If $H$ and $G/H$ are pseudocompact then is G also pseudocompact?

Question

Theorem 5.25 in Edwin, Hewitt Abstract Harmonic Analysis Part 1, says if $H$ is a subgroup of a topological group $G$ and $H$ and $G/H$ are compact (resp. locally compact), then so is $G$.

For my purpose I can assume topological groups to be Hausdorff, if necessary. My question is does this property extends to weaker forms of compactness, like pseudocompact or quasicompact (if every covering of G by co-zero sets admits a finite subcovering). Is such a result already known to be true or false? If not which way should I look for? Any help is appreciated.

Henno Brandsma · Answer 1 · 2019-05-11T11:28:14.910

2

Yes for pseudocompact, e.g. It's a so-called three-space property according to the references and introduction of this paper by Tkachenchko and Bruguera.

Quasicompactness is not considered there. (Normally we only consider Hausdorff/$T_0$ groups and then there is no difference between compact and quasicompact.

edited May 11 '19 at 11:28

answered Mar 16 '19 at 07:21

Henno Brandsma

242,131

Where can I find a proof that quasicompactness is equivalent to compactness for Hausdorff topological groups? – Bhaskar Vashishth Mar 16 '19 at 07:50
@BhaskarVashishth it’s trivial because in a Tychonoff space the co-zero sets are a base. – Henno Brandsma Mar 16 '19 at 07:51
"There is a continuous map from $H\times G/H$ onto $G$" This is wrong. I've asked a question on this. – YuiTo Cheng May 11 '19 at 11:15
@YuiToCheng Note that the counterexamples given there are not pseudocompact e.g. – Henno Brandsma May 11 '19 at 11:19
I know. But we need an alternative idea anyway. – YuiTo Cheng May 11 '19 at 11:20
@YuiToCheng I found a relevant reference. See the updated answer. – Henno Brandsma May 11 '19 at 11:30
Okay, I will check it out. – YuiTo Cheng May 11 '19 at 11:33

If $H$ and $G/H$ are pseudocompact then is G also pseudocompact?

1 Answers1