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It is known that on a hyperelliptic surface the set of Weierstrass points and the set of ramification points of the extension of the projection map $(x,y)\mapsto x$ to $\mathbb{P}^1$ coincide.

However, I am not sure if this is the case for a general Riemann surface. Maybe $p$ a Weierstrass point iff there is a ramified covering of $\mathbb{P}^1$ with $p$ as a ramification point.

It seems like I can't do much in either direction to prove this, but it also seems like something similar to this proposition would be nice to have since ramification provides a nice geometric intuition.

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    For future reference, this overflow post contains what I was looking for http://mathoverflow.net/questions/9204/what-do-weierstrass-points-look-like2011-06-02

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