Existence of a holomorphic map from Riemann Surface to an algebraic curve .

Let $C$ be an algebraic curve in $\mathbb P^2( \mathbb C)$ with singular points $p_i : \{1 \le i \le n \}$ . Then there exists a holomorphic map $\Phi : S \to C$ , where $S$ is a Riemann surface.

How to i construct such a map , considering the fact that its a map from a surface to a curve , it doesn't look obvious since the open sets are different on a surface than on a curve . Will the constructed surface have Hausdroff and compactness property ? Your help is appreciated . Thanks

• Note that Riemann surfaces are complex manifolds of (complex) dimension $1$, and algebraic curves in $\mathbb{P}^2(\mathbb{C})$ also have complex dimension $1$. – Daniel Fischer Jul 4 '13 at 16:12
• @DanielFischer : But aren't the open sets different ? I somehow don't seem to capture the situation here i.e relating curve with the surface . aren't open sets on the curve similar to the open sets in $\mathbb R$. I am confused . – Theorem Jul 4 '13 at 16:16
• No, curve here mean complex dimension $1$, which is real dimension $2$, as for Riemann surfaces (the real dimension is where the "surface" part of the name comes from). – Daniel Fischer Jul 4 '13 at 16:20
• @DanielFischer : So indeed the curve actually looks like a surface say for example $f(z)=z^2$ – Theorem Jul 4 '13 at 16:25
• What does "with singular points $p_i$" have to do with the rest of the question? – Matt Jul 4 '13 at 18:41

The normalization $S$ of $C$ is a smooth projective algebraic curve over $\mathbb C$, so it is a Riemann surface. The normalization map $S\to C$ is holomorphic because it is even algebraic.