A morphism $f: X \to Y$ is an epimorphism if for all $g, h: Y \to Z$, if $g \circ f = h \circ f$ then $g = h$. The epimorphisms in the category of groups are the surjective group homomorphisms. The epimorphisms in the category of topological spaces are the continuous surjections. Are the epimorphisms in the category of topological groups the continuous surjective homomorphisms?
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1Yes, according to this paper: http://www.numdam.org/article/RSMUP_1969__42__135_0.pdf – Cheerful Parsnip Apr 06 '22 at 02:49
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Thank you for the reference! – user46484 Apr 06 '22 at 02:51
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Is there any reason to suppose things are different in the category of topological abelian groups? – user46484 Apr 06 '22 at 18:27
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Interestingly, the epimorphisms in the category of Hausdorff topological groups are different. – Jochen Apr 09 '22 at 11:16
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More precisely, unlike in the category of Hausdorff space where the epis are the continuous functions with dense image, in the category of Hausdorff topological groups an epimorphism doesn't need to have dense image – Alessandro Codenotti Apr 17 '22 at 07:17
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A cool answer is this one. Sorry if this the same proof given in the reference in the commentaries.
Consider the forgetful functor
$$U: TopGrp \rightarrow Grp$$
For any group $G$ let $L(G)$ be the the same group with the discrete topology and $R(G)$ be with the codiscrete topology. Then
$$TopGrp(L(G),H)=Grp(G,U(H))$$
since every function $L(G)\rightarrow H$ is continuous. We also have that
$$TopGrp(G,R(H))=Grp(U(G),H)$$
since every function $G \rightarrow R(H)$ is continuous.
Thus we conclude that $U$ is both right and left adjoint. Right adjoints preserve monomorphisms and left adjpints preserve epimorphisms. We conclude that a morphism of topological groups is an (mono)epimorphism if, and only if, its underlying morphism of groups is.
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1$U$ having a (left and) right adjoint shows only the necessity for the underlying group homomorphism to be a (mono)epimorphism. For sufficiency, you need an additional property like $U$ being faithful since then $fe=ge$ for $Ue$ epimorphic ($mf=mg$ for $Um$ monomorphic) implies $Uf=Ug$, This is in fact necessary as $U$ being faithful is equivalent to the counit $\epsilon_G\colon LU(G)\to G$ being an epimorphism of topological groups, but $U\epsilon_G\colon ULUG\to UG$ is a split epimorphism with section the unit $\eta_{UG}\colon UG\to ULUG$ (dually for the unit $G\to RU(G)$ being monomorphic). – Vladimir Sotirov Apr 09 '22 at 04:47