Questions for understanding fiber bundle definition 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.


I was wondering why the local triviality condition (the second
paragraph) is initiated from &quot;every $x$ in $E$&quot;? 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.


What does &quot;the first factor&quot; mean?

Generally, what does &quot;a mapping carries over to another mapping onto
another thing&quot; mean?


Thanks and regards!
(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".
Tim's suggested definition of a fibre bundle is in fact not equivalent to the standard definition. For a counterexample -

Let $C_0$ be the standard Cantor discontinuum.
Let $C_n=C_0+n$ - The Cantor discontinuum translated by $n$.

Consider the following subspaces of $\mathbb{R}^2$

$$A_0=\{0\} \times C_0$$
$$A_{n,(n\geq1)}=\left\{\frac{1}{n}\right\} \times C_0$$
$$D_{n,(n\geq1)}=\left\{\frac{1}{n}\right\} \times C_{n+1}$$

Let $E$ be the union of $A_0$ and all of the $A_n$ and $D_n$.

Let $B = \{(0,0)\} \cup \left\{\left(\frac{1}{n},0\right)|n\geq1\right\}$

$E$ and $B$ are taken with the subspace topologies from $\mathbb{R}^2$.

Let $F$ be the standard Cantor discontinuum.

Now define $\pi: E \to B$ as the projection $(x,y) \mapsto x$

Note that the disjoint union of two copies of the Cantor discontinuum is
homemorphic to the Cantor discontinuum. So all fibres of $\pi$ are homeomorphic.

Then $E$,$B$,$\pi$,$F$ satisfy Tim's suggested definition of a fibre bundle.

But $\pi$ in fact is not locally trivial at $(0,0)$.