It is well know that the category of locales is not a regular category, that is the pullback of a regular epimorphism is not always a regular epimorphism: for example, the classical counterexample given on the nlab for cat/top/poset also work for locales.

Now, in all the counterexamples I know of, the pullback of a regular epimorphism is still an epimorphism.

My question is hence: is every regular epimorphism of locale a stable epimorphism (i.e. a map whose pullbacks are all epimorphisms) ?

Although this might not seems really interesting it has, if it is true, several interesting consequences for the category of locales, for example it imply that every morphism of locale factor as a regular epimorphism followed by a monomorphism and that every strong epimorphism of locales is a regular epimorphism. (so counterexample to these properties would also answer my question).

**Edit : Re-Formulaton in terms of frames**

Let me first recall the tensor product of frame (or sup-lattice), which corresponds to the fiber product of locales.

let $A$ be a frame, and $B$ and $C$ two frame endowed with morphism $f,g:A \rightrightarrows B,C$. Then one defines the frame $C \otimes_A B$ as the sup-lattice freely generated by the symbole $c \otimes b$ for $c \in C$ and $b \in B$ subject to the relations:

$\bullet$ for all $b$, $c \mapsto b \otimes c$ is a sup-lattice morphism.

$\bullet$ for all $c$, $ b \mapsto b \otimes c$ is a sup-lattice morphism.

$\bullet$ $ b \wedge g(a) \otimes c = b \otimes f(a) \wedge c$.

One can for example describes it as the set of subset $X \subset B \times C$ which satisfies the following conditions: if $(b,c) \in X$ and $b' \leqslant b$, $c' \leqslant c$ then $(b',c') \in X$; $(b, g(a) \wedge c) \in X$ iff $(b \wedge f(a),c) \in X$; if for all $i$, $(b_i,c) \in X$ then $(\sup b_i,c) \in X$; the same thing with $c_i$ instead of $b_i$. In this description, the element $b\otimes c$ corresponds to the smallest such subset $X$ containing $(b,c)$.

The "natural frame morphism" from $B$ and $C$ to $B \otimes_A C$ are the map $b \mapsto b \otimes \top_C$ and $c \mapsto \top_B \otimes c$ ($\top$ denoting the maximal element of a frame)

For more details see for example Borceux's handbook of categorical algebra tome 3 section 1.4, or I think any other book talking about frames. There is an other completely different (but equivalent) description of the tensor product of frames using pre-frames instead of sup-lattices, I have no reasons to think that one of the two presentations will be more suited for this question than the other.

A morphism of frame $f:A\rightarrow B$ is a regular monomorphism if it identifies $A$ with the equalizer of some pair of frame morphisms $u,v : B \rightrightarrows C$ (i.e. the subframe of $b \in B$ such that $u(b)=v(b)$), or equivalently with the equalizer of the two maps $B \rightrightarrows B \otimes_A B$.

The question is then: if $f:A \rightarrow B$ is a regular monomorphism of frames, and $g : A \rightarrow C$ is any frame morphism, is the natural map $C \rightarrow B \otimes_A C$ a monomorphism ? (it is known that it does not have to be a regular monomorphism).

I don't know if this help, but one can restrict ourselves to the case where $C$ is a complete boolean algebra.