I was reading Lee's smooth manifold,there is a proposition in page 262:
Let $E,E'$ be two smooth vector bundle over $M$,if the bundle homorphism between them $F:E\to E'$ is bijective smooth then it's diffeomorphism.
The problem is :Can we generalize it to $E$ be bundle over $M$ and $E'$ be bundle over $M'$ with bundle homorphism covers $f:M\to M'$.Can we claim that if $F:E\to E'$ is bijective smooth then it's differomorphism?I think it still holds.
Consider a smooth function $f\colon M\to N$. If $X$ is an arbitrary manifold, then the map $F\colon X\times M\to X\times N$ given by $F(x,p)=(x,f(p))$ is a fiber bundle homomorphism. But if $f$ is smooth, bijective, and not a diffeomorphism, then $F$ is smooth, bijective homomorphism, but it isn't an isomorphism.
In order to give an specific example: take the smooth manifold $\mathbb{R}$ and think of $\mathbb{R}\times \mathbb{R}$ as a trivial bundle. A possible counterexample can be $F(x,y)=(x,y^{3})$.
The reason the argument works for a vector bundle is that the bijective condition implies that, in fibers, $F$ is a linear isomorphism (hence a diffeomorphism). When you remove the vector structure, you only know that it is a smooth bijective map on the level of fibers.
FOR THE CASE WHEN OUR BUNDLES ARE VECTOR BUNDLES
Suppose $F\colon E\to E'$ is a vector bundle isomorphism over $f\colon M\to M'$. This means that there is a vector bundle homomorphism $G\colon E'\to E$ over some smooth function $g\colon M\to M'$ such that $G=F^{-1}$ as maps. In particular, $g=f^{-1}$, so $f$ is a diffeomorphism. On the other hand, $F_{p}\colon E_{p}\to E'_{f(p)}$ is a linear isomorphism with inverse $G_{f(p)}\colon E'_{f(p)}\to E_{p}$.
Now suppose $f$ is a diffeomorphism and $F_{p}$ is an isomorphism for every $p$ (in the case that $F$ is bijective, the second part is true). Then $F$ is bijective, and by using local trivializations you can see that $F_{*}$ has maximal rank at every point of $E$, so that $F^{-1}$ is a diffeomorphism (and a homomorphism).
EDIT: For completeness sake, I'm adding a proof of the fact that $F$ is a local diffeomorphism.
Let $p\in M$, $U\subseteq M$ and $V\subseteq M'$ trivial coordinate open subsets for $E$ and $E'$, with coordinate systems $\varphi\colon U\to \varphi(U)\subseteq \mathbb{R}^{n}$, $\psi\colon V\to \psi(V)\subseteq \mathbb{R}^{n}$ such that $f(U)\subseteq V$, let $\pi$, $\pi'$ be the projections, and $\alpha\colon E_{U}\to U\times \mathbb{R}^{k}$, $\beta\colon E'_{V}\to V\times \mathbb{R}^{k}$ trivializations for both bundles (with second components $\alpha_{2}$ and $\beta_{2}$). Let $\tilde{f}\colon \varphi(U)\to \psi(V)$ be the coordinate representation of $f$ (that is, $\tilde{f}=\psi\circ f \circ \varphi^{-1}$). Then $E_{U}$ and $E'_{V}$ are coordinate open subsets with the maps $\Phi\colon x\in E_{U}\mapsto (\varphi(\pi(x)),\alpha_{2}(x))$ and $\Psi\colon z\in E'_{V}\mapsto (\psi(\pi'(z)),\beta_{2}(z))$. A coordinate representation of $F$ with respect to $E_{U}$ and $E'_{V}$ is therefore given by $\tilde{F}=\Psi\circ F \circ \Phi^{-1}\colon \varphi(U)\times \mathbb{R}^{k}\to \psi(V)\times \mathbb{R}^{k}$.
We now compute $\tilde{F}$: let $(x,v)=(x^{1},...,x^{n},v^{1},...,v^{k})\in \varphi(U)\times \mathbb{R}^{k}$. We get:
$$ \tilde{F}(x,v)=\Psi \circ F \circ \Phi^{-1}(x,v)=(\psi(\pi'(F(\Phi^{-1}(x,v)))),\beta_{2}(F(\Phi^{-1}(x,v))))=(\psi(f(\pi(\Phi^{-1}(x,v)))),\beta_{2}(F(\Phi^{-1}(x,v))))=(\tilde{f}(x),\beta_{2}(F_{\varphi^{-1}(x)}(\Phi^{-1}(x,v))))=(\tilde{f}(x),A(x)v) $$
where $A(x)=\beta_{2}\circ F_{\varphi^{-1}(x)}\circ \Phi^{-1}(x,\cdot)$ is a linear isomorphism by hypothesis. The Jacobian matrix is therefore
$$ D\tilde{F}(x,v)=\begin{pmatrix} D\tilde{f}(x) & 0 \\ * & A(x) \end{pmatrix} $$
which is nonsingular since both diagonal blocks are nonsingular.
Hope this helps!