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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

1 Answers

One 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.

Answered by Simon Henry on February 1, 2021

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