# Topological covering + local diffeomorphism gives smooth covering

I got stuck at some point while working on this part of an exercise from Lee's Introduction to Smooth Manifolds, 2nd edition.

The part which I am stuck on is to prove (one of the directions of proposition 4.33(c)): A topological covering map is a smooth covering map if it is a local diffeomorphism.

What I have is as follows. Let $\pi : E \to M$ be a topological covering map. Let $q\in M$. By assumption, we have some open subset $V \subset M$ and collection of disjoint open subsets $\{ U_\alpha \}$ of $E$ such that each $U_\alpha$ is homeomorphic to $V$ under $\pi$. Further, let $p_\alpha \in U_\alpha$ be such that $\pi(p_\alpha)=q$. Now, for each $\alpha$, I can find open neighbourhoods $\hat{U}_\alpha$ of $p_\alpha$ and $\hat{V}_\alpha$ of $q$ such that $\hat{U}_\alpha$ and $\hat{V}_\alpha$ are diffeomorphic. We may demand that $\hat{U}_\alpha \subset U_\alpha$.

The problem is I need one single neighbourhood of $q$ that lifts diffeomorphically, but all I have is the collection $\hat{V}_\alpha$ and taking their intersection does not necessarily give an open set.

How should I correctly approach the problem?

An alternative idea I have: Repeat the first three lines of the first proof. Now for some $\alpha '$, define $\hat{U}_{\alpha '}$ and $\hat{V}_{\alpha '}$ as above. Then lift $\hat{V}_{\alpha '}$ using $\pi^{-1}$ to all other $U_\alpha$ to obtain $\hat{U}_\alpha$. This time I know that each $\hat{U}_\alpha$ is smoothly homeomorphic to $\hat{V}_{\alpha '}$.

The key is that a bijective local diffeomorphism is a diffeomorphism. So once you're finished those first three lines of your proof, use this fact, and then actually each $U_\alpha$ is diffeomorphic to $V$ under $\pi$. See here or here on math.SE for that fact.
Using the reference added in the comment and aes suggestion I found this way to prove it. Let $\pi:E\to M$ be a (topological) covering map. Take $q\in M$, there exists a neighborhood $U\subset M$ such that each connected component of $\pi^{-1}(U)$ is homeomorphic to $U$. Let $V$ be an arbitrary component. The map $\pi|_V$ is bijective (because homeomorphic) and locally diffeomorphic, therefore it is a diffeomorphism $\pi|_V:V\to U$. This shows that $\pi$ is diffeomorphic on each component and then it is a smooth covering map.