Let us recall that a smooth bundle is a triple $(E,\pi,M)$ where $E,M$ are smooth manfiolds and $\pi : E\to M$ a smooth surjection. Consider the following two definitions of a fibre bundle:
A fibre bundle with typical fibre $F$ is a bundle $(E,\pi,M)$ such that for every $x\in M$ there is a diffeomorphism $\phi_x : \pi^{-1}(x)\to F$.
A fibre bundle with typical fibre$F$ is a bundle $(E,\pi,M)$ such that for every $x\in M$ there is an open set $U\subset M$ with $x\in U$ and a diffeomorphism $\varphi : U\times F\to \pi^{-1}(U)$ such that $\pi\circ \varphi(x,a)=x$.
The first definition simply states that every fibre is diffeomorphic to a standard manifold $F$.
The second definition states that the bundle is localy trivial: around every point there is a neighborhood such that the bundle is isomorphic to the trivial bundle $U\times F$.
I've read that these two are equivalent.
That (2) implies (1) is obvious, by just restricting $\varphi$ to $\{x\}\times F$.
The issue is with (1) implying (2). I mean, what I don't get, is how do we get around each $x$ a neighborhood with the required property. What the map will be seems (at least by now) clear. We must set $\varphi : U\times F\to \pi^{-1}(U)$ to be
$$\varphi(x,v)=\phi_{x}^{-1}(v).$$
But where the $U$ comes from seems tricky.
Is it true that these two definitions are equivalent? How can we prove that (1) implies (2)?
They are not equivalent. For instance, let $M=\mathbb{R}$ and let $E$ be the disjoint union of all open intervals in $\mathbb{R}$ with rational endpoints, with $\pi:E\to M$ the inclusion map on each such interval. Then $(E,\pi,M)$ satisfies definition (1) with $F=\mathbb{Z}$ (each fiber is countably infinite and discrete) but does not satisfy definition (2) (this takes a bit of work to prove).
Definition (2), in my experience, is far more commonly used than definition (1).