Is the extension of a locally compact group by a compact group a locally compact group?

Question

Let $1 \to N \overset{\iota}{\to} G \overset{\pi}{\to} G/N \to 1$ be a short exact sequence of topological groups. Here, $\iota$ is the inclusion of a closed, normal subgroup and $G/N$ has the quotient topology.

Question 1: If $N$ and $G/N$ are locally compact (Hausdorff) groups, is $G$ also a locally compact group?

In fact, I would be perfectly satisfied with an answer to:

Question 2: If $N$ is a compact group and $G/N$ is a locally compact group, is $G$ a locally compact group?

If the quotient map admits a continuous local section $s :W \to G$ defined on a neighbourhood $W$ of $1 \in G/N$, then one easily checks that $G$ is locally homeomorphic to $N \times (G/N)$, so the answer to both questions is "yes". However, as noted here, the existence of such a section is not guaranteed, even if all of $G,N,G/N$ are compact. One may take $G = \mathbb{T}^\mathbb{N}$, $N = \{\pm 1 \}^\mathbb{N}$.

It's at least easy to check that $G$ is Hausdorff. Since $\{1\}$ is closed in $N$, and $N$ is closed in $G$, we know that $\{1\}$ is closed in $G$. Thus, the diagonal $\Delta = \{ (x,x) : x \in G\}$ is closed in $G$, because it is the primage of $\{1\} \subseteq G$ under the continuous map $(x,y) \mapsto xy^{-1} G \times G \to G$.

Answer 1

It seems to me the answer to Question 2 (the title question, and the simpler of the two) is "Yes". Suppose that $N$ is compact and $Q = G/N$ is locally compact. It's easy to check $G$ is Hausdorff (done in the post), so we just need to show that some point of $G$, say $1_G$, has a compact neighbourhood.

Notes to get started:

• It is important to know that the projection $\pi : G \to Q$, a priori just a topological quotient map, is actually an open map. Indeed, if $U \subseteq G$ is open, then $\pi^{-1}(\pi(U)) = \bigcup_{x \in N} xU$ is open, whence $\pi(U)$ is open by definition of the quotient topology.
• Since the cosets $\pi^{-1}(q)$, $q \in Q$ are compact (all are homeomorphic to $N$), one suspects the map $\pi : G \to Q$ should be proper.
• In fact, if $\pi$ is proper, then we will be done: If $K$ is a compact neighbourhood of $1_{G/N}$ and $\pi$ is a proper map, then $\pi^{-1}(K)$ is a compact neighbourhood of $1_G$.

So, we will have shown the answer to Question 2 is "Yes" once we prove the following proposition.

Proposition: If $N$ is compact and $Q = G/N$ is locally compact, then $\pi :G \to Q$ is a proper map.

Proof of proposition: Suppose that $K \subseteq Q$ is compact. Let $\{U_i : i \in I\}$ be an open cover of $\pi^{-1}(K)$. In particular, for each $q \in K$, we've an open cover of the compact coset $\pi^{-1}(q)$, and so there is a finite set $F_q \subset I$ such that $\{ U_i : i \in F_q\}$ covers $\pi^{-1}(q)$.

For each $q \in K$, define $W_q = \bigcup_{i \in F_q} U_i$, an open set containing $\pi^{-1}(q)$. We would like a kind of a "tube lemma" showing that $W_q$ actually contains all sufficiently near cosets as well. Fix $x \in \pi^{-1}(q)$. Since multiplication is continuous and $x = 1 \cdot x \in W_q$, there are open sets $A_x,B_x \subseteq G$ with $1 \in A_x$ and $x \in B_x$ such that $A_xB_x \subseteq W_q$. Note $\{B_x : x \in \pi^{-1}(q)\}$ is an open cover of $\pi^{-1}(q)$, which is compact, so there exist $x_1,\ldots,x_n \in \pi^{-1}(q)$ so that the corresponding $B_1,\ldots, B_n$ cover $\pi^{-1}(q)$. Define $$T_q = (A_1 \cap \ldots \cap A_n) \cdot \pi^{-1}(q).$$ This set $T_q$ is open and satisfies $$\pi^{-1}(q) \subseteq T_q \subseteq W_q.$$ The rationale behind calling it a "tube" is that it is saturated with respect to the quotient map, because $\pi^{-1}(q) \cdot N = \pi^{-1}(q)$.

Now, since $\pi$ is open, $\{ \pi(T_q) : q \in K\}$ is an open cover of $K$. So, there are finitely many $q_1,\ldots,q_n \in K$ such that $\pi(T_{q_1}),\ldots,\pi(T_{q_n})$ cover $K$, whence $T_{q_1},\ldots,T_{q_n}$ cover $\pi^{-1}(K)$ (because the tubes are saturated). Thus, the sets $W_{q_1},\ldots,W_{q_n}$ cover $\pi^{-1}(K)$ which is the same as saying the finite family of open sets $\{U_i : i \in F_{q_1} \cup \ldots \cup F_{q_n}\}$ covers $\pi^{-1}(K)$. We have shown every open cover of $\pi^{-1}(K)$ has a finite subcover, so $\pi^{-1}(K)$ is compact, as was needed.

I suspect that one can push these ideas a little further to get a positive answer to the (more general) Question 1, and will attempt to do so. The immediate obstacle is the lack of an obvious candidate for a compact neighbourhood in $G$.

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Added: I found a reference answering the more general Question 1. On pp.39 of Abstract Harmonic Analysis I by Hewitt and Ross, one finds the following theorem.

(5.25) Theorem: Let $G$ be a topological group and $H$ a subgroup of $G$. If $H$ and $G/H$ are compact, then $G$ itself is compact. If $H$ and $G/H$ are locally compact, then $G$ is also locally compact.

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Added: This is discussed at the following mathoverflow post, where user27920 gives an argument using nets.