# Every submanifold is orientable (co-dimension 1)?

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?

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.
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.