Equivalences of categories of sheaves vs categories of $\infty$-Stack

Question

Let say I have two different sites $(\mathcal{C},I)$ and $(\mathcal{D},J)$ for an ordinary topos $\mathcal{T}$. I.e.

$$Sh(\mathcal{C},I) \simeq \mathcal{T} \simeq Sh(\mathcal{D},J)$$

And we want to know if one also have an equivalence of the $\infty$-topos of sheaves of spaces on these sites:

$$ Sh_{\infty}(\mathcal{C},I) \overset{?}{\simeq} Sh_{\infty}(\mathcal{D},J)$$

This question is about finding an explicit counter example where this is not the case. But I'll give a bit more context.

If my understanding is correct, there are two cases where one can say something:

1) If $ Sh_{\infty}(\mathcal{C},I)$ and $Sh_{\infty}(\mathcal{D},J)$ are hypercomplete, or more generally if one only care about their hypercompletion. Indeed, one always has:

$$ Sh_{\infty}(\mathcal{C},I)^{\wedge} \simeq \ Sh_{\infty}(\mathcal{D},J)^{\wedge}$$

(the $^{\wedge}$ denotes hypercompletion) this is because one can show that they are both equivalent to the $\infty$-category attached to category of simplicial objects of $\mathcal{T}$ with the Jardine-Joyal model structure on simplicial sheaves. And the equivalences of this model structure only depends on the underlying ordinary topos.

2) If $\mathcal{C}$ and $\mathcal{D}$ have finite limits, then one gets the equivalence:

$$ Sh_{\infty}(\mathcal{C},I) \simeq Sh_{\infty}(\mathcal{D},J)$$

This follows from J.Lurie Lemma 6.4.5.6 in Higher topos theory. This lemma assert that under these assumptions, there is a natural equivalence between

$$ Geom( \mathcal{Y}, Sh_{\infty}(\mathcal{C},I)) \simeq Geom( \tau_0 \mathcal{Y} , Sh(\mathcal{C},I) ) $$

Where $\mathcal{Y}$ is any $\infty$-topos, $Geom$ denotes the spaces of geometric morphsisms, either of $\infty$-topos or ordinary toposes, and $\tau_0 \mathcal{Y}$ is the ordinary topos of homotopy sets of $\mathcal{Y}$. THis in particular gives a universal property to $Sh_{\infty}(\mathcal{C},I)$ which only depends on the $Sh(\mathcal{C},I)$ and so this implies the isomorphisms above.

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So what about the general case ? I have quite often heard that this was not true in general, and I'm willing to believe it. But I would really like to see a counter-example !

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In some of the places where I have seen asserted that this is not true in general, people were pointing out to the examples where $Sh_{\infty}(C,J)$ is not hypercomplete, and often these examples fall under the assumption of the second case.

In a comment on this old closely related question of mine, David Carchedi mention a counter-example due to Jacob Lurie which indeed avoids both situation... But I havn't been able to understand how it works, and it does not seem to appears in print. If someone can figure it out I'll be very interested to see the details.

Also a counter example to this lemma 6.4.5.6 mentioned above (without the assumption that the site has finite limit) would probably produce an answer immediately. Moreover (assuming such a counterexample exists) there must exists one where the topology is trivial, so I guess there has to be some relatively simple counter examples.

Answer 1

I just found an example, so I thought it would be good to post it here, but if anyone knows other examples, or a more general way to construct some I would be interested to see them as well.

This example comes relatively immediately from the example of $1$-site producing a non Hypercomplete $\infty$-topos in the paper of Dugger, Hollander and Isaksen Hypercovers and simplicial presheaves.

They are looking at the site whose underlying category is the posets $I$:

$$ \require{AMScd} \begin{CD} V_0 @<<< U^l_0 \\ @AAA @AAA \\ U^r_0 @<<< V_1 @<<< U_1^l \\ @. @AAA @AAA \\ @. U^r_1 @<<< V_2 @<<< U^l_2 \\ @. @. @AAA @AAA \\ @. @. \vdots @<<< \vdots @<<< \dots \end{CD} $$

And where for each $i$, $U_i^r, U_i^l$ forms a cover of $V_i$. It is not too hard to check that this defines a topology. They show that the $\infty$-topos of sheaves of spaces on this site is not Hypercomplete.

But if one now look at the category of sheaves of sets. Then One can apply the comparison lemma. Let $J \subset I$ be the full subcategory on the $U^{l/r}_i$, then as each $V_i$ is covered by the $U^{l/r}_i$ so by the comparison lemma, the category $Sh(I)$ and $Sh(J)$ (for the induced topology) are equivalent. But the topology induced on $J$ is trivial so one has:

$$Prsh(J) \simeq Sh(I) $$

One the other hand $Prsh_{\infty}(J)$ is obviously hyperconnected while $Sh_{\infty} (I)$ has been proved to not be hyperconnected in the paper mentioned above, so:

$$ Prsh_{\infty}(J) \not\simeq Sh_{\infty} (I) $$

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Also note that this example of Dugger-Hollander-Isaksen is closely related to J.Lurie's example involving the Hilbert cube. Indeed if $Q= [0,1]^{\mathbb{N}}$ is the Hilbert Cube, then defining:

$$ U_i^l = ]0,1[^i \times [0,1[ \times Q $$ $$ U_i^r = ]0,1[^i \times ]0,1] \times Q $$ $$ V^i = U_{i-1}^r \wedge U_{i-1}^l = ]0,1[^{i} \times Q $$

Gives a full subcategory of $\mathcal{O}(Q)$ (the category of opens of $Q$) stable under intersection and isomorphic to the $I$ above, with the induced topology on $I$ being the topology described by Dugger-Hollander-Isaksen. So one has a geometric morphisms from $Sh(Q)$ to $Sh(I)$. This somehow show how something along the line of the comment of David Carchedi linked in the question might produces an examples (though as I understand it one might needs to consider more open subset than what the comment said, but I might still be missing something)