$M$ is orientable iff $\bigwedge^n TM \setminus \{ \text{$0$-section} \}$ has $2$ connected components Let $M$ be an manifold of dimension $n$. We say that $M$ is orientable if it has an $n$-form that doesn't varnishes in any point if $M$. My question is about the following claim:

$M$ is orientable if and only if $\bigwedge^n TM \setminus \{ \text{$0$-section}\}$ has $2$ connected components.

Is it true? If yes, how can I see it?
 A: If $M$ is not connected, then it is obviuosly not true. However, in this case we can consider each component separately.
We know that $M$ is orientable if and only if the the line bundle $\bigwedge^n TM$ is trivial. See for example https://unapologetic.wordpress.com/2011/08/25/oriented-manifolds/ .
Now let $M$ be connected.
If $M$ is orientable, then $\bigwedge^n TM \setminus \{ \text{$0$-section}\} \approx M \times \mathbb{R} \setminus \{ \text{$0$-section}\} = M \times (\mathbb{R} \setminus \{ 0 \})$ has two connected components.
To show the converse, choose a Riemannian metric on $M$. This induces a metric on the line bundle $\bigwedge^n TM$ and provides a sphere bundle $S(M) = S(\bigwedge^n TM) \subset \bigwedge^n TM$. It consists of exactly two non-zero points in each fiber. The bundle projection $\pi : \bigwedge^n TM \to M$ restricts to a map $p : S(M) \to M$ which obviously is a $2$-sheeted covering. If $\bigwedge^n TM \setminus \{ \text{$0$-section}\}$ has two components, then so has $S(M)$. Let $S_+(M)$ be one of these components. Then $p$ restricts to a $1$-sheeted covering $p_+ : S_+(M) \to M$. This must be a homeomorphism. The inverse $s$ of $p_+$ gives you a non-vanishing section of $\bigwedge^n TM$.
