2
$\begingroup$

I am reading the paper "A universal characterization of higher algebraic K-theory" by Blumberg, Gepner, and Tabuada, and I am stuck on Corollary 4.25:

…the fact that we have accessible localizations provides the following corollary about the structure of $\mathrm{Cat}_\infty^\mathrm{ex}$ and $\mathrm{Cat}_\infty^\mathrm{perf}$.

Corollary 4.25. The ∞-categories $\mathrm{Cat}_\infty^\mathrm{ex}$ and $\mathrm{Cat}_\infty^\mathrm{perf}$ are compactly generated, complete, and cocomplete.

Here, $\mathrm{Cat}_\infty^\mathrm{ex}$ (resp. $\mathrm{Cat}_\infty^\mathrm{perf}$) is the ∞-category of small stable ∞-categories (resp. of small idempotent-complete stable ∞-categories).

The "accessible localizations" they are referring to are Theorem 4.22 and Theorem 4.23, which state that $\mathrm{Cat}_\infty^\mathrm{ex}$ and $\mathrm{Cat}_\infty^\mathrm{perf}$ are accessible localizations of the ∞-category of small spectral categories.

My thought is as follows: Dugger's theorem shows that the ∞-category of small spectral categories can be modelled by a combinatorial simplicial model category, so it is presentable. Therefore $\mathrm{Cat}_\infty^\mathrm{ex}$ and $\mathrm{Cat}_\infty^\mathrm{perf}$ are presentable. However, I cannot see how to prove that they are $\omega$-accessible, i.e. compactly generated. Can anyone help me fill in this gap?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
5
$\begingroup$

Proposition 4.20 in the same paper shows that the localization is in fact $\omega$-accessible, i.e. the fully faithful right adjoint in question preserves filtered colimits. This implies that the localized category is $\omega$-accessible, in particular compactly generated (because the category you're localizing from is compactly generated)

Alternatively, you can observe that the forgetful functor $Cat^{ex}_\infty\to Cat_\infty$ preserves filtered colimits and arbitrary limits, and that therefore its left adjoint preserves compacts. Because this forgetful functor is furthermore conservative, the claim also follows.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you for your answer. This may be a dumb question, but can you tell me why the ∞-category of spectral categories is compactly generated? $\endgroup$
    – Jun Koizumi
    Commented May 13, 2022 at 16:12
  • $\begingroup$ This is not a dumb question at all, and in fact not having to think about this is the reason I put the "alternatively" part of my answer :) I think to prove this you need to go deeper into the references that show that it is a combinatorial model category. Specifically, if you look at Corollary 2.4 in that paper, it gives an explicit set of generators, which I suppose you can check are compact (although you have to prove that the Bousfield localization that's involved is also compact-preserving, but this should follow from the weak equivalences being stable under filtered colimits or something $\endgroup$
    – Maxime Ramzi
    Commented May 13, 2022 at 16:24
  • $\begingroup$ The bottom line is: I'm not sure, and this is why I put the second half of my answer there $\endgroup$
    – Maxime Ramzi
    Commented May 13, 2022 at 16:25
  • $\begingroup$ Thank you so much! I never imagined there would be so much between the lines on that paper, but I will attempt to prove it in the way you suggest. The alternative argument was also very helpful. $\endgroup$
    – Jun Koizumi
    Commented May 13, 2022 at 16:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .