Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

• Forgot your password?
• Apply for membership

1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeJun 14th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItex[This question](http://mathoverflow.net/questions/99616/surjection-of-localic-infinity-toposes) 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?

   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?