Principal bundle map is fiber homeomorphism

Question

let $B_1(\mathcal{P}_1:P_1\rightarrow X_1)$ and $B_2$ be two principal G-bundles and let $\tilde f:P_1 \rightarrow P_2$ be a principal bundle map. I want to prove that $\tilde f$ carries each fiber of $\mathcal{P}_1$ homeomorphically onto a fiber of $\mathcal{P}_2$. I managed to prove that it is continuous and bijective but I don't know how to prove the last step.

Thanks,

Simon

P.S.: A principal bundle map from $B_1$ to $B_2$ is a continuous map $\tilde f:P_1\rightarrow P_2$ such that $\tilde f(p\cdot g) = \tilde f(p)\cdot g$ for all $p\in P_1$ and $g\in G$, i.e. it preserves fibers.

Answer 1

I believe that for a principle bundle the fibre is homeomorphic to $G$. Thus it is now easy to see that $f$ restricted to a single fibre is just multiplication by an element of $G$.

Answer 2

Consider $\{U_j\}_{j \in J}$ a trivializing cover of $X_1$ and $\{V\}_{k \in K}$ a trivializing cover $X_2$. Then there exist homeomorphisms $\Psi^1_{j} : \mathcal P_1^{-1} (U_j) \to U_j \times G $ and $\Psi^2_{k} : \mathcal P_2^{-1}(V_k) \to V_k \times G$ with $$\Psi^1_{j}= (\mathcal P_1 , \psi^1_j) \ \ \text{and} \ \ \Psi^2_{k}= (\mathcal P_1 , \psi^2_k)$$ where $\psi^1_j$ takes $\mathcal P_1^{-1}(\mathcal P_1 (p))$ homeomorphically to $G$. Now notice that $\tilde f (\mathcal P_1^{-1}(\mathcal P_1 (p))) = \mathcal P_2^{-1}(\mathcal P_2 (\tilde f (p))$ since $\tilde f$ preserves fibers. Thus we have the following commutative diagram

$$\require {AMScd} \begin{CD}\mathcal P_1^{-1}(\mathcal P_1 (p)) @>\tilde f>>\tilde f (\mathcal P_1^{-1}(\mathcal P_1 (p))) = \mathcal P_2^{-1}(\mathcal P_2 (\tilde f (p))\\@V \psi^1_j VV @AA(\psi^2_k)^{-1}A\\G @>>id_G >G\end{CD}$$ since $\psi^1_k,id_G$ and $(\psi^2_k)^{-1}$ are homeomorphisms it follows that $\tilde f$ restrict to $\mathcal P_1^{-1}(\mathcal P_1 (p))$ is a homeomorphism over $\mathcal P_2^{-1}(\mathcal P_2 (\tilde f (p))$.