8
$\begingroup$

Reading Scholze's notes on Condensed Mathematics it is mentioned that when considered as $\infty$-categories,

$$ D(\operatorname{Cond(Ab)}) \cong \operatorname{Cond}(D(\operatorname{Ab}))$$

and that this is not a feature of condensed abelian groups but rather of the category of sheaves on a site.

I have been looking for references on this fact, but haven't been able to find anything. This thread seems to suggest that we have some additional conditions on our site. Any references to read further on this and see why this is true for this example would be greatly appreciated.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ This holds when the site is hypercomplete (or, just every sheaf of HZ-modules is hypercomplete), but not in general. It looks like this is all explained in Marc's answer to your linked question; could you clarify what more you're looking for? $\endgroup$
    – Dustin Clausen
    Commented Apr 9, 2021 at 18:31
  • 2
    $\begingroup$ I guess another, more positive, way of saying things is that this is always true if you replace "sheaves with values in D(Ab)" by "hypersheaves with values in D(Ab)". The basic point is that it's only in the hypersheaf context that you can check isomorphisms on homology groups sheaves, which matches up with the equivalences you see in the derived category of sheaves. $\endgroup$
    – Dustin Clausen
    Commented Apr 9, 2021 at 18:33
  • 2
    $\begingroup$ As to why it holds in this example, the point is the simple structure of the site of extremally disconnected compact Hausdorff spaces: the finite disjoint union topology. This means that evaluation on extr. disconnecteds detects isomoprhism and is t-exact. Thus homology commutes with evaluation on extr. disconnecteds which implies hypercompleteness $\endgroup$
    – Dustin Clausen
    Commented Apr 9, 2021 at 18:39
  • $\begingroup$ Sorry if this all just sounds like a technical jumble. Let me know what further explanations you want. $\endgroup$
    – Dustin Clausen
    Commented Apr 9, 2021 at 18:40
  • 1
    $\begingroup$ @DustinClausen thanks a lot! I probably just need to make myself more familiar with Lurie's work to truly appreciate this... $\endgroup$
    – Sofía Marlasca Aparicio
    Commented Apr 30, 2021 at 20:24

1 Answer 1

Reset to default
6
$\begingroup$

It's true in any $1$-topos for hypercomplete sheaves, see Theorem 2.1.2.2 in Spectral Algebraic Geometry.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

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