Let $G, G\_1, G\_2$ be compact Lie groups with homomorphisms $f\_1:G\_1 \to G$ and $f\_2: G\_2\to G$. Let $P\_1, P\_2$ be principal bundles for $G\_1,G\_2$ and assume that the bundles $P\_i\times\_{G\_i} G$ are both isomorphic (by fixed isomorphisms) to a bundle $P$.


Let now $H$ be the pullback of the group diagram given by $f\_1$ and $f\_2$. Let $Q$ be the (topological) pullback of the induced diagram given by $P\_1$, $P\_2$ and $P$.


Now $Q$ has "fibres" equivalent to $H$, but does $Q$ always form a (locally trivial) principal $H$-bundle?



pullback diagram of principal bundles
In the stated generality, it is false; for example, suppose that $G\_1$ and $G\_2$ are trivial groups, $P = B \times G$ (here $B$ is the base), and the maps $P\_1 \to P$ and $P\_2 \to P$ are given by two disjoint sections $B \to P$. In this case $Q$ is empty.


On the other hand, it is easy to see that the answer is positive if one of the $f\_i$ is surjective (if $f\_1$ is surjective, choose a local section of $P\_2$, and locally lift the composite section of $P$ to a section of $P\_1$, obtaining a local section of $Q$, which gives an local isomorphism of $Q$ with $B \times H$).


[Edit] The point is that the map $f\_1: G\_1 \to G$, as a surjective map of Lie groups, is a fibration; hence $P\_1 \to P$ is also a fibration. If $B\to P$ is a section, the pullback of $P\_1$ to $B$ is a fibration, hence it has local sections.


