In my question, all the domain and codomain of any covering map is assumed to be connected and locally path-connected.
I'm trying to prove corollary 4.43 (whose proof is left as an exercise) of Lee's "Introduction to Smooth Manifolds" 2nd edition:
Corollary 4.43 (Existence of a Universal Covering Manifold). If $M$ is a connected smooth manifold, there exists a simply connected smooth manifold $\widetilde{M}$, called the universal covering manifold of $M$, and a smooth covering map $\pi:\widetilde{M}\to M$. The universal covering manifold is unique in the following sense: if $\widetilde{M}'$ is any other simply connected smooth manifold that admits a smooth covering map $\pi':\widetilde{M}'\to M$, then there exists a diffeomorphism $\phi:\widetilde{M}\to\widetilde{M}'$ such that $\pi'\circ\phi=\pi$.
I proved existence, put how do I prove uniqueness (the second sentence of the above corollary)? I know the following two theorems.
Theorem 80.3. (from Munkres's "Topology" 2nd ed.) Let $p:E\to B$ be a covering map, with $E$ simply connected. Given any covering map $r:Y\to B$, there is a covering map $q:E\to Y$ such that $r\circ q=p$.
Proposition 4.40 (Covering Spaces of Smooth Manifolds). (from Lee's "Introduction to Smooth Manifolds" 2nd ed.) Suppose $M$ is a connected smooth $n$-manifold, and $\pi:E\to M$ is a topological covering map. Then $E$ is a topological $n$-manifold, and has a unique smooth structure such that $\pi$ is a smooth covering map.
By theorem 80.3, I see that there are topological covering maps $q$ and $q'$ such that $\pi'\circ q=\pi$ and $\pi\circ q'=\pi'$. By proposition 4.40, I know there is a unique smooth structure for each $\widetilde{M}$ and $\widetilde{M}'$ which makes $q$ and $q'$ smooth covering maps. But I don't know if those smooth structures are the same as the original smooth structures. Also, I don't know how to get a diffeomorphism from $q$ and $q'$.
Best Answer
Disclaimer: I am not as familiar with differential topology as I am with algebraic topology.
I believe your question can be answered via the classification of covering spaces. I think it can also be done by invoking various lifting/uniqueness criteria regarding covers, but this way is less technical.
(I am being sloppy and ignoring basepoints/additional conditions)
Two covers $p_1 : X_1 \to X$, $p_2:X_2 \to X$ are said to be isomorphic if there is some homeomorphism $f:X_1 \to X_2$ such that $p_2 \circ f = p_1$. This agrees with the intuitive notion of two covers being "essentially the same".
The classification of covering spaces states (among other things) that there is a bijection
$$\{\text{isomorphism classes of path connected covers of }X \} \leftrightarrow \{\text{conjugacy classes of subgroups of } \pi_1(X,x_0) \}$$
Where an isomorphism class of a covering space $p:X_1 \to X$ is associated with the image of its fundamental group under the induced map $p_\ast$.
Furthermore, if $H \subseteq G \subseteq \pi_1(X,x_0)$, with corresponding covers $p_H:X_H \to X$, $p_G:X_G\to X$, then there is some cover $f:X_H \to X_G$ with $p_G \circ f = p_H$.
Let $p:E \to B$ be some cover. If $E$ is simply connected, then the image of its fundamental group is trivial (note that this makes 80.3 a special case of the classification). The trivial subgroup is of course its own conjugacy class, so it is associated to a single isomorphism class of covers. From this we can conclude that any two covers with simply connected domains must be isomorphic, and in particular your $q,q'$ are homeomorphisms.
Does this solve your problem?