This shows that Shv(C)^{hyp} cannot be recovered as sheaves on itself.

@MarcHoyois Ah, I see. This is similar to Denis-Charles' answer to Charles' related MO question. Why do we need the representables to be hypercomplete? Upon reflection, it seems to me that if $i^\ast: E \to LE : i_\ast$ is any cotopological localization, then $i_\ast$ preserves effective epimorphisms, so as you say, any object $X$ of $E$ not in $LE$ yields a functor $Hom(i_\ast -, X): LE^{op} \to Top$

which satisfies descent with respect to all effective epimorphsims but isn't representable.

@TimCampion Representables being hypercomplete ensures that Map(-,F) ≠ Map(-,G) on Shv(C)^{hyp} if F ≠ G on C.

I don't really understand Denis-Charles' answer: I'm not sure why we can assume all these things about the site of definition...

In fact I think Charles' example above is a couter-example to Denis-Charles' claim.

@MarcHoyois Why is $F$ hypercomplete iff $Map(-,F)$ takes po to pb?

If $\mathcal{X}$ is $1$-excisive functors from finite pointed spaces, then $\mathcal{X}^{\mathrm{hyp}}\approx \mathcal{S}$, which is certainly sheaves on a site.

@CharlesRezk The presheaf Map(-,F) on Shv(C)^{hyp} preserves products, so if it preserves pullbacks then it is representable by some G \in Shv(C)^{hyp}. This implies F=G (using the assumption that representables are hypercomplete).

Oh, $\mathcal{X}^{\mathrm{hyp}}\to \mathcal{X}$ preserves coproducts.

So I guess it's possible that you can have $PSh(C)\to Sh(C,T) \to Sh(C,T)^{\mathrm{hyp}}$, so that the first functor is a topological localization, the second functor is a "cotopological" localization, and yet the composite functor is a topological localization.

Let me complete my thoughts about the example chat.stackexchange.com/transcript/message/52487042#52487042 above. As this is sheaves on a topologicial space, it has "enough" points, so $F\to F'$ is $\infty$-connected iff the induced maps on all stalks are equivalences. But $U_n^\pm$ is the smallest open set containing $x_n^\pm$, so there is a canonical equivalence $F(U_n^\pm) \approx F_{x_n^\pm}$.

Let $\mathcal{O}$ be the open set lattice of $X$. Then we have a lex left-adjoint localization $a\colon PSh(\mathcal{O})\to Sh(X)$. Let $\mathcal{U}\subset \mathcal{O}$ be the full subcategory spanned by the $U_n^\pm$. Since this is a full subcategory, we have a lex (in fact, limit preserving) left-adjoint localization $r\colon PSh(\mathcal{O})\to PSh(\mathcal{U})$, given by restriction.

Stalks make sense for presheaves on $X$, and the above argument shows $(rF)(U_n^\pm)=F_{x_n^\pm}$. Since any map of presheaves which $a$ sends to an equivalence must induce an equivalence on stalks, there is a unique $f\colon Sh(X)\to PSh(\mathcal{U})$ so that $fa=r$. Formally, $f$ is also a left-adjoint localization, and by the above argument the class of maps it inverts is precisely the $\infty$-connected maps of $Sh(X)$.

So $f\colon Sh(X)\to PSh(\mathcal{U})$ is hypercompletion. But $Sh(X)$ is known not to be hypercomplete: $F$ defined by $F(V_n):=S^{n-1}$, $F(U_n^\pm)=D^n_{\pm}$ is a presheaf whose sheafification has $F(V_n)\approx \Omega^\infty\Sigma^\infty S^{n-1}$ for $n\geq 1$, but has contractible stalks.

Because there's a "formula" for sheafification: $aF$ is a countable direct limit of $F\to F'\to F''\to\cdots$, where $F'$ has $F'(U_n^\pm)\approx F(U_n^\pm)$, $F'(V_n)\approx F(U_n^+)\times_{F(V_{n+1})} F(U_n^-)$.

The composite $r=fa\colon PSh(\mathcal{O})\to PSh(\mathcal{U})$ is necessarily a topological localization, since presheaf categories are hypercomplete anyway.

@CharlesRezk I don't think that's possible. If L and L' are two consecutive left exact localizations and L' ∘ L is topological, then L' is topological, since the L'-equivalences are L of the (L' ∘ L)-equivalences and L preserves monomorphisms.

On the other hand I agree with your arguments above...

Actually I don't see why a left exact localization between ∞-categories of presheaves has to be topological.

@MarcHoyois I think, as Charles noted, parametrized spectra are a counterexample to that assertion

@DenisNardin I'm not quite following. My current beliefs are that (1) Denis-Charles' general claim that Shv(C)^{hyp} is not a sheaf ∞-topos (under some assumptions on C) is wrong because Charles' example just above contradicts it, and (2) left exact localizations between presheaf ∞-categories need not be topological.

and also (3) under Denis-Charles' assumptions, Shv(C)^{hyp} is not sheaves on itself with respect to the canonical topology.

@MarcHoyois Actually, isn't there a non-topological left exact localization between presheaves on {finite spaces}^{op} to spaces?

