Let $G$ be an abelian topological group and let $H$ be a non-Hausdorff closed subgroup (so that $G/H$ is Hausdorff). If $H$ and $G/H$ are second countable, is $G$ second countable?
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asked Oct 23, 2021 at 19:30
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1$\begingroup$ Your title is about spaces but the question is about topological groups. Anyway, a topological group is always extension of a indiscrete subgroup (the closure of $\{1\}$) by a Hausdorff, and in particular, $G$ is 2nd-countable iff its maximal Hausdorff quotient is second countable. The answer to your question hence thus clearly boils down to the case when $G$ is Hausdorff. $\endgroup$– YCorCommented Oct 23, 2021 at 21:35
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$\begingroup$ Dear YCor, many thanks for your answer. It is exactly what I needed. In Hewitt and Ross, Abstract Harmonic Analysis I, (5.38)(e), p.47, a proof is given ( independently of any separation hypotheses) of the stated result for first-countability $\endgroup$– Cristian D. Gonzalez-AvilesCommented Oct 24, 2021 at 14:04
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$\begingroup$ The proof alluded to is an adaptation of a 1950 argument by Graev (apparently only in Russian), cited 4, p.497 of the Hewitt-Ross book. As explained on p.30 of «Metric Geometry of Locally Compact Groups» (a book that you might know very well), the required property for second-countability then follows from certain other results which, unfortunately, depend on the Hausdorff hypothesis. I work in Number Theory, where non-Hausdorff group quotients appear often, so I often regret that many excellent texts make this Hausdorff blanket hypothesis right from the start. $\endgroup$– Cristian D. Gonzalez-AvilesCommented Oct 24, 2021 at 14:18
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