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Stream: deprecated: topos theory

Topic: 'colevel' of a topos?

Matteo Capucci (he/him) (Jun 07 2020 at 15:55):

A level $i$ of a topos $\Bbb H$ is an essential subtopos $\Bbb H' \overset{i_! \vdash i^* \vdash i_*}\hookrightarrow \Bbb H$, hence a geometric inclusion whose inverse part has a left adjoint $i_!$.

Does anyone know if the dual notion is studied? I'm talking about a geometric surjection of topoi whose inverse part has a left adjoint.

Morgan Rogers (he/him) (Jun 08 2020 at 10:24):

I wouldn't describe that as a "dual" notion, partly because surjections are very different in character than inclusions, and because toposes are definitely not well-copowered wrt these morphisms. Indeed, if $\mathbb{H}$ is a presheaf topos on a category $\mathcal{C}$, then any functor out of $\mathcal{C}$ which is essentially surjective on objects up to retracts gives an essential surjection between the corresponding presheaf toposes, which in particular means that each monoid $M$ produces a presheaf topos which is a "colevel" of $\mathrm{Set}$ in the sense you are describing.
You might want to consider a stronger concept.
Some relevant possibilities are connectedness and local connectedness (C3.3 of the Elephant) and total connectedness (half way through C3.6 of the Elephant). Johnstone doesn't spend much time on essential geometric morphisms which aren't locally connected, because they aren't sufficiently stable, but if they're what you want (ie if openness doesn't allow for enough interesting morphisms) you might want to weaken some of his definitions by replacing "locally connected" with "essential" and working out the consequences.

Jens Hemelaer (Jun 08 2020 at 10:52):

Because in topology quotient maps are always open, in topos theory open surjections are sometimes seen as "quotient maps".
But as @[Mod] Morgan Rogers says, the open surjections with fixed domain (up to equivalence) form a proper class.
For example: the geometric morphism $\mathbf{Sets} \to G\text{-}\mathbf{Sets}$ is a open surjection, for every group $G$. They are even locally connected, so the inverse image functor has a left adjoint.
Maybe hyperconnected geometric morphisms are strong enough so that the hyperconnected geometric morphisms with fixed domain (up to equivalence) form a set instead of a proper class.