TITLE: Line bundles of the circle
QUESTION [19 upvotes]: Up to isomorphism, I think there exist only two line bundles of the circle: the trivial bundle (diffeomorphic to a cylinder) and a bundle that looks like to a Möbius band. Although it seems obvious geometrically I did not find a good argument to justify it. Do you have an idea?

REPLY [10 votes]: Here is the sketch of an elementary proof:
Let $E \overset{\pi}{\longrightarrow} \mathbb{S}^1$ be a line bundle and $p : \mathbb{R} \to \mathbb{S}^1$ be the usual universal covering. 

Let $F= \coprod\limits_{x \in \mathbb{R}} E_{p(x)}$ be a line bundle $F \overset{q}{\longrightarrow} \mathbb{R}$ defined by the local trivializations $\varphi= (p_{|V}^{-1} \times \operatorname{Id}) \circ \phi \circ (p \times \operatorname{Id})$ where $\phi$ is a local trivialization for $E \overset{\pi}{\longrightarrow} \mathbb{S}^1$ and $V$ is an elementary neighborhood of the overing $\mathbb{R} \to \mathbb{S}^1$. Then $p \times \operatorname{Id} : F \to E$ is a morphism.
Because $\mathbb{R}$ is contractible, every vector bundle on it is trivial (see here). Let $\psi : F \to \mathbb{R} \times \mathbb{R}$ be a global trivialization.
Notice that $(p \circ \operatorname{Id}) \circ \psi$ defines a surjective smooth map between $[0,2\pi] \times \mathbb{R}$ and $E$, injective on $[0,2\pi) \times \mathbb{R}$. Moreover, it induces linear isomorphisms $f_1 : \{2\pi\} \times \mathbb{R} \to E_0$ and $f_0 : \{0\} \times \mathbb{R} \to E_0$. For convenience, let $f= f_2^{-1} \circ f_1$.
We deduce the diffeomorphism $$E \simeq [0,2\pi] \times \mathbb{R} / \{ f(2\pi,x) \simeq (0,x) \}.$$ Without loss of generality, we may normalize $f$ so that $f= \operatorname{Id}$ or $f=- \operatorname{Id}$. Therefore, two cases happen: either $$E \simeq [0,2\pi] \times \mathbb{R} / \{(0,x) \sim (2\pi,x) \} \simeq \mathcal{C},$$ or $$E \simeq [0,2\pi] \times \mathbb{R} / \{(0,x) \sim (2\pi,-x) \} \simeq \mathbb{M}.$$