Let $f:X\to Y$ be proper holomorphic map of the connected Riemann surface.Let $R$ be the set of ramification points.
Prove $f:X\setminus R \to Y$ needs not to be proper,however the map $f:X\setminus (f^{-1}(f(R))) \to Y$ is a proper map.
My attempt to prove the second statement, since $f$ is proper map from $X\to Y$ therefore it's closed map, since the remification set $R$ is closed and discrete, the image of the set is also closed and discrete therefore $f(R)$ is closed and $f^{-1}(f(R)) = C$ is also closed subset.
To prove $$f:X\setminus C \to Y$$ is proper, we need to show that for any compact set $K\subset Y$ we have $f^{-1}(K) \cap X\setminus C$ is compact.
It will be helpful to check it on a simple example for exmaple consider the map $$f:\Bbb{S^2}\to \Bbb{C}\\z\mapsto z^2(z-1)$$ where $\Bbb{S^2}$ is Riemann sphere. This seems not to be a good one since it has pole at infinity.