I believe that if I have a diagram X : P --> Top of spaces indexed by a poset P and it has the property that for every p in P the map colim_{q<p} X(q) --> X(p) is a cofibration, then the ordinary colimit of X is also a homotopy colimit. For finite P, this is exactly the condition that the obvious Reedy structure on P give you, but I believe it's also true for non-Reedy posets P. Is that right? Does anyone know a reference for this fact (or worse, a counterexample :D)?

@OmarAntolín-Camarena so take an arbitrary sequence of spaces Y_0 -> Y_1 -> Y_2 -> .... define your poset P to be ℕ x ℤ with lexicographic order, and define X(n,m) = Y_n.

then for any (n,m), the set of elements less than (n,m) has a final object: (n,m-1), and so the map colim X(p,q) -> X(n,m) is identified with the isomorphism X(n,m-1) -> X(n,m); that's a cofibration.

however, the colimit over the whole diagram is colim(Y_n). this can't be the hocolim (this construction doesn't even preserve levelwise homotopy equivalences)

if it's, say, a well-order, then you're probably OK because you can show the diagram is cofibrant in the projective model structure

the argument I'm thinking of for that has to induct on "extending lifting, one object of P at a time" and it needs a method for choosing a "next" object of P to do. so maybe you want the poset version of a well-order -- every subset of P has a minimal element?

@TylerLawson Oh my! Thanks for the counterexample Tyler!

@TylerLawson If you define the "poset version of a well order" to mean P comes with a map to an ordinal that preserves strict <, then that map gives P a Reedy category structure with all maps increasing degree and the condition I gave for X is being cofibrant in the Reedy model structure on diagrams.

@OmarAntolín-Camarena Oh, shoot. I was thinking natural-numbers graded Reedy categories. But I think that the condition I wrote can be converted into a map to an ordinal, which means no extra generality from what you already knew. Sorry

It is not the case that a filtered colimit of spaces is a homotopy colimit?

(spaces = simplicial sets, for definiteness)

@TimCampion Thanks, I'll take a look at that. :)

What's an explicit example of an $\infty$-topos which is not equivalent to an $\infty$-category of sheaves of $\infty$-groupoids on some Grothendieck site (i.e., on some small $\infty$-category equipped with a Grothendieck topology)?

Hypersheaves on some non-hypercomplete site?

Like Fin_{BZ/2}, to be super explicit.

@CharlesRezk There's a theorem due to someone named Rezk I believe that the category of $n$-excisive functors $Top \to Top$ is an $\infty$-topos which is not hypercomplete (and so not a category of sheaves with respect to a Grothendieck topology) :)

Categories of sheaves with respect to a Grothendieck topology can fail to be hypercomplete.

Wait really?

Yes

Hypersheaves on finite sets with a Z/2-action (and coverings the surjective families) gives you spaces with a Z/2-action. I suspect this is not the same as sheaves on any site (its certainly not sheaves on finite Z/2-sets), but I don't know offhand how to prove that.

By "sheaf", I mean "has descent", not "has hyperdescent".

Every $\infty$-topos is a left exact localization of $Sh(C,T)$ for some Grothendieck site $(C,T)$, in particular via left exact localizations which kill some class of $\infty$-connected maps.

If $Sh(C,T)$ was always hypercomplete, then every $\infty$-topos would be hypercomplete, but we know that's not true.

@TomBachmann I would really like to see a proof of such a statement.

Me too :).

Wait -- doesn't Lurie at least define a Grothendieck topology to be a localization at a collection of monomorphisms?

@TomBachmann Simplicial sets are fine

And isn't a monomorphism which is $\infty$-connecctive an equivalence?

@TylerLawson Whew. But topological spaces are not?

Nope. E.g. if Y_n is the quotient of [0,1] by {0} union [1/n,1], with map the induced projection

Today's entry in "everything that's wrong with toplogical spaces"...

Thanks!

@TimCampion He shows that "topological localizations" of presheaves on $C$ correspond to Grothendieck topologies, and "topological localizations" are the left exact localizations generated by monomorphisms.

Right. So in his terminology at any rate, it seems that localizing at a collection of $\infty$-connective morphisms doesn't qualify as localization a "Grothendieck topology"

No. A Grothendieck topology is (basically) the classical notion: a collection of sieves in $C$ satisfying some properties.

That's why I'm confused by your claim that localizing at some $\infty$-connected maps is an example of a localization at a Grothendieck topology

I never said it was.

I said any $\infty$-topos $X$ is $Sh(C,T)_S$, where $Sh(C,T)$ is a localization of $PSh(C)$ (via a Grothendieck topology), and $X$ is a further localization by some class $S$ of $\infty$-connected maps.

Ah, ok

In the paragraph after [HTT 6.1.0.7], he says "We will introduce a theory of Grothendieck topologies on $\infty$-categories in 6.2.2 and show that every Grothendieck topology determines a left exact localization of $P(C)$. However, it turns out that not every $\infty$-topos arises via this construction."

However, I don't see any proof of the claim of the second sentence. What he does prove is that not every left exact localization of $PSh(C)$ comes from a Grothendieck topology, but that's not exactly the same thing.

So until someone tells me otherwise, I have to admit to the possibility that every $\infty$-topos might be equivalent to some $Sh(C,T)$.

Then $Sh(X)$ is not hypercomplete. (I learned this example from the paper of Dugger-Hollander-Isaksen).

Let $\mathcal{U}\subset Open_X$ be the subset of the open set lattice spanned by the $U_n^\pm$. Then $Psh(\mathcal{U})$ is a hypercomplete $\infty$-topos. I think it is the case that $Psh(\mathcal{U})$ is the hypercompletion of $Sh(X)$. So both $Sh(X)$ and its hypercompletion are sheaves on Grothendieck sites.

The reason that $Sh(X)^{hyp}=Psh(\mathcal{U})$: for $F\in PSh(Open_X)$ we have that $F_{x_n^\pm}=F(U^\pm_n)$, i.e., the stalk of $F$ at a point $x_n^\pm$ is the same as its evaluation at an open set $U^\pm_n$.