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I am interested in what the regular monomorphisms are in the category of locally compact (for me, always Hausdorff) groups (with continuous group homomorphisms).

It is easy to see that the equaliser (I am unsure how to typeset a fork in MathJax) of $f:G\rightarrow H$ and $g:G\rightarrow H$ is the closure of $\{ s\in G : f(s)=g(s) \}$ which is a closed subgroup $G'$, together with the inclusion $G'\rightarrow G$.

However, it is not true that every closed subgroup inclusion arises in this way; I know of a counter-example given by G.A. Reid, Proposition 8. As this is behind a paywall, let me say that he shows that if $G=SL(2,\mathbb R)$ and $G'$ is those matrices $\begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}$ with non-zero $a$, then the inclusion $G'\rightarrow G$ is an epimorphism. That is, if $f,g:G\rightarrow H$ agree on $G'$ they are equal, and so $G'$ does not arise as an equaliser.

Is it known what are the regular monomorphisms? (Perhaps this is hopeless?)

The coequaliser of $f:G\rightarrow H$ and $g:G\rightarrow H$ is the closure of the normal subgroup generated by $\{ f(s)^{-1} g(s) : s\in G\}$, say $N$ a closed normal subgroup of $H$. Then $H/N$ is a locally compact group, and with the quotient map gives the coequaliser.

Notice that every $N$ arises in this way. Indeed, given $N$ a closed normal subgroup of $H$ let $G=N, f:N\rightarrow H$ be $f(n)=1$ for all $n$, and $g:N\rightarrow H$ be the inclusion. Then the generated normal subgroup is $N$ itself. So we have proved:

Claim: The regular epimorphisms are the "quotient maps" $f:G\rightarrow H$, that is, surjective continuous group homomorphisms where the topology on $H$ is the quotient topology.

Is this correct? My doubt here is that this still works without topology, and is elementary, while here it's stated as a corollary of the result that (for groups without topology) subgroups inclusions are equalisers.

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    $\begingroup$ I also doubt of a complete description of epimorphisms in the category of locally compact groups. In a connected semisimple Lie group $G$, I think the inclusion of every Zariski-dense closed subgroup is an epimorphism of locally compact groups. $\endgroup$
    – YCor
    Commented Oct 28, 2019 at 12:37
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    $\begingroup$ In terms of diagrams, do you want just $\rightrightarrows$ $\rightrightarrows$? MathJax also supports AMScd (math.meta.stackexchange.com/questions/2324/…). $\endgroup$
    – LSpice
    Commented Oct 28, 2019 at 12:48
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    $\begingroup$ @LSpice: Can I put arrows above and below $\underset{h}{\overset{g}{\rightrightarrows}}$ Ah, yes, that works. Thanks! $\endgroup$
    – Matthew Daws
    Commented Oct 28, 2019 at 14:35
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    $\begingroup$ Is the category of locally compact groups cocomplete (I'm not sure, but my casual understanding of Gelfand duality suggests that it is)? If so, then note that in a finitely complete and cocomplete category, a monomorphism $i: G' \to G$ is regular iff it is the equalizer of the two maps $G' \rightrightarrows G \ast_{G'} G$, where $G \ast_{G'} G$ denotes the pushout of $i: G' \to G$ along itself (the "cokernel pair" of $i: G' \to G$). So if one can understand such pushouts, one will have a criterion. $\endgroup$
    – Tim Campion
    Commented Oct 28, 2019 at 18:01
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    $\begingroup$ @TimCampion One needs to be careful waving the phrase "Gelfand duality" around when dealing with locally compact groups and continuous homomorphisms between them, especially because many of the interesting homomorphisms are not proper maps. $\endgroup$
    – Yemon Choi
    Commented Oct 29, 2019 at 15:18

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