Is every frame monomorphism regular? Is every monomorphism in $\mathbf{Frm}$, the category of frames, regular?
 A: The answer is no. A category in which all monomorphisms are regular must be balanced, i.e., every map that is monic and epic is an isomorphism. (For, any equalizer map that is epic must be an isomorphism.) Intuitively, thinking of locales as "spaces", one should expect this fails badly, since there are many monic epimorphisms of spaces that are not isomorphisms (think of the identity function from a finer topology to a coarser topology). This intuitive argument requires some work though, as far as I can tell, because monomorphisms of locales (or epimorphisms of frames) are harder to get a handle on than in the case of topological spaces. In any case, the fact that the category of locales is not balanced is given in the book by Picado and Pultr, Frames and Locales: Topology Without Points (unfortunately I cannot extract a precise page number from Google books). Or, we can give a general construction as follows. 
For any frame $L$, let $C(L)$ be the lattice of congruences on $L$, i.e., the lattice of equivalence relations that are subframes of $L \times L$. Then $C(L)$ is also a frame; indeed there is an isomorphism $C(L) \to N(L)$ to the frame of nuclei, taking a congruence $\theta$ to the nucleus $a \in L \mapsto \bigvee \{x: (x, a) \in \theta\}$ (this is proved in Johnstone's Stone Spaces; I don't have a page number to hand). There is a canonical map $\nabla: L \to C(L)$ which takes each $a \in L$ to the congruence $\{(x, y) \in L \times L: x \vee a = y \vee a\}$. This is monic (if $\nabla(a) \leq \nabla(b)$, then $a \leq b$ by considering $(x, y) = (a, \top)$; therefore $\nabla(a) = \nabla(b)$ implies $a = b$). It is also epic. This is not hard to check from the following facts, after we introduce the congruence $\Delta(a) := \{(x, y) \in L \times L: x \wedge a = y \wedge a\}$ [N.B. we do not claim that $\Delta: L \to C(L)$ is a frame map]. 


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*Every $\theta \in C(L)$ is a join $\bigvee_{(x, y) \in \theta} \theta_{(x, y)}$, where $\theta_{(x, y)}$ is the smallest congruence containing the pair $(x, y)$. For any $(x, y) \in L \times L$, we have that $\theta_{(x, y)} = \theta_{(x \wedge y, x)} \vee \theta_{(x, x \vee y)}$ (in other words, $x \sim y$ iff $x \wedge y \sim x$ and $x \sim x \vee y$). 

*If $a \leq b$, then $\theta_{(a, b)} = \Delta(a) \cap \nabla(b)$ (this intersection is the meet in $C(L)$). Combined with the previous point, this means every $\theta \in C(L)$ is a join of congruences of the form $\Delta(a) \cap \nabla(b)$. 

*Each $\nabla(a)$ is complemented in $C(L)$, and its (necessarily unique) complement is $\Delta(a)$. Indeed, that $\nabla(a) \cap \Delta(a)$ is the bottom congruence (the diagonal in $L \times L$) just comes down to the fact that from $x \wedge a = y \wedge a$ and $x \vee a = y \vee a$, we can prove $x = y$; this is true in any distributive lattice. Furthermore, $\nabla(a) = \theta_{(0, a)}$ and $\Delta(a) = \theta_{(a, 1)}$, and we infer that $\nabla(a) \vee \Delta(a) = \theta_{(0, 1)}$ which is all of $L \times L$ (since $0 \sim 1$ implies $x \sim y$ for all $x, y \in L$). 

*If $f, g: C(L) \rightrightarrows A$ are two frame maps such that $f(\nabla(a)) = g(\nabla(a))$ for all $a \in L$, then also $f(\Delta(a)) = g(\Delta(a))$ for all $a \in L$ (since frame maps preserve complements). Thus $f, g$ agree on all elements of the form $\Delta(a) \wedge \nabla(b)$, and hence on all joins of such, which means $f(\theta) = g(\theta)$ for all $\theta \in C(L)$. 
The last point shows that $\nabla: L \to C(L)$ is epic. Finally, if $\nabla: L \to C(L)$ were an isomorphism, then every $a \in L$ must have already been complemented. Thus, $\nabla: L \to C(L)$ is an isomorphism only if $L$ is a complete Boolean algebra. 
