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

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?

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.
• Thanks! Where can I find the proof of the fact that $f$ is smooth if and only if $fp$ is smooth? – Marco All-in Nervo Jun 28 at 18:46