Induced map on associated bundles

Let $P\to M$ and $P'\to M'$ be principal G-bundles. A principal bundle morphism consists of a smooth map smooth map $f:M\to M'$ and a smooth map $\Phi:P\to P'$ such that

\pi'\circ \Phi = f\circ \pi,\qquad \Phi(pg)=\Phi(p)g\ \text{ for all }g\in G.

Let $F$ and $F'$ be left $G$ -spaces, and let $\psi:F\to F'$ be $G$ -equivariant.

Construction. Define

\Phi\times_G \psi : P\times_G F \to P'\times_G F',\qquad [p,u]\mapsto [\Phi(p),\psi(u)].

This is well-defined (independent of the representative $(p,u)$ ) and is a smooth bundle map covering $f$ :

\pi_{F'}\circ (\Phi\times_G\psi) = f\circ \pi_F.

When $F=F'$ and $\psi=\mathrm{id}$ , one gets the induced map on a fixed associated bundle functorially from $\Phi$ .

Examples

(Pullback trivialization maps) If $P'=P$ and $f=\mathrm{id}_M$ , then a gauge transformation $\Phi$ induces a bundle automorphism of any associated bundle $P\times_G F$ by $[p,u]\mapsto [\Phi(p),u]$ .
(Vector bundle case) For $F=V$ and $F'=V'$ linear $G$ -spaces, a $G$ -equivariant linear map $\psi:V\to V'$ yields a vector bundle morphism $P\times_G V\to P'\times_G V'$ covering $f$ .
(Adjoint functoriality) Taking $F=F'=G$ with conjugation and $\psi=\mathrm{id}$ gives a morphism of adjoint bundles induced from any principal bundle morphism.