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I am learning about free topological groups, and I am trying to understand whether the analogue of "every group is a quotient of a free group" holds in the continuous setting too. I am particularly interested in compactly generated groups, and so getting an analogue of "every finitely generated group is a quotient of a free group of finite rank".

Let $X$ be a Tychonoff space (for what I'm interested in, we may even assume compact Hausdorff), and let $F(X)$ be the free topological group generated by $X$. Let $\sigma : X \to G$ be a topological embedding of $X$ into a topological group $G$ such that $\sigma(X)$ algebraically generates $G$. This induces a continuous surjective homomorphism $f : F(X) \to G$. The first isomorphism theorem for topological groups yields a topological isomorphism $F(X) / \ker(f) \cong G$ only if $f$ is already an open map.

So my question is: when is $f$ open? In "Locally compact groups and related structures" (Arhangel’skii-Tkachenko) the only sufficient condition that is given for this setting is when $\sigma$ is a quotient map. However, I am thinking of $\sigma(X)$ as a (compact) generating set for $G$, so in the most interesting cases $\sigma$ will not be onto. Maybe the assumption that $\sigma$ is a topological embedding helps?

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  • $\begingroup$ Why you do not take the underlying topological space of $G$ for $X$? I mean, that's what I do if I want to write an arbitrary group as the quotient of a free group. $\endgroup$ – Paul K Apr 7 at 8:46
  • $\begingroup$ It is enough if $f$ is a quotient map, it doesn't have to be open. Not sure if that helps. $\endgroup$ – freakish Apr 7 at 9:40
  • $\begingroup$ @PaulK yes but this does not shed any light on compact generation. I'll edit the beginning to make that clearer $\endgroup$ – frafour Apr 7 at 9:45

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