Q:

1-dim Vector Bundle sufficient condition to be trivial

I'm a physics student studying differential geometry. I'm trying to understand how vector bundles work, I have the following exercise.

Let be $ L $ a $1$-dim vector bundle on $M$. Prove that if $\exists\sigma: M \to L$, a section of $L$, so that $\sigma(P) \neq 0 \; \forall P \in M$ then $L$ is trivial.

This is how I tried to proceed.
$L$ is a vector bundle so $ \forall P \in M, \; \exists U_P (\ni P) \subset M $ open
$\exists\chi$ a diffeomorphism so that $\chi: L_U = \pi^{-1}(U) \mapsto U \times R$ where $\pi: L \to M$ is the projection ($\pi^{-1}(P) = L_P$ is a 1-dim vector space).
I have to prove that $L = M \times \mathbb{R}$ so $U = M$.
Let be $P \in M$, then $\chi \circ \sigma (U_P) = U_P \times \mathbb{R}$. But I can't figure out how the value $0$ is important. 
Are there any propositions or theorems to prove that a vector bundle is trivial? And is it useful to know that is trivial?

A:

Here's a theorem that might help you.
Let $\pi:E\to M$ be a rank $k$ vector bundle. Then $E$ is trivial if and only if it admits a global frame.
Admitting a global frame means that there exists $k$ linearly independent sections. That is, there exists sections $s_1,\dots,s_k$ such that $\{s_1(p),\dots,s_k(p)\}$ form a basis for $E_p$.
Now look at your question where you are considering a line bundle. What does it mean to have a global frame in this case?

