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From Wikipedia:

A fiber bundle consists of the data $(E, B, π, F)$, where $E, B, $and $F$ are topological spaces and $π : E → B$ is a continuous surjection satisfying a local triviality condition outlined below:

for every $x$ in $E$, there is an open neighborhood $U ⊂ B$ of $π(x)$ (which will be called a trivializing neighborhood) such that $π^{-1}(U)$ is homeomorphic to the product space $U × F$, in such a way that $π$ carries over to the projection onto the first factor.

  1. I was wondering why the local triviality condition (the second paragraph) is initiated from "every $x$ in $E$"? In other words, can it be instead initiated from $B$ as follows:

    there is an open cover of $B$ such that each open subset in the cover is homeomorphic to the product space $U × F$, in such a way that $π$ carries over to the projection onto the first factor.

  2. What does "the first factor" mean?
  3. Generally, what does "a mapping carries over to another mapping onto another thing" mean?

Thanks and regards!

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up vote 2 down vote accepted

(1) Yes, your formulation is easily seen to imply and be implied by (hence be equivalent to) Wikipedia's one.

(2) In this case "the first factor of $U\times F$" means simply $U$ -- the first of the two factors that were multiplied to get the product space. So the condition is that there is a $\psi: \pi^{-1}(U) \to U \times F$ which is a homeomorphism and satisfies that if $\psi(e)=\langle u,f\rangle$ then $\pi(e)=u$. One might also phrase the condition as: "... there exists $\phi: \pi^{-1}(U) \to F$ such that the map $e\mapsto \langle\pi(e),\phi(e)\rangle$ is a homeomorphism $\pi^{-1}(U) \to U \times F$.

(3) I don't think "carries over" is a technical term with a distinct meaning here, apart from "such that the obvious diagram one might draw commutes".

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Thanks! +1. (3) I think it is one of the jargons used by mathematicians, which I have never been able to understand. –  Tim Jan 22 '12 at 20:26
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