# Free topological groups and quotients

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?

• 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. – Paul K Apr 7 at 8:46
• It is enough if $f$ is a quotient map, it doesn't have to be open. Not sure if that helps. – freakish Apr 7 at 9:40
• @PaulK yes but this does not shed any light on compact generation. I'll edit the beginning to make that clearer – frafour Apr 7 at 9:45