The inverse of a smooth bundle map is smooth. Consider smooth fiber bundles $P_i:E_i\to B_i,\quad i=1,2\quad$ with fiber $F$. Let $\tilde f:E_1\to E_2$ be a smooth bundle map. That is a smooth map which preserves the fiber $F$. Let $f:B_1\to B_2$ be the unique map such that $P_2\circ\tilde f=f\circ P_1$.
If $f$ is a diffeomorphism, we can see that $\tilde f$ is a bijection. How can we see that $\tilde f^{-1}$ is smooth?
Moreover, if $\tilde f$ is a diffeomorphism we can see that $f$ is a bijection. How can we see that $f^{-1}$ is smooth?
Observation: If $g:X\to Y$ is a continuous bijection between a compact space $X$ and a Hausdorff space $Y$, then $f^{-1}$ is continuous. However, neither $E_1$, nor $B_1$ are compact.
Addendum: I am reading Morita's "Geometry of Differential Forms." His definitions are like the following.
Def $1$: (Differentiable fiber bundle) Let $F$ be a smooth manifold. Suppose $E$ and $B$ are smooth manifolds and $P:E\to B$ is a smooth map. We say that $K=(E,P,B,F)$ is a differentiable fiber bundle with fiber $F$ is it satisfies the local triviality condition. That is $(\forall b\in B) (\exists b\in U_{\text{open}}\subset B)(\exists \phi:P^{-1}(U)\tilde = U\times F)(\forall u\in P^{-1}(U))(P(u)=P_1\circ\phi(u))$, where $P_1$ is the projection onto the first factor.
Def $2$: (Bundle map) Let $P_i:E_i\to B_i,\quad i=1,2\quad$ be differentiable fiber bundles with the same fiber. By a bundle map from $P_1$ to $P_2$ we mean smooth maps $\tilde f:E_1\to E_2$ and $f:B_1\to B_2$ such that $P_2\circ\tilde f=f\circ P_1$ and such that the restriction of $\tilde f$ to any fiber $P_1^{-1}(b)$ is a diffeomorphism.
 A: In a trivializing neighborhood $U \times F$ (using the notation of your comment), we may write
$$
\tilde{f}(b, e) = \bigl(f(b), \phi(b, e)\bigr)
$$
for some smooth mapping $\phi:U \times F \to F$ that is a diffeomorphism on each fibre $\{b\} \times F$. The Jacobian has a block decomposition
$$
D\tilde{f}(b, e)
  = \left[\begin{array}{@{}cc@{}}
  Df(b) & D_{1}\phi(b, e) \\
  0     & D_{2}\phi(b, e) \\
  \end{array}\right],
$$
in which $D_{2}\phi$ is invertible at each point $u = (b, e)$ by the definition of a bundle map. It follows that $D\tilde{f}$ is invertible at each point if and only if $Df(b)$ is invertible at each point.
You may be able to express this in matrix-free form, but matrices do highlight the main idea: $\tilde{f}^{-1}$ "is a diffeomorphism in the fibre direction" by hypothesis, so any failure of smoothness must occur "in the base direction". This happens if and only if $f^{-1}$ is not smooth, since $P_{1} \circ x_{1}^{-1} \circ \tilde{f}^{-1} = f^{-1} \circ P_{2} \circ x_{2}^{-1}$.
