Converse To Quotient Manifold Theorem [Exercise in Lee Smooth Manifolds] I would like help with the following problem (chapter 9, #4) from Lee's Smooth Manifolds [its not homework, I'm reading it and I got stuck on this one]
If a Lie group $G$ acts smoothly and freely on a smooth manifold $M$ and the orbit space $M/G$ has a smooth manifold structure such that the quotient map $\pi: M\to M/G$ is a smooth submersion, then $G$ acts properly. 
Its kind of a converse to the standard theorem about quotienting a manifold by a group action.
Any hints/help?
 A: I'm pretty sure this is true. Here's my suggested approach. There's a step I don't know how to do, so hopefully someone can fill in the gap, or you can figure it out as an exercise. (And please post a comment if you do, as I'm curious!)
First, show that $M$ is a principal fiber bundle with structure group $G$. That is, locally on $M/G$, $M$ looks like $(M/G)\times G$. To do this, use that $M\to  M/G$ is a submersion to find neighborhoods $U$ on $M/G$ and $V$ on $M$ where $M\to M/G$ looks like the projection onto several coordinates. The coordinates you "forget" in this projection give a chart on some neighborhood of $G$. (This is why the fibers of submersions are smooth manifolds; in this case, the fiber is $G$ since the action is free.)  Now use translation by the group law to "spread $V$ out" so that the entire preimage of $U$ in $M$ is diffeomorphic to $U\times G$, with the restriction of $M\to M/G$ to this open being the projection onto the first factor. This ensures that $M$ is a principal $G$-bundle.
This reduces the problem to showing that $G$ acts properly and freely on any principal fiber bundle with structure group $G$ (freeness being clear). To show that $G\times M\to M\times M$ is proper, it suffices to show that it is closed and that the preimage of a point in compact. The latter condition is obvious, by working locally in a trivializing neighborhood of that point. I don't immediately see how to get the closedness of the map, but I've convinced myself it's true in some examples.
A: The following is an addendum to Sam's answer, meant to fill-in missing details.
Part 1. Suppose that $G\times M\to M$ is such that the quotient $B=M/B$ has structure of a smooth manifold such that the quotient map $q: M\to B$ is a submersion. Then $q: M\to B$ is a topological principal $G$-fiber bundle.
Proof. Pick a $G$-orbit $Gx\subset M$. Our goal is to find a $G$-invariant neighborhood $W$ of $Gx$ in $M$ such that $W$ is $G$-equivariantly homeomorphic to a product $U\times G$, $U$ is an open subset of $B$, and $G$ acts trivially on the first factor and by left multiplication on the second factor. Equivariance of a homeomorphism $f: U\times G\to U$ means:
$$
f(g(u,h))=g f(u,h)
$$
for all $u\in U, g\in G, h\in G$.
Since $q$ is a submersion, there exists, $V$, a neighborhood of $x$ in $M$ and a diffeomorphism $g: V\to U\times {\mathbb R}^n$, where $U=q(V)$ and $q|_V= p_U\circ g$, where $p_U: U\times {\mathbb R}^n\to U$ is the projection to the first factor, $g(x)=(q(x), 0)$. Here $n+dim(B)=dim(M)$.
Set $U':= g^{-1}(U\times \{0\})$. Such a subset $U'$ of $V$ is called a slice through $x$ for the $G$-action on $M$. By the construction, every $G$-orbit intersects $U'$ in at most one point and $U'$ is a smooth submanifold of dimension $d=dim(B)$ in $M$. Now, consider the orbit map
$$
f: G\times U'\to M, \quad f(g,y)= gy. 
$$
This map is smooth and 1-1 (since each $G$-orbit intersects $U'$ at most once). Moreover, $dim(G\times U')=dim(M)$. Hence, by the invariance of domain theorem, $f$ is a homeomorphism to its image, which is an open subset $W\subset M$. Clearly, $W$ contains $Gx$. Moreover, by the construction, $f$ is $G$-equivariant.
Remark. i. With more work, one can prove that $f$ is also a diffeomorphism, but I do not need this.
ii. The map $f$ as above plays the role of "spreading $V$ out'' in Sam's answer.
This concludes the proof of Part 1.
Part 2. If $M\to B$ is a principal $G$-fiber bundle (where $G$ is a Lie group and $M, B$ are topological manifolds), then the $G$-action on $M$ is proper.
This properness property holds even in much greater generality, see Lemma B in my answer here.
