If a surface has a differentiable Gauss map, then it has an orientation?

Question

If a surface $S\subset R^3$ has a differentiable Gauss map $N:S\rightarrow S^2$, then $S$ has an orientation? How can I prove this statement? (Here, orientation is defined by a choice of equivalence class, specified by a choice of ordered basis; a Gauss map is "a map $N:S\rightarrow S^2$, such that $T_xS=N(x)^\perp$")

Answer 1

Orient $\mathbb{R}^3$ by some choice of basis. Given any other basis of $\mathbb{R}^3$ we can say if it is positive or negative with respect to this orientation.

Let's suppose we have a Gauss map of our surface $S$. At each point $x\in S$ we can, using the oriention of $\mathbb{R}^3$ and the choice of Gauss map $N$ define an orientation of basis of $T_xS$: We say a basis $e_1,e_2\in T_xS$ is a positive basis, if $(e_1,e_2)$ is a basis of $T_xS$ and $(e_1,e_2,N(x))$ is a positive basis of $\mathbb{R}^3$.