Quotient of a Hausdorff topological group by a closed subgroup - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:33:40Z http://mathoverflow.net/feeds/question/9022 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9022/quotient-of-a-hausdorff-topological-group-by-a-closed-subgroup Quotient of a Hausdorff topological group by a closed subgroup Dyke Acland 2009-12-15T18:41:30Z 2009-12-16T10:19:00Z <p>Sorry if this question is below the level of this site: I've read that the quotient of a Hausdorff topological group by a closed subgroup is again Hausdorff. I've thought about it but can't seem to figure out why. Is it obvious? A simple yes or no (with reference is possible) is all I need.</p> http://mathoverflow.net/questions/9022/quotient-of-a-hausdorff-topological-group-by-a-closed-subgroup/9023#9023 Answer by jvp for Quotient of a Hausdorff topological group by a closed subgroup jvp 2009-12-15T18:46:20Z 2009-12-16T10:19:00Z <p>Edit: Below I expand my crude original answer "<a href="http://en.wikipedia.org/wiki/Hausdorff%5Fspace" rel="nofollow">Yes</a>" as requested by the community.</p> <p><hr /></p> <p>Yes. Let $G$ be the group and $H$ be the closed subgroup. The <a href="http://en.wikipedia.org/wiki/Kernel%5Fof%5Fa%5Ffunction" rel="nofollow">kernel</a> of the quotient map $G \to G/H$ is equal to $\Delta^{-1}(H)$ where $\Delta : G \times G \to G$ is the continuous function $\Delta(x,y)= x- y$. Hence the kernel is closed. According to <a href="http://en.wikipedia.org/wiki/Hausdorff%5Fspace#Properties" rel="nofollow">this</a> $G/H$ is Hausdorff.</p> http://mathoverflow.net/questions/9022/quotient-of-a-hausdorff-topological-group-by-a-closed-subgroup/9028#9028 Answer by S1 for Quotient of a Hausdorff topological group by a closed subgroup S1 2009-12-15T19:09:50Z 2009-12-15T19:09:50Z <p>In fact, an even stronger statement holds: If $G$ is a topological group and $H$ is an (abstract) subgroup, then $G/H$ is Hausdorff if and only if $H$ is closed (cf Bourbaki, General Topology, III.2.5, prop 13). It's not hard to prove.</p>