Proposition 3.5, page 5: Suppose $P \rightarrow B$ is a principal $G$-bundle, and let $H$ be an admissible subgroup of $G$. Then the quotient map $P \rightarrow P/H$ is a principal $H$-bundle.
Definition: Call a subgroup $H$ of $G$ admissible if the quotient map $G \rightarrow G/H$ is a principal $H$-bundle.
The proof given goes as follows: we can identify the map with the map, $$P \times_G G \rightarrow P \times_G (G/H)$$ and we utilize the fact that $H$ of $G$ is admissible.
Now I could not really spell out the maps of local trivilization, we want open set of $P \times_G (G/H)$ such that its preimage, in $P \times_G G$ has a local trivilization.
Justifying this seems like a mess. I would like to see some neat approach.
My attempt goes as follows.
Since the quotient map is an open map, take a local trivilization, open set $U$ of $G/H$, so $P \times U$ has open image in $P \times_G(G/H)$, where $\pi:G \rightarrow G/H$, under quotient $t:P \times G/H \rightarrow P \times_G (G/H)$
Consider its preimage, which I believe should be the image of $P \times \pi^{-1}(U)$ under the quotient. $s:P \times G \rightarrow P \times_G G$
So we show that we have a $G$-homeomorphism,
$$ s(P \times \pi^{-1}(U)) \rightarrow t(P \times U) \times (P \times_G G/H)$$
Everything here seems messy. But is this right?