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Let $S\subset\mathbb{R}^3$ be an orientable surface, but not oriented (yet).

Let $X:(u_1,u_2)\in U\subset \mathbb{R}^2\longrightarrow X(U)\subset S$ be a local parametrization of the surface $S$. We call $N=\dfrac{X_{u_1}\times X_{u_2}}{\lVert X_{u_1}\times X_{u_2} \rVert}$ a normal vector and we take the positive orientation.

Suppose $R\subset X(U)\subset S$ is a simple region on the surface, with $\partial R$ its boundary, and suppose it is a closed, piece-wise, smooth curve. We wish to give $\partial R$ a positive orientation.

Let $A=X^{-1}(R)\subset U\subset\mathbb{R}^2$. We compute $\partial A$. We give $\tilde{\alpha}\equiv A$ the positive orientation on the plane. Now we take $\alpha(t)=X(\tilde{\alpha}(t))$ the boundary of $\partial R$. Is it positively oriented this way?

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Yes, because you make the two necessary choices to introduce orientations both in the region $R$ and in its boundary $\partial R$, and you may take these choices as "positive".


A more detailed answer follows for the interested.

Surface $S$ being orientable is equivalent to $S$ having a globally defined unit normal field. There are two options. "Not oriented" means a choice just has not been made.

Introducing a local parametrization implies a choice of the unit normal via the formula for $N$ that you have mentioned. This singles out a global (!) orientation of $S$ (since $S$ is orientable).

An orientation of a plane curve $\alpha(t)$ can be similarly seen as a choice of the unit normal $n$ along this curve (when it is piecewise smooth, you need to understand what is happening in the corners, but perhaps a continuity argument may sort it out?). It may be more convenient to define the orientation of the curve by the direction of the velocity vector, so that the pair $(\dot{\alpha},n)$ is a positive basis at each point of the curve. This clearly depends on the parametrization $\alpha(t)$ of the boundary $\partial A$. This is the second choice, and you have freedom!

Using your method you merely impose the orientations on $R$ and $\partial R$ by translating the standard orientation in $\mathbb{R}^2$ and your choice of orientation on $\partial A$ using a diffeomorphism $X$. This orientation should be called positive, if you want the thing to agree with each other.

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