I am starting to learn about fiber bundles, and wanted to understand the following.
We consider a fiber bundle $(E,B,\pi,F)$ where $E,B,F$ are smooth manifolds, and $\pi : E \to B$ is a continuous surjection satisfying the local triviality condition.
We say a fiber bundle $(E,B,\pi,F)$ is trivial if $E$ is isomorphic to $B\times F$. I wanted to understand how we may prove that triviality of a fiber bundle occurs if and only if we have a smooth map $f : E \to F$ where the restriction of $f$ to any fiber $E_b = \pi^{-1}(b)$ with $b \in B$ gives that $f\mid_{E_b} : E_b \to F$ is a diffeomorphism.
Triviality of the fiber bundle gives us a isomorphism from $E$ to $B\times F$, but how might one reduce this to a smooth function from $E$ to $F$ with the given property? Even further, how might one deduce the converse?
If $E$ is trivial, and $\varphi : E \to B\times F$ is a bundle isomorphism, then the map $f := \operatorname{proj}_2\circ\,\varphi : E \to F$ has the property that $f|_{E_b} : E_b \to F$ is a diffeomorphism for every $b \in B$.
Conversely, if $f : E\to F$ is a map with the property that $f|_{E_b} : E_b \to F$ is a diffeomorphism for every $b \in B$, then $\varphi := (\pi, f) : E \to B\times F$ is an isomorphism of bundles.
In summary, the maps mentioned above are related by the following commutative diagram.
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