It is known (see for example Corollary 2.6 of The Topology of Fiber Bundles Lecture Notes by Ralph Cohen) that a fiber bundle $\pi : E \rightarrow B$ with fiber $F$ over a contractible space $B$ is trivial i.e. there exists a homeomorphism $\varphi : E \rightarrow F \times B$ such that $\pi = p_2 \circ \varphi$ where $p_2 : F \times B \rightarrow B$ is the natural projection.
My question is, if $B$, $E$ and $F$ are smooth manifolds (without boundaries) and $\pi$ is a smooth fiber bundle (i.e. $\pi$ is a smooth map and the local trivialisations can be chosen to be smooth diffeomorphism), can we choose $\varphi$ to be a smooth diffeomorphism (or at least a $\mathcal{C}^1$ diffeomorphism) ?
The question was asked for vector bundles here : Smooth vector bundles over contractible smooth manifolds with or without boundary but I need the answer for any smooth fiber bundle. John Lee answered "yes" in this post (for the case of vector bundles) by saying that we can transform the homeomorphism $\varphi$ into a close smooth diffeomorphism $\tilde{\varphi}$ using locally the Whitney approximation theorem. I have his book and I don't see how to use this theorem ; in particular how can we be sure that $\tilde{\varphi}$ is still injective ?
Here is the Whitney theorem from Lee's book,
Theorem 6.21 (Whitney Approximation Theorem for Functions).
Suppose $M$ is a smooth manifold with or without boundary, and $F : M \rightarrow \mathbb{R}^k$ is a continuous function. Given any positive continuous function $\delta : M \rightarrow \mathbb{R}$, there exists a smooth function $\tilde{F} : M \rightarrow \mathbb{R}^k$ that is $\delta$-close to $F$. If $F$ is smooth on a closed subset $A \subset M$; then $\tilde{F}$ can be chosen to be equal to $F$ on $A$.
Thank you for your help !