Every submanifold is orientable (co-dimension 1)?

Question

Suppose I have a submanifold $M \subset \mathbb{R}^{n}$, of dimension $n-1$. Apparently it's orientable if and only if there exists a unit normal vector field on $M$. Where a unit normal vector field is a section $\nu$ of the normal bundle $ TM^{\bot} \to M$. So the fibers are all the vectors that are perpendicular to the tangent space of the same base point. With the addition that $\| \nu(p) \| =1$, for all $p$ in $M$.

However, since the codimension is 1, can I not simply identify $T_{P}M^{\bot}$ with $\mathbb{R}$, and consider a smooth section $\nu$ of the bundle $\mathbb{R} \times M \to M$ where every point is mapped to either 1 or -1 by $\nu$, therefore having the section and therefore there exists a unit normal vector field and an orientation. Of ccourse this should be wrong, but what goes wrong in my reasoning and why?

Answer 1

For each $p$ you have that $T_pM^\perp \cong \Bbb R$, indeed. But these isomorphisms need not be natural. In some loose sense, for each $p$ you have an isomorphism $\Phi(p)\colon T_pM^\perp \to \Bbb R$, however the map $$TM^\perp \ni (p,v) \mapsto \Phi(p)(v) \in \Bbb R$$need not even be continuous, since $\Phi(p)$ and $\Phi(q)$ can be completely unrelated for $p \neq q$. One possible way to make a consistent choice of isomorphisms (ensuring good properties of the "coupled" map) is via a nonvanishing normal field defined globally along the manifold.

Answer 2

A dimension one real vector bundle is not always trivial, (consider the tautological bundle over the projective space) therefore, you cannot identify $TM^{\perp}$ to $\mathbb{R}\times M$. The existence of a unit normal vector field is equivalent to saying that $TM^{\perp}$ is trivial.