Toposes with only preorders of points

Question

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?

Answer 1

$(i) \Leftrightarrow (ii)$ 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.