Prove that orientable surface has differentiable normal vector Prove that:

a regular surface
  $S\subset \mathbb{R}^3$ is an
  orientable manifold if and only if
  there exists a differentiable mapping
  of $N:S\rightarrow \mathbb{R}^3$ with
  $N(p)\perp T_p(S)$ and $|N(p)|=1$, for
  all $p\in S$. 

If part:
I guess I have to first let $X_a,X_b$ be parametrization of $S$, and $<dX_a,N>=<dX_b,N>=0$. Differentiability of $N$ imply (not sure) $N\circ X_a$ and $N\circ X_b$ are also differentiable, which their differentials are linear function. I then have no idea how to show that $\det(d(X_b^{-1}X_a))>0$.
Only if part: Zero idea.
Please give me some insight!!
 A: Hint : I guess your definition is the following: A regular surface $S\subset \mathbb R^3$ is orientable if one can find a family $(X_a)_{a\in \alpha}$ so that $\det(dX^{-1}_bX_a)) >0$. 
So if $S$ is orientable, then you are given that family of charts $X_a$'s. For each chart, one can define locally the normal vector 
$$N_a = \frac{\partial_u X_a\times \partial_vX_a}{|\partial_u X_a\times \partial_vX_a|}$$
So is it true that you can extends this map to the whole $S$? Or, in another chart $X_b$, do you have $N_a  = N_b$? If you can show that, then $N: S\to \mathbb R^3$ is the mapping you want. 
On the other hand, if such a $N$ is given, then you can consider all charts $X_a$ of $S$ so that  
$$N = \frac{\partial_u X_a\times \partial_vX_a}{|\partial_u X_a\times \partial_vX_a|}. $$
Will this families of charts $X_a$'s satisfy $\det(dX^{-1}_bX_a)) >0$?
A: Let $\{(U_{\alpha},x_{\alpha})\}$ be an atlas of a regular surface  $S\subset\mathbb{R}^3$. We will construct a differentiable mapping $N:S\to\mathbb{R}^3$ as its "unit normal vector bundle". Arbitrarily choose
$$
  p=(y^1,y^2,y^3)\in x_{\alpha}(U_{\alpha})\subset\mathbb{R}^3
  $$
and write
$$
  x_{\alpha}^{-1}(p)=(x_{\alpha}^1,x_{\alpha}^2)\in U_{\alpha}\subset\mathbb{R}^2.
  $$
Then a normal vector of $S$ at $p$ can be expressed as
$$
  m=\left(
  \frac{\partial(y^2,y^3)}{\partial(x_{\alpha}^1,x_{\alpha}^2)},
  \frac{\partial(y^3,y^1)}{\partial(x_{\alpha}^1,x_{\alpha}^2)},
  \frac{\partial(y^1,y^2)}{\partial(x_{\alpha}^1,x_{\alpha}^2)}
  \right),
  $$
which follows that $n:=m/|m|$ is a unit normal vector of $S$ at $p$. Define $N(p)=n$. Clearly $N:S\to\mathbb{R}^3$ is differentiable in each $U_{\alpha}$. We only need to testify that it is well-defined. Let $q\in x_{\alpha}(U_{\alpha})\cap x_{\beta}(U_{\beta})$. Through direct computations, we have
$$
  \left(
  \frac{\partial(y^2,y^3)}{\partial(x_{\beta}^1,x_{\beta}^2)},
  \frac{\partial(y^3,y^1)}{\partial(x_{\beta}^1,x_{\beta}^2)},
  \frac{\partial(y^1,y^2)}{\partial(x_{\beta}^1,x_{\beta}^2)}
  \right)=
  \det(x_{\alpha}^{-1}x_{\beta})
  \left(
  \frac{\partial(y^2,y^3)}{\partial(x_{\alpha}^1,x_{\alpha}^2)},
  \frac{\partial(y^3,y^1)}{\partial(x_{\alpha}^1,x_{\alpha}^2)},
  \frac{\partial(y^1,y^2)}{\partial(x_{\alpha}^1,x_{\alpha}^2)}
  \right).
  $$
Hence $N:S\to\mathbb{R}^3$ is globally differentiable if and only if $S$ is orientable.
