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Is there a simple 'topological' condition to detect whenever a morphism of locales $f : X \rightarrow Y$ induces a surjection of infinity-toposes $f : \mathrm{Sh}_{\infty}(X) \rightarrow \mathrm{Sh}_{\infty}(Y)$ (i.e. such that $f^*$ is conservative)?

It's not enough to assume that f is a surjection of locales: indeed, if we take a topological space $X$ such that $\mathrm{Sh}_{\infty}(X)$ is not hypercomplete, and $X^{\mathrm{disc}}$ is its space of points endowed with the discrete topology, then $\mathrm{Sh}_{\infty}(X^{\mathrm{disc}}) \rightarrow \mathrm{Sh}_\infty (X)$ can't be a surjection, because the pullback of an $\infty$-connected map in $\mathrm{Sh}_\infty (X)$ is a weak equivalence in $\mathrm{Sh}_{\infty}(X^{\mathrm{disc}})$...

Thank you!

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Actually, the notion of "surjection" for $(\infty,1)$-toposes seems slippery! For 1-toposes, it is equivalent for $f^*$ to be (1) comonadic, (2) conservative, and (3) faithful. Of course (2)$\Leftrightarrow$(3) doesn't categorify (what does faithfulness even mean for $(\infty,1)$-categories?). But I don't see that (1)$\Leftrightarrow$(2) categorifies either, because the $(\infty,1)$-comonadicity theorem (unlike the 1-categorical one) is not about finite limits, but $f^*$ is still only assumed to preserve finite limits. –  Mike Shulman Jun 14 '12 at 21:14
as I said, by a surjection I just mean $f^*$ being conservative. But yes, maybe i shouldn't have used the term surjection ... –  Simon Henry Jun 15 '12 at 12:43
@Mike: an infinity functor is "faithful" if and only if it is 0-truncated ( (-2)-truncated=equivalence, (-1)-truncated=mono) –  David Carchedi Jun 16 '12 at 3:11
@David: I guess that's one reasonable definition. –  Mike Shulman Jun 16 '12 at 4:36
@Simon: I didn't mean to imply that you said anything wrong or should have said it differently (you did clarify exactly what you meant). I just wanted to share a thought that occurred to me after reading your interesting question! –  Mike Shulman Jun 16 '12 at 4:37

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