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While reading about connections I stumbled on this entry of the encyclopedia of math. In the second paragraph of the comments section it's written that if a fiber bundle admits coherent isomorphisms of its fibers, where "coherent" means "satisfying the cocycle condition", then the fiber bundle is actually trivial.

What's the geometric intuition behind this fact, and how is it proved?

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The notion that a bundle is "trivial" just means it is isomorphic to the product bundle. In your case what is involved is the fiber over the base space changing continuously with the choice of the base. The author is basically saying that if one can show isomorphism of the fiber from one point in the base "continuously" or "transitively" to the other, then the bundle is the same as the product bundle.

To prove it suffice to take a neighborhood over a fixed point $b$,using Zorn's lemma to find a largest open set $U$ containing $b$ over which the bundle is trivial. Then pick up a point on the boundary of $U$ and a small neighborhood around it. By the cocycle condition all the fibers above points over $U$ which has been identified vertically over the base can be "moved" to the new small neighborhood. Thus if we incorporate the small neighbhorhood into $U$ to enlarge the fiber bundle using the isomorphism of fibers, we face no difficulty. This contradicts the maximality of $U$. So $U=B$ and $E$ is isomorphic to the product bundle as desired.

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  • $\begingroup$ Is there perhaps a proof which does not require dealing with boundaries or the axiom of choice? $\endgroup$
    – Arrow
    Jan 13, 2017 at 12:17
  • $\begingroup$ @Arrow: The argument will work by adjoining one open set at a time, but I think some inductive procedure (and taking maximal elements) will generally be needed. $\endgroup$ Jan 15, 2017 at 0:15

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