In a principal $G$-bundle $\pi: P \to M$ there is a right action on $P$ such that $M \cong P/G$

Question

Suppose we have a principal $G$-bundle $\pi: P \to M$. The definition of principal bundle is: a fiber bundle with a $G$-structure, fiber $G$ and action on fiber given by left multiplication. We can define a right action on $P$ and prove that this is free and proper. So $P/G$ is a manifold. I want to prove that $P/G \cong M$. I can define the bijection $[x] \to \pi(x)$, but I don't how to prove that it is $\mathcal{C}^\infty$. Any help?

Answer 1

From the action being free and proper, you have a smooth structure on $P/G$ such that the standard projection $p:P\to P/G$ is a submersion.

With $p$ being a surjective submersion, you have the property that a function $f:P/G\to M$ is smooth if and only if $f\circ p$ is smooth. If $f$ is given by $f([x])=\pi(x)$, then $f\circ p=\pi$, which is smooth.

Furthermore, $\pi$ is also a surjective submersion (if this is not clear to you, you can check out this question), so you can apply the same argument for the inverse.