Suppose $\alpha:A\to B$ is a fiber bundle. Suppose for every pair of points $b_1,b_2\in B$ there's an isomorphism $\phi_{b_2b_1}:\alpha^{-1} \left\{ b_1 \right\} \cong \alpha^{-1} \left\{ b_2 \right\} $ and that these isomorphisms satisfy the cocycle condition $\phi_{b_3 b_2}\circ \phi_{b_2 b_1}=\phi_{b_3 b_1}$. Why does it follow that $\alpha$ is a trivial bundle?

@Arrow: This seems unlikely unless you have some sort of smoothness in these isomorphisms as the points vary. The usual thing is to show that the Čech 1-cocycle (for an open cover of trivializations) is a coboundary.

@TedShifrin Thank you for the reply. I'm trying to understand the second paragraph of the comment section here, that's all. What's the correct statement and how do I prove the assertion?

Well, you certainly would have isomorphisms between a fixed fiber and every other. But to have a trivial bundle, you need a smooth diffeomorphism $B\times F\to E$, and I don't see how they arrive at that from compatible pointwise isomorphisms.

P.S. What a horrible site for math symbol graphics.