I am trying to show the hint to this problem. The set-up is as follow
Suppose 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 $\pi:M\rightarrow M/G$ is a smooth submersion.
and the hint is
show that for any smooth local section $\sigma: U \rightarrow M$ of $\pi$, the map $\left(g, x\right)\mapsto g\cdot\sigma(x)$ is a diffeomorphism from $G\times U$ to $\pi^{-1}(U)$.
I am able to show that $\left(g, x\right)\mapsto g\cdot\sigma(x)$ is a smooth (left-) equivariant bijection so I know what follows is that its inverse $g\cdot\sigma(x) \mapsto \left(g, x\right)$ is well-defined and equivariant. However, I am not sure how to show the smoothness of this inverse.
In my solution leading up to the smoothness and equivariance of $\left(g, x\right)\mapsto g\cdot\sigma(x)$, I have used smoothness and freeness of the action $G\times M\rightarrow M$. As such my guess would be to use the smooth structure on $M/G$ that makes $\pi$ a submersion. But besides admitting the existence of such section $\sigma$, I am not sure how else to manipulate this smooth structure of $M/G$'s.
Best Answer
Given $p\in M$, you have $G\to M\to M/G$, where the first map is $g\mapsto g\cdot p$ (call this map $f$). Taking the differential $$\mathfrak{g} \to T_pM\to T_{\pi(p)}(M/G).$$ Since $\pi$ is a submersion the second map is surjective, and the composition is obviously zero, so the first map is injective, and its image is in $\ker d\pi$, and so is $\ker d\pi$ by dimensional considerations.
Differentiating a local section of $\pi$ at $\pi(p)$ you get a section of $d\pi$, whose image must be transversal to $\ker d\pi= df(\mathfrak{g})$. So the map $(g,x)\mapsto g\cdot\sigma(x)$ has a differential at $(e,\pi(p))$ that takes $\mathfrak{g}$ to $f(\mathfrak{g})$ and $T_{\pi(p)}(M/G)$ to a subspace transversal to it. So it's an isomorphism, and therefore the function is a local diffeomorfism