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NNN
NNN
3:17 PM
Why is the definition of flat $(\infty, 1)$-functor in nlab (=left exact in HTT, meaning that the pullback along $f: A\to B$ preserves left fibration with cofiltered total space) correct? I'm interested in the case $B=\Spaces$, and I believe that the general case is defined representably using this. So I mean why is it better than the competing weaker (HTT 5.3.2.11) property that the 'category of elements' = unstraightening
of the functor $f: A \to \Spaces$ is cofiltered? Which one is equivalent to being a filtered colimit of representable functors, or its Yoneda extension being left exact?
I thought that the proof of equivalence "category of elements is cofiltered <=>filtered colimit of representable functors<=>yoneda extension is left exact" for 1-categories works equally well for infinity-categories, but I feel I need to be careful around pointwise Kan extension, so my guess is that there I need to use all comma categories instead of just category of elements, and this may result in requiring the stronger definition, though I'm having trouble making this precise.
 
 
2 hours later…
Adrian Clough
Adrian Clough
5:49 PM
It is a classical theorem in topos theory that for any ordinary topos $E$, if $A$ is a small subcategory, such that for any object $X \in E$, there exists an effective epimorphism $\coprod U \to X$, where the coproduct consists of objects in $A$, then $A$ is dense in $E$. Is the same true for $\infty$-toposes? Note that I'm not asking for every $\infty$-topos to be a category of sheaves.
 
Adrian Clough
Adrian Clough
6:20 PM
Ok, I just discovered that my question may be answered in the affirmative in the case when $E$ is hypercomplete, as shown in SAG Proposition 20.4.5.1.
 
S. carmeli
S. carmeli
6:36 PM
@AdrianClough note that hypercompleteness is necessary here. For example, if X is a topological space of infinite homotopy dimension and B is a basis for the topology, if B is not closed under intersections then restriction to B of sheaves might fail to be conservative. The point is exactly that you can find a hypercover of every open sets by members of B but of course not a cover.
 

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