Is the projection onto a quotient by a compact normal subgroup proper?

Question

My ultimate goal is to prove the following: If $G$ is a locally compact group and $K_\alpha$ a net of compact normal subgroups with trivial intersection, then the inverse limit $\projlim G/K_\alpha$ is locally compact. (I want to prove this because it's all I have left in order to prove $G$ is topologically isomorphic to this inverse limit).

I attempted to prove it as follows: every identity neighbourhood in our inverse limit contains some neighbourhood of the form $V_\beta K_\beta \times \prod_{\alpha\ne \beta}G/K_\alpha$, and since $G/K_\beta$ is locally compact $V_\beta K_\beta$ contains some compact neighbourhood $U_\beta K_\beta$ of the identity (in $G/K_\beta$). Consider $U_\beta K_\beta$ as a subset of $G$, i.e. as $\pi_\beta^{-1}(U_\beta K_\beta)$ (where $\pi_\beta:G\to G/K_\beta$ is the projection); if this subset is compact, then $U_\beta K_\beta \times \prod_{\alpha\ne \beta}G/K_\alpha$ is a compact identity neighbourhood contained in our original one.

Assuming I didn't make some mistake, this means what I have left to prove is this:

If $G$ is locally compact and $K$ is a compact normal subgroup, then the projection $p:G\to G/K$ is proper, i.e. its preimage of a compact set is compact.

Can anyone help proving this statement? Or help prove the inverse limit is locally compact in some other way?

Answer 1

Theorem $1.5.7$ Suppose that $G$ is a topological group and that $H$ is a compact subgroup of $G$. Then the quotient mapping $\pi$ of $G$ onto the quotient space $G / H$ is perfect.

Proof: Take any closed subset $P$ of $G$. Then, by Theorem $1.4.30$, $PH$ is closed in $G$. However, $PH$ is, obviously, the union of a certain family of left cosets, that is, $PH = \pi^{-1} \pi(P)$. It follows by the definition of a quotient mapping that the set $\pi(P)$ is closed in the quotient space $G / H$. Thus $\pi$ is a closed mapping. In addition, if $y \in G/H$ and $\pi(x) = y$ for some $x \in G$, then $\pi^{-1}(y) = xH$ is a compact subset of $G$. Hence the fibers of $\pi$ are compact and $\pi$ is perfect. $\square$

Source: Alexander V. Arhangel'skii, Mikhail G. Tkachenko, Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.

$3.7.2$ Theorem: If $f : X \to Y$ is a perfect mapping, then for every compact subspace $Z \subset Y$ the inverse image $f^{-1}(Z)$ is compact

Proof: As $f^{-1}(Z)$ is clearly a Hausdorff space, it suffices to show that for any family $\{ U_s \}_{s \in S}$ of open subsets of $X$ whose union contains $f^{-1}(Z)$ there exists a finite set $S_0 \subset S$ such that $f^{-1}(Z) \subset \bigcup_{s \in S_0} U_s$. Let $\mathcal{T}$ be the family of all finite subsets of $S$ and $U_T = \bigcup_{s \in T} U_s$ for $T \in \mathcal{T}$. For each $z \in Z$ the fiber f$^{-1}(z)$ is compact and thus contained in the set $U_T$ for some $T \in \mathcal{T}$; it follows that $z \in Y \setminus f(X \setminus U_T)$, and thus $Z \subset \bigcup_{T \in \mathcal{T}} (Y \setminus f(X \setminus U_T))$. The sets $Y \setminus f(X \setminus U_T)$ being open, there exist $T_1, T_2, \ldots, T_k \in \mathcal{T}$ such that $Z \subset \bigcup_{i=1}^k (Y \setminus f(X \setminus U_{T_i}))$. Hence

$$f^{-1}(Z) \subset \bigcup_{i=1}^k f^{-1}(Y \setminus f(X \setminus U_{T_i})) = \bigcup_{i=1}^k (X \setminus f^{-1} f (X \setminus U_{T_i})) \subset$$ $$\subset \bigcup_{i=1}^k (X \setminus (X \setminus U_{T_i})) = \bigcup_{i=1}^k U_{T_i} = \bigcup_{s \in S_0} U_s$$

where $S_0 = T_1 \cup T_2 \cup \ldots \cup T_k$. $\square$

Source: Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.