I want to solve the following exercise from Lee's book on smooth manifolds. 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.
This question was asked before somewhere but the answer is not satisfactory. Can someone give some hint on how to solve this problem?
We show this by using the characterization of proper actions via sequences:
Let $(g_{i})\subseteq G$ and $(p_{i})\subseteq M$ be sequences such that $p_{i}\to p$ and $g_{i}\cdot p_{i}\to q$. We need to find a convergent subsequence of $(g_{i})$.
Observe that $\pi\colon M\to M/G$ is a smooth submersion. This means that there is an open set $V\subseteq M/G$ and a smooth section of $\pi$ defined on $V$ (that is, a map $\chi\colon V\to M$ such that $\pi\circ \chi=\operatorname{Id}_{V}$ and $p\in \chi(V)$). Define a map $\Phi\colon G\times V\to \pi^{-1}(V)$ by letting $\Phi(g,z)=g\cdot\chi(z)$. This map is bijective and $G$-equivariant, so it has constant rank along the orbits $G\times \{s\}$ for all $s\in V$. Therefore, if we can prove that the rank of $\Phi$ is globally constant, we get that $\Phi$ is a diffeomorphism. This is because, by taking a point of the form $(e,s)$, we get that $\Phi_{*(e,s)}(T_{(e,s)}(G\times V))=T_{\chi(s)}(G\cdot \chi(s)))+\chi_{*s}(T_{s}(M/G))$. Since $\chi$ is an immersion and those two summands have trivial intersection, we get that $\Phi$ has constant rank in $\{e\}\times V$, so $\Phi$ has constant rank everywhere and is therefore a diffeomorphism.
Now, since $p_{i}\to p$, we can suppose that $\pi(p_{i})\in W$ for all $i$, where $W\subseteq V$ is an open subset containing $p$ and having compact closure in $V$. Therefore, we can write $\Phi^{-1}(p_{i})=(h_{i},s_{i})$ for $h_{i}\in G$, $s_{i}\in V$. Notice that $\pi(p_{i})=\pi(\Phi(h_{i},s_{i}))=\pi(h_{i}\cdot \chi(s_{i}))=\pi(\chi(s_{i}))=s_{i}$, so that $p_{i}=h_{i}\cdot \chi(\pi(p_{i}))$. Also, $p=\chi(\pi(p))=e\cdot \chi(\pi(p))$. Since $p_{i}\to p$, $h_{i}\to e$, because $\Phi$ is a diffeomorphism.
Furthermore, $g_{i}\cdot p_{i}\to q$, so that $\pi(p_{i})\to \pi(q)\in \overline{W}\subseteq V$. We can therefore write $q=h\cdot \chi(\pi(q))$ for some $h\in G$. Also, $g_{i}\cdot p_{i}=(g_{i}h_{i})\cdot \chi(\pi(p_{i}))\to q=h\cdot \chi(\pi(q))$. Again, since $\Phi$ is a diffeomorphism, this implies that $g_{i}h_{i}\to h$, so $g_{i}=g_{i}h_{i}h_{i}^{-1}\to h$. We get that $(g_{i})$ is a convergent subsequence.
Hope this helps!