The following is a definition of a vector bundle isomorphism.
Let $E,F$ be vector bundles over a base manifold $M$. A smooth map $u:E\rightarrow F$ is a an isomorphism of vector bundles if
$u$ respects the fibres, that is for all $x\in M$ it follows that $u_x:=u|_{E_x}:E_x\rightarrow F_x$.
for each $x\in M$ the map $u_x$ is a vector space isomorphism.
It is not obvious to me how this implies that there is a smooth inverse $u^{-1}:F\rightarrow E$.It's clear that it has something to do with vector bundles being locally trivial but I am not sure how to complete my arguement.
Let $U$ be a neighborhood on which $E$ and $F$ are trivial. Then there are diffeomorphism $\phi:E_U\rightarrow U\times \mathbb{R}^n$ and $\psi:F_U\rightarrow U\times \mathbb{R}^n$. but since clearly $U\times \mathbb{R}^n \cong U\times \mathbb{R}^n$ it follows that $E_U \cong F_U$. So we have shown that $E$ and $F$ as manifolds are locally diffeomorphic?
I'm not sure if this is correct, and I haven't shown that $u$ has a smooth inverse.
Best Answer
This is essentially the same as Showing map is vector bundle isomorphism. The difference is that you work in the smooth category.
$GL(n,\mathbb R)$ is smooth submanifold of $End(\mathbb R^n) \approx \mathbb R^{n^2}$ and it is well-known that for product bundles the assignment $\phi \mapsto \phi''$ given in my answer to the linked question establishes a bijection between smooth bundle maps $\phi$ ansd smooth maps $\phi'' : B \to GL(n,\mathbb R)$ (provided $B$ is a smooth manifold).
Since inversion in $GL(n,\mathbb R)$ is smooth, we see that $\phi^{-1}$ is smooth.
This transfers to smooth bundle maps between trivial bundles.
Thus, if $u : E \to F$ is a general smooth bundle map, we see that $u^{-1} : F \to E$ is locally smooth. But local smoothness implies global smoothness. This follows from the definition of a smooth map - it is a local property. You can also find it in any good textbook.