Let $G$ be a Hausdorff topological group with center $Z$ and closed subgroup $H$. Suppose that $H.Z = G$. Is the product map
$$H \times Z \rightarrow G$$
necessarily an open map? That is, can we identify with $G$ as a quotient group of $H \times Z$?
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5$\begingroup$ If $G$ is second-countable and locally compact (so $H \times Z$ is too) then this is affirmative since for any transitive continuous action of a second-countable group $\mathscr{G}$ (e.g., $H \times Z$) on a locally compact Hausdorff space $X$ (e.g., left multiplication on $G$ through the surjective homomorphism) the continuous bijective orbit map $\mathscr{G}/{\rm{Stab}}_{\mathscr{G}}(x) \to X$ for $x \in X$ is a homeomorphism. The role of second-countability is to apply the Baire category theorem. See section 5 in Ch. IX of Bourbaki's General Topology for a proof in a wider context. $\endgroup$– nfdc23Commented Feb 24, 2018 at 3:25
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$\begingroup$ Thank you, this was exactly the kind of result I was hoping for $\endgroup$– D_SCommented Feb 24, 2018 at 3:29
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$\begingroup$ I meant to require $\mathscr{G}$ to also be locally compact and Hausdorff (not just second-countable) in the general assertion of my previous comment. $\endgroup$– nfdc23Commented Feb 24, 2018 at 4:35
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2$\begingroup$ The same is true if $G$ is Polish, more generally, Lindelof and Cech-complete. $\endgroup$– Taras BanakhCommented Feb 24, 2018 at 6:27
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1$\begingroup$ Btw the results mentioned by nfdc and Taras are fine for $Z$ central, not with the artificial assumption that the center is reduced to $Z$ (and probably $Z$ normal is enough; I'm only sure in the locally compact 2nd countable case). An easy counterexample, with non-Polish separable metrizable groups (and when one assume $Z$ central) is when $H,Z$ are two mutually irrational cyclic subgroups of the reals and $G$ is their sum, with the induced topology. With $Z$ equal to the center we can produce more complicated examples as other dense countable subgroups of Lie groups. $\endgroup$– YCorCommented Feb 24, 2018 at 19:02
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