That can be factored as the topological localization to the universal topos with an hypercomplete object followed by hypercompletion

Yes, by "need not be" I meant there are such examples.

Ah ok... Then I think we are in complete agreement :). Sorry for the noise

Ok, maybe my argument that $r$ is a topological localization is wrong.

Do we know that if an ∞-topos is sheaves on something, then it is sheaves on something with a subcanonical topology?

"Sheaves on something" meaning a Grothendieck site?

Yes.

That would be useful to try to prove that something is not sheaves on a site.

Yes.

I guess the standard argument just shows that any $\infty$-topos is a "subcanonical left-exact localization": i.e., that there exists $C\to PSh(C) \twoheadrightarrow \mathcal{X}$ so that the composite is fully faithful.

Let $\mathcal{X}$ be an $\infty$-topos. Then there's a category whose objects are $f_*\colon \mathcal{X}\rightarrowtail PSh(C)$, where $C$ is a small $\infty$-category, and $f_*$ is a fully faithful right-adjoint of a geometric morphism. We can let the morphisms be compatible fully faithful geometric morphisms of the presheaf categories. (Are these called "embeddings"?)

What can we say about this category, even for "easy" examples like $\mathcal{X}=\mathcal{S}$?

Is there a place where the theory of classifying objects $\mathbb{B}\mathcal{G}$ of an internal groupoid $\mathcal{G}$ in an $\infty$-topos $\mathcal{X}$ is worked out?

I would think the descent property (in particular, the effectivity of groupoids) takes care of that, no?

I guess I'm asking: what is the "universal bundle" over $\mathbb{B}\mathcal{G}$, and what is the class of objects ("bundles") for which it is universal? In particular, what type of object is it? I guess it should be some kind of functor $\mathbb{E}\colon \mathcal{G}\to \mathcal{X}_{/\mathbb{B}\mathcal{G}}$, but which is in some sense "internal to $\mathcal{X}$".

Maybe it's obvious, but I'm not sure how you say these things.

If $\mathcal{X}=\mathcal{S}$, then you just get the "usual" theory of principal bundles (for $\infty$-groupoids $\mathcal{G}$), as an application of descent. Which you can then extend to a theory of $\mathcal{G}$-bundles in any $\infty$-topos, but $\mathcal{G}$ is still just a groupoid object in $\mathcal{S}$.

Consider the usual proof that in ordinary topos theory, sheaves on a topos in the canonical topology are all representable. I believe the proof generalizes to show that if $\mathcal X = Sh(C,J)$ for some subcanonical site $(C,J)$, then $\mathcal X$ coincides with sheaves on itself in the canonical topology.

So this means that Marc's argument (that $\mathcal X$ doesn't coincide with canonical sheaves on itself under certain conditions) actually shows that (under the same conditions) $\mathcal X$ is not sheaves on any subcanonical site.

@MarcHoyois I'm referring to Marc's argument back here.

The main thing you need to know is that if $F \in Sh(C,J)$, then $F$ admits a canonical covering sieve of representables, which is true.

@TimCampion I don't see that. The usual proof uses the comparison lemma which we don't have. What it shows is that hypercomplete sheaves on X is X^{hyp}, I think.

Also in Charles' example, X^{hyp} is certainly sheaves on a subcanonical site.

I'm just thinking through the proof on the nlab. There is a Yoneda embedding $Sh(C,J) \to Sh(Sh(C,J),can)$, which is right adjoint to the restriction along Yoneda $Sh(Sh(C,J),can) \to Sh(C,J)$. This exhibits $Sh(C,J)$ as a reflective subcategory of $Sh(Sh(C,J),can)$, so you just have to show that restriction along Yoneda is conservative, i.e. that you can recover a sheaf $F: Sh(C,J)^{op} \to Top$ from its restriction to representables.

This follows because any $G \in Sh(C,J)$ is a colimit of the canonical diagram of representables. The corresponding sieve of all maps into $G$ which factor through a representable is a cover of $G$ in the canonical topology, so the value of $F(G)$ is determined by the restriction of $F$ to the representables.

I'll have to try to get my head around Charles' example though.

@CharlesRezk Descent tells you BG classifies cartesian transformations to G, which might be a reasonable way to define an internal functor G → X.

@MarcHoyois Doesn't Charles' example fail one of your hypotheses? The representables on $Open_X$ aren't all hypercomplete.

They are (-1)-truncated.

@TimCampion I don't see why F(G) is determined by the restriction of F to C. Your sieve also contains non-representables.

Hm. I suppose you're right on all counts, of course.

Is there a good example of a compact Hausdorff space X, and a filtered diagram of compact Hausdorff spaces {Y_\alpha}, such that Maps(X, ) doesn't preserve the filtered colimit of the diagram? I'm thinking about the claim in Example 2.1.6 of arxiv.org/pdf/1904.09966.pdf which says that the Yoneda embedding from compactly generated spaces to pyknotic sets doesn't preserve colimits.

There's an example like this on the nLab page for "compact object", but X is the two-point indiscrete space (so not Hausdorff).