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I have this lemma from "Kazhdan’s Property (T)" book (page 343 in the link).

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Here, $G$ is locally compact and $H$ is a closed subgroup of $G$.

Can't really understand why $K$ is compact. The union is obviously compact, but $p^{-1}(Q)$ not necessarily.

Also, I understand why $p(K) \subseteq Q$, but the other direction is not clear to me.

Thanks!

N17Math
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2 Answers2

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You can just show this by hand:

Let $q\in Q$. Since $Q\subset\bigcup_{i=1}^n p(x_iU)$ there exists $i\in\{1,\dots,n\}, a\in U$ such that $p(x_ia)=q$. Hence, $x_ia\in p^{-1}(Q)$. Since clearly $x_ia\in x_iU$, we deduce that $$x_i a\in K=p^{-1}(Q)\cap\bigcup_{i=1}^n x_iU$$ Therefore $q=p(x_ia)\in p(K)$.

This shows that $p(K)\supseteq Q$.

tth2507
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(Thank you Izaak, brain wasn't online) Quotients of Hausdorff topological groups by closed subgroups are themselves Hausdorff, so $Q$ is a compact subset of a Hausdorff space, and thus is closed. By continuity of $p$, $p^{-1}(Q)$ is hence a closed set, so the intersection of it with a compact set is thus compact. You are correct that $p^{-1}(Q)$ is not necessarily compact.

Andrew Turner
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    Hi Andrew. I'm not sure it's true that $p^{-1}(Q)$ will be compact (consider eg $\Bbb R/\Bbb Z$ or $G/G$). I think the problem is that $\mathcal V$ covering $Q$ doesn't imply that the corresponding subfamily in $G$ covers $p^{-1}(Q)$. However the book assumes $G$ is Hausdorff, from which it does follow that $p^{-1}(Q)$ is closed, so $K$ is the intersection of a closed set with a finite union of compact sets, which does make $K$ compact. – Izaak van Dongen Mar 15 '24 at 12:35
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    Thanks, yes that is slightly subtle point. I have edited my answer now. – Andrew Turner Mar 15 '24 at 12:47
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    Thanks a lot guys! – N17Math Mar 15 '24 at 14:25