# Toposes with only preorders of points

For a Grothendieck topos $$mathcal{E}$$, are the following assertions equivalent?

$$(i)$$ $$mathcal{E}$$ is localic.
$$(ii)$$ The diagonal geometric morphism $$mathcal{E} to mathcal{E} times mathcal{E}$$ is an embedding. (Here $$mathcal{E} times mathcal{E}$$ is the product topos, not the product category.)
$$(iii)$$ For every Grothendieck topos $$mathcal{E}’$$, $$mathrm{Geom}(mathcal{E}’, mathcal{E})$$ is a preorder (no parallel geometric transformations).

The implications $$(i) Rightarrow (ii)$$ and $$(ii) Rightarrow (iii)$$ do hold:

• $$(i) Rightarrow (ii)$$: Any diagonal morphism $$X to X times X$$ (in any category) is a split mono and a split mono of locales is an embedding. The (forgetful) functor from locales to toposes preserves the product and turns embeddings of locales into geometric embeddings.
• $$(ii) Rightarrow (iii)$$: If $$mathcal{E} to mathcal{E} times mathcal{E}$$ is an embedding, then the diagonal functor $$mathrm{Geom}(mathcal{E}’, mathcal{E}) to mathrm{Geom}(mathcal{E}’, mathcal{E} times mathcal{E}) simeq mathrm{Geom}(mathcal{E}’, mathcal{E}) times mathrm{Geom}(mathcal{E}’, mathcal{E})$$ must be fully faithful. But this means precisely that $$mathrm{Geom}(mathcal{E}’, mathcal{E})$$ is a preorder.

So in summary, is a topos with only a preorder of $$mathcal{E}’$$-based points for every $$mathcal{E}’$$ already localic?

MathOverflow Asked by Matthias Hutzler on February 1, 2021

$$(i) Leftrightarrow (ii)$$ is true and is Proposition C.2.4.14 in Peter Johnstone's Sketches of an elephant. More generally he shows that a bounded geometric morphism $$f: mathcal{E} to mathcal{S}$$ is localic if and only if $$mathcal{E} to mathcal{E} times_{mathcal{S}} mathcal{E}$$ is an embedding.

$$(ii)$$ and $$(iii)$$ are not equivalent: there is a large gap between "the diagonal is a monomorphism" and "the diagonal is an embedding"

For a typical example, take a free but non-proper action of a group $$G$$ on a locale (or space) $$X$$. To fix the idea, take $$G = mathbb{Z}$$ acting on $$X=S^1$$ the unit circle by rotation by an irrational angle.

The topos of equivariant sheaves $$X//G$$ classifies "orbits" for the action of $$G$$ on $$X$$, that is a $$G$$-torsor $$T$$ (a principale $$G$$-bundle) together with a $$G$$-equivariant map $$T to X$$. Because the action is free, the category of point in any topos will have no non-trivial morphisms.

But that topos is not localic at all: its subterminal objects are the $$G$$-invariant open subset so in our concrete example its only $$emptyset$$ and $$1$$.

One can also compute the diagonal map. $$mathcal{T} times mathcal{T}$$ can be shown to be the topos corresponding to the action of $$G times G$$ on $$X times X$$. Subtopos of this would corresponds to $$G times G$$-equivariant sublocales of $$X times X$$ and the diagonal is not $$G times G$$-equivariant.

To make more explicit construction, we can use that for a discrete group $$G$$, a topos localic over $$BG$$ (the topos of $$G$$-set) is the same as a local with a $$G$$-action. $$mathcal{T}$$ corresponds to $$X$$ with its $$G$$ action. $$mathcal{T} times mathcal{T} to BG times BG$$ is also localic (product of localic map), and the corresponding locale is obtained by pulling back along the point $$* to BG times BG$$, which allows to see that the corresponding locale is indeed $$X times X$$. Now if I see $$mathcal{T}$$ over $$BG times BG$$ as $$mathcal{T} to mathcal{T} times mathcal{T} to BG times BG$$, then it corresponds to the local $$X times G$$ where $$G times G$$ acts on $$X$$ and $$G$$ separately (to be more symetric it is the the locale of triplets $$(x,x',g)$$ where $$x'=gx$$).

As locale with $$G times G$$ action, the diagonal map of $$mathcal{T}$$ hence corresponds to the map $$X times G to X times X$$ that sends $$(g,x)$$ to $$(x,gx)$$. Which is a mono because $$G$$ acts freely, but is not en embedding.

Of course some of the claim I made above would require a proof... but that might be a bit too long for MO.

Answered by Simon Henry on February 1, 2021

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