I am having a little trouble with Theorem 2.3 in Professor Johnstone's book on "Stone Spaces". The theorem depends on a Lemma (which I am not struggling with; I only include it for context) which refers to a situation in which we have a locale A and a nucleus $j : A \rightarrow A$, and we have defined $A_j = \text{Im}\;j = \{ a \in A \;\mid\; j(a) = a \}$, and claim the following in Lemma 2.2:
$A_j$ is a frame, and $j : A \rightarrow A_j$ is a frame homomorphism, whose right adjoint is the inclusion $A_j \rightarrow A$.
Now Professor Johnstone follows this with the Theorem of section 2.3:
For any locale $A$, there is a bijection between nuclei on $A$ and regular subobjects of $A$ in Loc (i.e. isomorphism classes of regular monomorphisms $B \rightarrow A$)
Now in the proof for this Theorem, we are told that "given a nucleus $j$, Lemma 2.2 gives us a regular monomorphism $A_j \rightarrow A$". My issue with this statement is that, although the Lemma clearly shows that we have a homomorphism $j : A_j \rightarrow A$ in Loc, and one can quickly see that it is monic (because its dual in Frm is surjective by construction and so epi) it is not clear that this is a regular monomorphism in Loc (i interpret "$j$ is a regular monomorphism" to mean that $j$ is to be the equaliser of a pair of arrows in Loc).
After a bit of pondering and a small amount of research on nlab, I guessed that the work needed to show $j$ is a regular monomorphism in Loc is to take its pushout along itself and then to prove that $j$ is the equaliser of those coprojections. I think this can be done in an elementary fashion by looking at the dual constructions in Frm (which seems to have pullbacks at least). Is that the correct / quickest approach or am I missing something obvious or more categorical in style? I'm wondering whether I should have been paying attention to something that Professor Johnstone might have written earlier in the book.
Use that $j^2=j$. The diagram below is thus a coequalizer: $$ A\underset{j}{\overset{{\rm id}}{\rightrightarrows}} A\to A_j\,.$$
Update: This answer is wrong, as commented below because $j:A\to A$ is not necessarily a frame morphism.