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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeJun 14th 2012

This question on MO made me realize that I don’t even know the right definition of surjective geometric morphism for $(\infty,1)$-toposes. For 1-toposes, it is equivalent for $f^*$ to be (1) comonadic, (2) conservative, and (3) faithful. Of course (2)⇔(3) doesn’t categorify (what does faithfulness even mean for (∞,1)-categories?). But I don’t see that (1)⇔(2) categorifies either, because the (∞,1)-comonadicity theorem (unlike the 1-categorical one) is not about finite limits, but $f^*$ is still only assumed to preserve finite limits.

Of course we want a (surjection, inclusion) factorization. For that, it seems almost certain that $f^*$ being conservative is the right definition, at least if we take the obvious definition of “inclusion” as $f_*$ being fully faithful. We can then produce the desired factorization of $f: \mathcal{F} \to \mathcal{E}$ by localizing $\mathcal{E}$ at the class of morphisms inverted by $f^*$.

But what happens if instead we construct the category of coalgebras of the comonad $f^* f_*$? Is the category of coalgebras of an (accessible) lex comonad on an $(\infty,1)$-topos again an $(\infty,1)$-topos? If so, what sort of factorization does this produce?