Let $M$ be a manifold, $G,H$ be some Lie groups, $\sigma:G\to H$ be a Lie group homomorphism, $K\subset H$ a maximal compact subgroup of $H$ and $\tilde K:=\sigma^{-1}(K)\subset G$ . Let further
- $\tilde P\to M$ be a principal $G$-bundle,
- $P\to M$ be a principal $H$-bundle,
- $Q\to M$ be a reduction of $P$ to $K$ and
- $\varphi:\tilde P\to P$ be a fiber-preserving map, such that $$ \varphi(\tilde p.g)=\varphi(\tilde p).\sigma(g),\qquad\forall\tilde p\in\tilde P,\ g\in G.$$
Since I can think of $Q$ as a submanifold of $P$, my question is whether the preimage $\varphi^{-1}(Q)\subset \tilde P$ is a principal $\tilde K$-bundle?
$\tilde K$ acts on $\phi^{-1}(Q)$: let $k\in \tilde K$, $y\in \phi^{-1}(Q)$, since $k\in G$ and $y\in P$, we can define $yk$, $\phi(yk)=\phi(y)\sigma(k)$ since $\phi(y)\in Q$ and $\sigma(k)\in K$, we deduce that $\phi(y)\sigma(k)\in Q$ and $yk\in \phi^{-1}(Q)$.
Let $p:\phi^{-1}(Q)\rightarrow M$ be the restriction of $r:\tilde P\rightarrow M$, $q:Q\rightarrow M$ and $y,y'\in \phi^{-1}(Q)$. Suppose $p(y)=p(y')$, this implies $q(\phi(y))=q(\phi(y'))$. Thus there exists $k\in K$ such that $\phi(y')=\phi(y)k$ since $p(y)=r(y)=p(y')=r(y'),$ we deduce that $y'=yg, g\in G$. This implies that $\phi(yg)=\phi(y)\sigma(g)=\phi(y)k$. Since the action of $H$ is free, we deduce that $\sigma(g)=k$ and $g\in \tilde K$.
We have shown that $\phi^{-1}(Q)$ is endowed with a free action of $\tilde K$ and this action is transitive on the fibres of $\phi^{-1}(Q)\rightarrow M$, thus it is a principal $\tilde K$ bundle.