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Perhaps someone can help me with this:

For simple closed curves on an orientable compact surface, if they form a bigon, then is it true that at the intersection points the orientations must be different?

To put this in context see http://www.math.uchicago.edu/~margalit/mcg/mcgv50.pdf ("Primer on Mapping Class Groups" by Farb and Margalit) page 73 end of the proof of prop.3.2. which states "in a true bigon, the orientations at the two intersection points are different,...".
Cheers!

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Call the two curves forming the bigon $\alpha$ and $\beta$, and the vertices $x_1$ and $x_2$. In just the region containing the bigon, the oriented intersection number of the curves is the sum of the orientations of $x_1$ and $x_2$. Since a bigon bounds a disk in the surface by definition, we can homotope $\alpha$ and $\beta$ off of each other so that they don't intersect at all. Intersection numbers are homotopy invariant, so this implies that the sum of the orientations of $x_1$ and $x_2$ is equal to zero, or in other words the orientation of $x_1$ is the opposite of the orientation of $x_2$.

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    @Sean: Yes, what I call the "oriented intersection number" is what Farb and Margalit call the "algebraic intersection number."2012-06-01