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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 (,1)(\infty,1)-toposes. For 1-toposes, it is equivalent for f *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 *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 *f^* being conservative is the right definition, at least if we take the obvious definition of “inclusion” as f *f_* being fully faithful. We can then produce the desired factorization of f:f: \mathcal{F} \to \mathcal{E} by localizing \mathcal{E} at the class of morphisms inverted by f *f^*.

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