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Suppose $\pi:P\rightarrow M$ is a principal $G$ bundle.

Let $H$ be a Lie group acting freely and properly on $P$ and on $M$ so that $P/H$ and $M/H$ are manifolds. Further assume this action is such that it defines a map $P/H\rightarrow M/H$.

Is it always true that the induced map $P/H\rightarrow M/H$ is also a principal $G$ bundle?

I think it is true, may be with some extra conditions.

Can some one please say what extra conditions (if at all) I need to confirm $P/H\rightarrow M/H$ is a principal $G$ bundle?

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In order for $P/H\to M/H$ to be a principal $G$-bundle, we need to have a well-defined, free $G$-action on $P/H$. This is the case if and only if $H$ acts on $P$ by morphisms of $G$-spaces, that is, the $H$ action commutes with the $G$ action.

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  • $\begingroup$ Ok. I am assuming $G$ action commutes with $H$ action... Then, it is true right? With no extra conditions it is true.. $\endgroup$ – user537667 Feb 26 at 11:31

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