I am crossposting this question from math stack exchange (link below) as it has not received any comments or answers over the past month. A regular epimorphism is a morphism $f: X \to Y$ that is the coequalizer of some parallel pair $a,b: Z \to X$. I believe regular epimorphisms in the category of topological groups are precisely the surjective open maps. Is this also true for the category of topological rings? If not, is there some other characterization of the regular epimorphisms in the category $\mathbf{TopRing}$? A reference would be very much appreciated.
1 Answer
It is true and hods more generlaly for models of any algebraic theory with a Mal'tsev operation, i.e. a trinary operation $f$ such that $f(x,x,y)=y=f(y,z,z)$. For groups, you can take $f(x,y,z)=xy^{1}z$. This is Theorem 10 in A.I. Mal'tsev, On the general theory of algebraic systems, Mat. Sb. (N.S.), 1954, Volume (35)(77), Number 1, 320.
A proof in English can be found on page 396 of V.V. Uspesnki$\check{\text i}$. The Ma'tsev operation on countably compact spaces. Comment. mat. Univ. Carolinae30 (1989) 395402. available at https://dml.cz/bitstream/handle/10338.dmlcz/106758/CommentatMathUnivCarol_03019892_22.pdf
The idea is to first simplify the situation of a regular epimorphism $X\to Y$ by using the fact it is the coequalizer of its kernel pair, hence the quotient of an equivalence relation on $X$, whence we want to show that the set of equivalence classes intersecting an open subset $U$ of $X$ is open, i.e. that any $y\in X$ equivalent to $x\in U$ has an open neighborhood of points $z$ equivalent to points in $U$.
But if this relation is kernel pair in the category of models of a theory, then whatever operations the theory has, the equivalence relation has to respect. In particular, $x\sim y$ implies $f(x,y,z)\sim f(x,y,z)\sim f(x,x,z)=z$. Then because $f$ is continuous, there is an open that contains $z$ if and only if $f(x,y,z)\in U$, in which case $z\sim f(x,y,z)\in U$ for any $z$ in that open. Moreover, $x=f(x,y,y)$ implies the open contains $y$ and so is the desired neighborhood.

$\begingroup$ Is there a similar result for proper surjections instead of open ones? $\endgroup$ Feb 22, 2022 at 17:56

$\begingroup$ @PaulTaylor $g(f(a,b,c))=f(g(a),g(b),g(c))=\begin{cases}g(a)&g(b)=g(c)\\g(c)&g(a)=g(b)\end{cases}$ so you have continuous maps from each to every other nonempty fiber. I think more would be necessary (a kind of semiassociativity) for these to be homeomorphisms, which would allow you to require compactness of only one fiber (like in groups). Still more I think would be required to make the quotient map closed when it has compact fibers (like in groups). Perhaps it all works for Topological Protomodular Categories, as in the paper by Borceux and Clementino. $\endgroup$ Feb 22, 2022 at 21:12