Showing smooth structure of a covering space of a smooth manifold is unique

Question

Let $M$ be a smooth manifold, and let $p:E\to M$ be a topological covering map. Then $E$ has a smooth structure so that $p$ becomes a smooth covering map. (This is shown in Proposition 4.40 of Lee's Introduction to Smooth Manifolds). Next I want to show that such a smooth structure of $E$ is unique. How do I have to start to do this? I should suppose $E$ has two different smooth structures and smooth covering maps $p,q:E \to M$ with respect to the two smooth structures, respectively, and then what I have to do? A smooth structure is by definition a maximal smooth atlas. Do I have to show that two atlases are the same by showing inclusions?

Answer 1

You should not introduce a new covering map $q$.

What you are asked to prove is that if $\mathcal T_1,\mathcal T_2$ are maximal smooth atlases for $E$, if $p : E \to M$ is a smooth map with respect to $\mathcal T_1$, and if $p$ is also smooth with respect to $\mathcal T_2$, then $\mathcal T_1 = \mathcal T_2$.

To prove that, choose $x \in E$, and choose an evenly covered open neighborhood $U \subset M$ of $p(x)$ which is the domain of a chart $\phi : U \to \mathbb R^n$ for the given smooth atlas of $M$.

Let $V \subset E$ be an open neighborhood of $x$ such that $p$ restricts to a homeomorphism $p \mid V : V \mapsto U$.

It follows that $\phi \circ (p \mid V) : V \to \mathbb R^n$ is a smooth map with respect to atlas $\mathcal T_1$ and with respect to atlas $\mathcal T_2$, and therefore $\phi \circ (p \mid V)$ is a chart in both of those atlases. One therefore has a cover of $E$ by charts that are in both atlases, so the atlases are equal.