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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.

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  • $\begingroup$ Excuse me for a basic question, but I prefer to think in the "original" directions of maps. So in terms of frames and the category $\textbf{Frm}$, your question would be: Given frames $X,Y$, and a regular frame monomorphism $\iota: X\to Y$, is the pushout of $\iota$ necessarily mono? $\endgroup$ – Dominic van der Zypen Apr 14 '15 at 9:04
  • $\begingroup$ Yes. and locale monomorphism are just injective map, regular mono are equalizer of a pair of map (I don't know any nice characterisation), and by pushout I mean pushout (I.e. tensor product) along any frame morphism $X \rightarrow X'$. $\endgroup$ – Simon Henry Apr 14 '15 at 9:21
  • $\begingroup$ Sorry for bothering you with stupid questions. Let $L$ be a frame, $S\subseteq L$ be a subframe and $\iota:S\to L$ be the inclusion map. So $\iota$ is regular. Q1: How do we construct the pushout object $P$ along $\iota$, or in other words, what is $P$? Q2: The "pushout morphism", what does it look like, does it go from $P$ or into $P$? Many thanks. $\endgroup$ – Dominic van der Zypen Apr 17 '15 at 9:27
  • $\begingroup$ no you construct the pushout of $\iota$ along a map from $S$ to $S'$. And it will be the map from $L$ to the tensor product $L \otimes_{S} S'$ (the tensor product is the tensor product of sup-lattice). also note that not any frame monomorphism is a regular monomorphism. $\endgroup$ – Simon Henry Apr 17 '15 at 11:35
  • $\begingroup$ What is $S'$? Can you write down $P$ in terms of $S$ and $L$, and also the pushout map? Thanks! $\endgroup$ – Dominic van der Zypen Apr 17 '15 at 14:15

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