Let $G$ be a lie group acting on a smooth manifold $M$ smoothly, properly and freely. Then, there exists a unique smooth structure on the orbit space $M/G$ such that the map $\pi:M\rightarrow M/G$ is a smooth submersion.
I am trying to show that this map has local trivialization property with $G$ as fiber space i.e., given $q\in M/G$ there exists an open subset $U\subseteq M/G$ containing $q$ and a diffeomorphism $$\varphi:U\times G\rightarrow \pi^{-1}(U).$$
Local section theorem says that any smooth submersion has abundant local sections. In particular, given $p\in M$ there exists a local section of $\pi$ whose image contains $p$. Let $q=\pi(p)$ and $\sigma:U\rightarrow M$ be a local section of $\pi$ with $\sigma(q)=p$.
As $\pi\circ \sigma=1_U$ we have $\sigma(a)\in \pi^{-1}(a)\subseteq \pi^{-1}(U)$ for all $a\in U$. So, we have ,$U\rightarrow \pi^{-1}(U)$ a smooth map. Define $\varphi:U\times G\rightarrow \pi^{-1}(U)$ by $(a,g)\mapsto g\sigma(a)$.
As $a=\pi(\sigma(a))=\pi(g\sigma(a))$, we have $g\sigma(a)\in \sigma^{-1}(a)\subseteq \sigma^{-1}(U)$. So, we have a well defined map $U\times G\rightarrow \pi^{-1}(U)$ given by $(a,g)\mapsto g\sigma(a)$.
Let $(a,g),(a',g')\in U\times G$ be such that $\varphi(a,g)=\varphi(a',g')$ i.e., $g\sigma(a)=g'\sigma(a')$.
So, $\pi(g\sigma(a))=\pi(g'\sigma(a'))$ i.e., $\pi(\sigma(a))=\pi(\sigma(a'))$ i.e., $a=a'$.
As the action of $G$ on $M$ is free, $g\sigma(a)=g'\sigma(a')=g'\sigma(a)$ implies $gg'$.
Thus, $\varphi :U\times G\rightarrow \pi^{-1}(U)$ is injective.
Let $x\in \pi^{-1}(U)$ then, $\pi(x)\in U$. This suggests to consider $(\pi(x),g)\in U\times G$ such that $\varphi(\pi(x),g)=x$ (This is with gut feeling that $\varphi$ is surjective.) i.e., $g\sigma(\pi(x))=x$. But, how do we know there exists $g\in G$ such that $g\sigma(\pi(x))=x$? We have not assumed that the action is transitive. Am I missing something?
Suppose such $g$ exists then it is clear that $\varphi$ is smooth bijective map. Its inverse $\pi^{-1}(U)\rightarrow U\rightarrow G$ is given by $x\mapsto (\sigma(x),g)$ where $g\in G$ is such that $g\sigma(\pi(x))=x$ and I do not see how one can show that this inverse is smooth. It is smooth in first coordinate as it is just the map $\pi$. How do we prove that $\pi^{-1}(U)\rightarrow G$ given by $x\mapsto g$ where $g$ is such that $g\sigma(\pi(x))=x$ is a smooth map.
Any suggestions are welcome.
You need to check that $d\phi_{(u,g)}$ is injective for all $u\in U, g\in G$. Consider the restriction of this map to $T_uU$ and notice that it is injective (since it is the differential of a local section), then restrict to $T_gG$ and notice its injectivity (due to the constant rank theorem), lastly notice transversality of $d\phi(T_uU)$ and $d\phi(T_gG)$ (again due to the fact that the map comes from a local section).