I know that given a continuous map $f: X \to Y$ and a bundle $F \to Y$ then the map $f^*F \to F$ is a fibrewise homeomorphism.
I vaguely remember reading a converse to this: if we have a fibre bundle map $\phi: E \to F$ covering $f: X \to Y$ on the base and $\phi$ is a fibrewise homeomorphism, then this is a pullback square, i.e. $E \cong f^*F$.
If this is indeed true, does anyone happen to know of a reference for this? Husemoller is my go-to reference for fibre bundle theory, but I couldn't find anything in there. If this is false, is there a straightforward counterexample?
I have played around and figured out a proof, but I would still appreciate a textbook reference for this, if anyone knows of one.
As usual, we check that $E \to X$ satisfies the universal property of the pullback.
Given another fibre bundle $V \to X$, with a bundle morphism $\psi: V \to F$ covering $f: X \to Y$ on the base, a bundle morphism $\theta: V \to E$ covering $1_X: X \to X$ on the base which makes the diagram commute (if it exists) must satisfy $$\phi_x \circ \theta_x = \psi_x: V_x \to F_{f(x)}$$ on the fibre over any point $x \in X$. But since $\phi_x: E_x \to F_{f(x)}$ is a homeomorphism then we must have $$\theta_x = \phi_x^{-1} \circ \psi_x.$$ Hence, such a morphism $\theta$, if it exists, is necessarily unique. To show existence, it suffices to check that the fibrewise formula for $\theta$ defines a continuous map. I haven't checked continuity yet, but I'm confident it will work.