# 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?

## 1 Answer

$$(i) \Leftrightarrow (ii)$$ 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.