is a well-generated triangulated category in the sense of Neeman's book. Must $\mathcal{C}$ be a presentable $\infty$-category?

Rosicky proved that if $\mathcal{M}$ is a combinatorial stable model category then $Ho(\mathcal{M})$ is well-generated, so I'm asking for a kind of converse, and I'm happy for an answer either about model categories being combinatorial or about $\infty$-categories being presentable.

The starting point is probably trying to figure out whether or not the isomorphisms of $Ho(\mathcal{M})$ form an accessible subcategory of the arrow category, which seems like something you'd need (since in order for $\mathcal{M}$ or $\mathcal{C}$ to be presentable you need to know the weak equivalences are an accessible subcategory of the arrow category, see A.2.6.6 of Lurie's HTT). It feels like the sort of thing that should be true, but I also don't know if it's enough to then deduce $\mathcal{C}$ is presentable (though Lurie's A.2.6.9 is surely relevant).

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    $\begingroup$ Sounds like you want Higher Algebra, Corollary 1.4.4.2. $\endgroup$
    – Dylan Wilson
    Commented Jan 4, 2018 at 20:48
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    $\begingroup$ @DylanWilson I guess it might be worthwhile to point out that, for a stable ∞-category C, hC has all coproducts iff C does (since if $X$ is the coproduct of $X_i$ in hC then for all $Y\in C$, $\mathrm{Map}(X,Y)\to\prod_i\mathrm{Map}(X_i,Y)$ is an equivalence on homotopy groups) $\endgroup$
    – Denis Nardin
    Commented Jan 4, 2018 at 21:13
  • $\begingroup$ @DylanWilson, thanks for the fast response! Right above the corollary, Lurie defines X to be a generator if $\pi_0 Map_C(X,Y)\simeq 0 \Rightarrow Y\simeq 0$. But for a triangulated category $T$ I thought I needed X such that the smallest localizing subcategory containing X is $T$, also called a generator. In this context, do these two notions of "generator" have to agree? $\endgroup$
    – David White
    Commented Jan 4, 2018 at 21:28
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    $\begingroup$ @DavidWhite In Krause's definition of well-generated triangulated category (which he proves to be equivalent to Neeman's) he just asks $[X,Y]=0\Rightarrow Y=0$, which is exactly Lurie's condition. $\endgroup$
    – Denis Nardin
    Commented Jan 4, 2018 at 21:38
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    $\begingroup$ @DavidWhite Sorry to resurrect this so late, but is it clear to you that having a $\kappa$-compact generator in the triangulated sense is the same as having a $\kappa$-compact generator in the $\infty$-categorical sense? Lurie goes to great lengths in 1.4.4.1 (3) that to prove that an $\aleph_0$-compact object in the homotopy category is $\aleph_0$-compact in the $\infty$-category, but doesn't do this for uncountable $\kappa$. I had assumed this was because $\kappa$-compact is something freaky and much more complicated than $\kappa$-small in triangulated land. $\endgroup$
    – Kevin Carlson
    Commented Sep 4, 2018 at 22:13

2 Answers 2

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I don't think this is actually fully answered by HA 1.4.4.2, which says that a stable $\infty$-category is presentable if and only if it's got coproducts (hence is cocomplete) and a $\kappa$-compact generator together with a locally small homotopy category (so that it's locally small.) That last condition sounds as if it would mean the homotopy category was $\kappa$-well generated, but there's no clear implication between being $\kappa$-compact in a stable $\infty$-category and being $\kappa$-compact in its underlying triangulated category. That's because being $\kappa$-compact in the homotopy category means being in the largest class of $\kappa$-small objects that is "$\kappa$-perfect" in the sense of Neeman, and it's only $\kappa$-smallness that has any clear connection to $\kappa$-filtered colimits.

However, a nearby claim is known, and was proved via a different model by Heider in his paper Two results from Morita theory of stable module categories. Heider shows that the homotopy category of a spectral model category is well generated if and only if it's triangulated equivalent to the localization of the derived category of a small spectral category at a set of objects. The derived category of a small spectral category models an arbitrary finitely presentable stable $\infty$-category, and its localizations at small sets model arbitrary accessible localizations, thus arbitrary $\alpha$-presentable stable $\infty$-categories.

Now, I'm pretty sure it's unknown precisely which stable $\infty$-categories underlie spectral model categories, so this doesn't quite answer the original question. I think it would be possible to imitate Heider's proof in general, though. Given an $\alpha$-compact generator $\mathcal G$ of the homotopy category of a stable $\infty$-category $\mathcal T$, seen as a full additive subcategory, take a small spectral category $\mathcal E$ with homotopy coherent nerve equivalent to the full subcategory of $\mathcal T$ spanned by the objects of $\mathcal G$. Then as far as I can tell, Heider's argument goes through without change to exhibit $\mathcal T$ as (the homotopy coherent nerve of) an $\alpha$-accessible localization of $D(\mathcal E^{\mathrm{op}})$.

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On any complete and cocomplete category, there is a model structure where trivial fibrations = weak equivalences = all morphisms. The homotopy category is equivalent to a single-object groupoid. So, I think that one cannot expect the converse for model categories.

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  • $\begingroup$ Thanks! Last night I typed up a proposition that contains my thinking on this question, and it reads "Suppose M is a stable model structure on a locally presentable category, not necessarily cofibrantly generated. Suppose Ho(M) is well-generated as a triangulated category. Then the $\infty$-category C associated to M is presentable." - so, I guess I was already thinking M should be a locally presentable category (but neglected to say this in the question), in which case I think your example actually would be combinatorial (since cofibrations are isomorphisms, so should be generated by a set) $\endgroup$
    – David White
    Commented Jan 5, 2018 at 18:11
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    $\begingroup$ Yes, they are and my example is then combinatorial. $\endgroup$
    – Jiří Rosický
    Commented Jan 5, 2018 at 18:58
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    $\begingroup$ Surely the correct statement is that $\mathcal{M}$ is Quillen-equivalent to a combinatorial model category (otherwise all model categories that are not combinatorial but that are equivalent to a combinatorial one are counterexamples), in which case the statement follows from the ∞-categorical version discussed in the comments. $\endgroup$
    – Denis Nardin
    Commented Jan 5, 2018 at 21:04

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