Any vector bundle on $\mathbb R$ is a trivial bundle

Question

How to prove that any vector bundle on a Euclidean space is a trivial bundle? It is enough to prove it for the case of dimension $1$ and I hope it will be a nice exercise for me to generalize to the higher dimensional case.

What is the underlying topological reason that all bundles are trivial? Certainly simply connectedness will not do, as a counterexample is given by the Hairy ball theorem(of which I know only the statement; not the proof). Is it enough that the space is contractible?

Answer 1

Let $E \to \mathbb{R}$ be a vector bundle and $\nabla$ be a connection. For all $x \in \mathbb{R}$, let $\tau_x$ denote the parallel transport along the path $\gamma_x : t \mapsto (1-t)x$, $t \in [0,1]$. Then $$\Phi : \left\{ \begin{array}{ccc} E & \to & \mathbb{R} \times E_0 \\ (x,v) & \mapsto & (x, \tau_x(v)) \end{array} \right.$$ defines a global trivialization. You may find more information in my answer.