6
$\begingroup$

A (1,1)-topos (i.e. an ordinary Grothendieck topos) is called localic if the following two equivalent conditions hold:

  • It is the category of sheaves on a (0,1)-site with finite limits$^*$ (i.e. a meet-semilattice with a Grothendieck topology).
  • It is a left exact localization of the category of presheaves on a (0,1)-category with finite limits (= a meet-semilattice).

The category of localic topoi is reflective in the category of all topoi, and equivalent to the category of locales.

For $0\le m\le n \le \infty$, Higher topos theory 6.4.5.7-8 shows that an $(n,1)$-topos is $m$-localic iff the analogue of the first condition holds: it is the category of $(n-1)$-sheaves on an $(m,1)$-site with finite limits. In 6.4.5.9 it is remarked that this is equivalent to it being a topological localization of the category of $\infty$-presheaves on an $(m,1)$-category with finite limits. The category of $m$-localic $(n,1)$-topoi is reflective in all $(n,1)$-topoi, and equivalent to the category of $(m,1)$-topoi.

When $n<\infty$, every localization of an $(n,1)$-topos is topological, so the two conditions are still equivalent, but this fails when $n=\infty$. So what can be said about the larger class of $(\infty,1)$-topoi that are merely a left exact localization of $\infty$-presheaves on some $(m,1)$-category with finite limits? I believe this is equivalently the class of cotopological localizations of $m$-localic $(\infty,1)$-topoi, or the class of $(\infty,1)$-topoi that have a generating set consisting of $(m-1)$-truncated objects. Does it have a name? Has it been studied?

$^*$ Edit: In the original version of this question I omitted the assumption of finite limits. This is irrelevant when $n<\infty$ but makes a difference when $n=\infty$; see the comments. I do want to ask about the case where finite limits are assumed, so I edited the question.

Improve this question
$\endgroup$
14
  • $\begingroup$ Is it really true that presheaves on a $n$-category form an $n$-localic $\infty$-topos? The statement in HTT assumes that the $n$-category has finite limits. $\endgroup$
    – Marc Hoyois
    Commented Jan 22, 2023 at 8:56
  • $\begingroup$ @MarcHoyois Good question. It's true when $n=0$, right? I would be surprised if it weren't true; I assumed the assumption of finite limits was just to avoid talking about flat functors. $\endgroup$
    – Mike Shulman
    Commented Jan 22, 2023 at 17:48
  • 1
    $\begingroup$ Since $(n-1)$-truncated objects are stable under finite limits, if we restrict ourselves to hypercomplete $(\infty,1)$-topoi, it looks like being $n$-localic really means being an hypercomplete left exact localization of a category of presheaves on an $(n,1)$-category. In particular, presheaves of $\infty$-groupoids on a $(n,1)$-category form a hypercomplete $(\infty,1)$-topos generated by its $(n-1)$-truncated objects and thus define a $n$-localic topos. $\endgroup$
    – D.-C. Cisinski
    Commented Jan 23, 2023 at 10:25
  • 1
    $\begingroup$ It seems that it fails even when $n=0$: there is an example of Charles Rezk of a topological space whose hypercomplete sheaves are presheaves on a poset, so that the latter cannot be $n$-localic for any finite $n$. So it seems "flatness" for functors to $\infty$-topoi is strictly stronger than for $n$-topoi, which is similar to what happens with "dense" subsites. $\endgroup$
    – Marc Hoyois
    Commented Jan 23, 2023 at 16:46
  • 2
    $\begingroup$ If $Shv(X)^{hyp}$ is $n$-localic then it must agree with $Shv(X)$ since they have the same $(n-1)$-truncated objects and the latter is $0$-localic. But in Rezk's example $Shv(X)$ is not hypercomplete. $\endgroup$
    – Marc Hoyois
    Commented Jan 24, 2023 at 19:02

0

Reset to default

You must log in to answer this question.