Is the quotient map of an Abelian $G$-space a locally trivial fiber bundle without assuming compactness Let $X$ and $G$ be locally compact, Hausdorff and second-countable spaces such that $G$ is an Abelian topological group with respect to that topology. Suppose that we have a continuous, free, and proper action of $G$ on $X$. That is, a continuous map $X\times G \to X$, $(x,g)\mapsto x+g$ such that for all $x\in X$ we have $\{g\in G: g+x=x\}=\{0_G\}$. Let $\pi:X\to X/G$ be the quotient map (with $X/G$ endowed with the quotient topology).
Is it true that this bundle map is locally trivial? That is: Is there a covering of $X/G$ by open sets $\{U_i\}_{i\in I}$ such that for every $i\in I$ there is a homeomorphism $\phi_i:\pi^{-1}(U_i)\to U_i\times G$ that is equivariant (i.e., that $\phi_i(x+g)=\phi_i(x)+g$ where addition on $U_i\times G$ is understood in the $G$ coordinate)?
There is a result of Gleason https://www.ams.org/journals/proc/1950-001-01/S0002-9939-1950-0033830-7/S0002-9939-1950-0033830-7.pdf that implies this if $G$ is a compact Lie group. But can we drop those assumptions to those that I wrote before? (maybe we need the Lie property?)
 A: The version you propose does not work. The question whether we can drop the Lie property seems interesting, but also too difficult to me, so I will take the cheap way out and illustrate that we cannot drop the compactness property (that said, I believe the result will be true again once you add the assumption that the action should be proper).
Let $X=S^1$ and $G=\mathbb{Z}$. Choose an irrational $r\in[0,1]$ and let $\mathbb{Z}$ act on $S^1$ via $n.z=e^{2\pi nr}z$. Certainly, $S^1$ and $\mathbb{Z}$ are second-countable LCH spaces ($S^1$ is compact, even) and $\mathbb{Z}$ is an abelian topological group. The action of $\mathbb{Z}$ on $S^1$ is free (exercise; this is where it matters that $r$ is irrational) and clearly continuous ($S^1$ is itself a topological group and this is just the left translation action of $S^1$ on itself restricted to the cyclic subgroup generated by $e^{2\pi r}\in S^1$). The quotient $S^1/\mathbb{Z}$ is very ugly, e.g. it is not even $T_1$. I claim that $\pi\colon S^1\rightarrow S^1/\mathbb{Z}$ is not a covering space, which contradicts your claim. Assume to the contrary that it is a covering space. Now, every covering space trivializes over some non-empty closed subspace of the base (exercise), so there is a closed $\emptyset\ne X\subseteq S^1/\mathbb{Z}$, such that $\pi^{-1}(X)\cong X\times\mathbb{Z}$. However, $\pi^{-1}(X)\subseteq S^1$ is closed and $S^1$ is compact, whence $\pi^{-1}(X)$ is compact, but $\pi^{-1}(X)\cong X\times\mathbb{Z}$ and $X\ne\emptyset$ then implies that $\mathbb{Z}$ is compact, which is absurd.
A: There is a partial positive answer to this question the case the group is Lie, https://www.jstor.org/stable/1970335 (Theorem 2.3.3) and slightly stronger condition is asked. It is required that the "difference" map that goes from the set $\{(x,y)\in X\times X:\exists g\in G \text{ such that }g+x=y\}$ to $G$ (by giving back the unique element $g$ such that $x+g=y$) is continuous. Still, I have not found a proof nor an counterexample of the question posted without assuming that the group is Lie.
