# Is an open map with open relative diagonal necessarily a local homeomorphism?

Let $f : X \to Y$ be an open (and continuous) map of locales. Suppose the relative diagonal $\Delta_f : X \to X \times_Y X$ is an open embedding of locales. Does it follow that $f : X \to Y$ is a local homeomorphism?

The answer is yes if I replace "locale" with "topological space". Indeed, by the definition of the product topology, there must exist an open covering of the image of $\Delta_f$ by "rectangles", i.e. open subspaces of the form $U \times_Y V \subseteq X \times_Y X$ for open subspaces $U \subseteq X$, $V \subseteq X$; but if $U \times_Y V \subseteq \operatorname{im} \Delta_f$, then the restriction $f : U \cap V \to Y$ must be injective, hence is an open embedding (because $f : X \to Y$ is an open map). But $\{ U \cap V : U \times_Y V \subseteq \operatorname{im} \Delta_f \}$ is an open covering of $X$, so we are done.

In fact, everything in the above argument goes through for locales, except for the very last step where I assert that we have an open covering of $X$. There, I have used the fact that a collection of open subspaces is an open covering if and only if every point is contained in some member of the collection. So the argument would work if $X$ is a spatial locale. But can we avoid the use of points?

For reference, here are some standard definitions:

The image of a locale morphism $f : X \to Y$ is the locale $\operatorname{Im} f$ corresponding to the frame $$\Omega (\operatorname{Im} f) = \{ v \in \Omega (Y) : f_* (f^* (v)) = v \}$$ where $f^* : \Omega (Y) \to \Omega (X)$ is the frame homomorphism corresponding to $f : X \to Y$ and $f_* : \Omega (X) \to \Omega (Y)$ is the right adjoint; the "inclusion" $\operatorname{Im} f \hookrightarrow Y$ corresponds to the frame homomorphism $v \mapsto f_* (f^* (v))$.

An open sublocale of a locale $Y$ is a locale $Y_v$ that corresponds to a frame of the form $$\Omega (Y_v) = \{ v' \in \Omega (Y) : v' \le v \}$$ for some $v \in Y$; the "inclusion" $Y_v \hookrightarrow Y$ corresponds to the frame homomorphism $v' \mapsto v' \land v$.

An open embedding of locales is a morphism $f : X \to Y$ that is isomorphic to the inclusion of some open sublocale of $Y$.

An open map of locales is a morphism $f : X \to Y$ such that the image of every open sublocale of $X$ is an open sublocale of $Y$.

A local homeomorphism of locales is a morphism $f : X \to Y$ for which there is a set $\mathfrak{U} \subseteq \Omega (X)$ such that:

• $\sup \mathfrak{U} = \bigvee_{u \in \mathfrak{U}} u = \top$.
• For each $u \in \mathfrak{U}$, the composite $X_u \hookrightarrow X \to Y$ is an open embedding.
• If $f:X\to Y$ is a continous map of locales, then it is "really" a frame homomorphism $f': Y \to X$, right? And what does it mean that $f$ is open? – Dominic van der Zypen Apr 9 '15 at 12:02
• There is a definition, which can be found in e.g. Stone spaces. – Zhen Lin Apr 9 '15 at 12:18
• For those who don't have access to Stone spaces maybe you could include the definition in the post such that it is more clear. – Dominic van der Zypen Apr 9 '15 at 13:09
• Apparently the definition does not appear explicitly in Stone spaces, so I have added it. – Zhen Lin Apr 10 '15 at 0:53
• Thanks for including the definitions. I think the problem you posted is a very interesting and natural one. – Dominic van der Zypen Apr 10 '15 at 6:31

Roughly, it can be proved by working in the internal logic of the target (hence assuming that the target is a point) and considering open subspaces $U$ of $X$ such that $U \times U \subset \Delta$, on can then show that such $U \times U$ cover the diagonal, hence the $U$ cover $X$ and that if they are positive they correspond to isolated points of $X$.
Edit : I didn't read your post well enough and didn't see you already proposed a proof. The trick you are missing is I think the following: If the $U \times V$ form a covering of $\Delta$ then the $\Delta^{-1}(U \times V)$ form a covering of $X$, and $\Delta^{-1}(U \times V) = U \cap V$.