# Orbit space of a free, proper G-action principal bundle

Let $G$ be a topological group and let $r \colon E \times G \to E$ be a continuous right-action on a topological space $X$.

If $p\colon E \to B$ is a continuous map into a topological space $B$ such that $(p, r)$ is a principal $G$-bundle, then it follows that $B \cong E/G$ where $E/G$ is the orbit space of the action, and that $p$ is essentially the projection map $\pi\colon E \to E/G$. It follows further that the action must be free.

I'm interested in the opposite direction, i.e. the question when the projection $E \to E/G$ determines a principal $G$-bundle.

In the beginning lines of this article I found the assertion that it is sufficient for the action of $G$ on $E$ to be free (this is obviously necessary) and proper. However, no proof is provided.

If $G$ is discrete, then one can show that the action is proper precisely if every $x \in E$ has a neighbourhood $U$ such that $Ua \cap U = \emptyset$ whenever $a \neq 1$, $a \in G$. Indeed, $V := \pi(U)$ is then the sought-after open neighbourhood of $xG \in E/G$ such that $\pi^{-1}(V) \cong V \times G$. It's not difficult to define a suitable $G$-homeomorphism.

If $G$ is locally compact, $X$ is Hausdorff, and the action is proper, then there is a similar result, for then every $x \in E$ has a neighbourhood $U$ such that $\{a \in G \mid Ua \cap U \neq \emptyset\}$ is contained in a compact set $K \subset G$.

I was wondering if this could be used for a proof. If anyone could help me with other ideas, I'd be very glad to hear them.

PS: I edited this post a bit as I didn't want to submit essentially the same question again.

-
One certainly has to be extra-careful when $G$ and $X$ are non-compact. – Neal Nov 10 '13 at 14:18