I'm reading John Lee's Introduction to Smooth manifolds 2nd edition. Consider the problem 10-1:
Define an equivalence relation on $\mathbb{R}^2$ by $(x,y) \sim (x',y')$ if and only if $(x',y')=(x+n,(-1)^n y)$ for some $n \in \mathbb{Z}$. Let $E=\mathbb{R}^2 / \sim$ and $q:\mathbb{R}^2 \rightarrow E$ be the quotient map. Let $\pi_1:\mathbb{R}^2 \rightarrow \mathbb{R}$ be the projection onto the first factor and $\epsilon:\mathbb{R} \rightarrow S^1,x \mapsto e^{2\pi i x}$ be the smooth covering map. Then there exists a map $\pi:E \rightarrow S^1$ such that $\pi \circ q=\epsilon \circ \pi_1$ since $\epsilon \circ \pi_1$ is constant on each equivalence class.
(a) Show that $E$ has a unique smooth structure such that the quotient map $q:\mathbb{R}^2 \rightarrow E$ is a smooth covering map.
(b) Show that $\pi:E \rightarrow S^1$ is a smooth rank-1 vector bundle.
(c) Show that it is not trivial.
My attempted solution:
proposition 4.40 in the book says
Suppose $M$ is a connected smooth $n$-manifold and $f:E \rightarrow M$ is a topological covering map. Then $E$ is a topological $n$-manifold and has a unique smooth structure such that $f$ is a smooth covering map.
This proposition is similar to part(a), however, the map $q$ is from $\mathbb{R^2}$ to $E$ not the other way around. How should I manipulate $q$ to satisfy this proposition?
For part(b), the only thing I know is that part (a) gives $E$ is smooth but I have no clue about the rest.
For part(c), Lee defines a trivial bundle to be a bundle with a global trivialization. Suppose $E$ has a global trivialization $\Phi:\pi^{-1}(S^1) \rightarrow S^1 \times \mathbb{R}$.
If I can show $\pi^{-1}(S^1)=E$, then since $\Phi$ is a homeomorphism we have $E \cong S^1 \times \mathbb{R}$. Then if I can show they can't be homeomorphic to each other, the result follows.
But is it true that $\pi^{-1}(S^1)=E$ and mobius strip can't be homeomorphic to the unit cylinder?
