The following is proposition 1.6 from chapter XII in the book 'Real and Functional Analysis' by S. Lang. In this context $G$ is a topological group, $H$ a subgroup and $\pi :G\to G/H$ the quotient map, where $G/H$ is the set of left congruence classes mod $H$.
If $K'$ is a compact subset of $G/H$, then there exists a compact subset $K$ of $G$ such that $K'=\pi(K)$
Proof. Let $A$ be a compact neighborhood of $e$ in $G$. Then $\pi(A)$ is a compact neighborhood of the unit coset in $G/H$. Let $x_1,\dots,x_n$ be elements of $G$ such that the sets $\pi(x_iA)$ cover $K'$. Let $K=\pi^{-1}(K')\cap (x_1A\cup\dots\cup x_nA)$. Then $K$ is compact and satisfies our requirements.
Questions:
- For the existence of $A$ we need to assume that $G$ is locally compact, right? I think this is also assumed before the proposition but it isn't really stated anywhere.
- Why is $K$ compact? Clearly $x_1A\cup\dots\cup x_nA$ is compact. If $\pi^{-1}(K')$ was closed then $K$ is compact but I don't see why $\pi^{-1}(K')$ should be closed. This should be the case if for example $G/H$ is Hausdorff, i.e. $H$ is closed. But this is not assumed here.
