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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?

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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.

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  • $\begingroup$ Thanks! Where can I find the proof of the fact that $f$ is smooth if and only if $fp$ is smooth? $\endgroup$ – Marco All-in Nervo Jun 28 at 18:46
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    $\begingroup$ This question is about that property and mentions that it is stated in Jack Lee's Introduction to Smooth Manifolds. Maybe you can find a proof there. $\endgroup$ – Paulo Mourão Jun 28 at 18:54
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    $\begingroup$ If you don't, then this question discusses the fact that, given a submersion, you can always find coordinates such that its local representation is the canonical submersion (also known as the submersion theorem). That local representation should be enough to prove the result. $\endgroup$ – Paulo Mourão Jun 28 at 19:08

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