Compact Riemann surface of genus 2 as the normalization of a quartic curve. I want to show that given a compact Riemann surface $X$ of genus 2, there is a map $\sigma:X\rightarrow \mathbb{CP}^2$, such that the followings are satisfied:


*

*$\sigma(X)$ is a quartic curve with a double point $s$

*$\sigma:X-\sigma^{-1}(s)\rightarrow \sigma(X)-s $ is biholomorphic.



What I have tried so far:
If we have found such map $\sigma$, then we can make a projection, $x = \pi \circ \sigma : X\rightarrow \mathbb{CP}^1$. And we can choose such projection that $\pi(s)$ is not a branch point of $x$. 
Apply Riemann-Hurwitz formula, $\operatorname{deg}{R_x} = 6$. And we also have $\operatorname{deg}x = 4$. 
I think there are more constraints on $x$, so that I can apply Riemann-Roch theorem to find such $x$ and then the quartic curve itself. But I don't know what to do next.

Any hints are welcome.
Thanks in advance!
 A: Consider any divisor $D$ of degree 4 on $X$ (with $D \ne 2K_X$) and consider map $f \colon X \to \mathbb{P}^2$ given by the linear system $|D|$. It will have all the required properties.
Let me show that $f$ is almost injective. Assume $f(P) = f(Q)$ for a pair of points $P,Q \in X$. This means that $h^0(X,O_X(D-P-Q)) = h^0(D)-1 = 2$, hence $D - P - Q$ is a $g^1_2$, hence $D - P - Q = K_X$. This means that $P + Q \sim D - K_X$, and since $D - K_X \ne K_X$ (by the assumption $D \ne 2K_X$), the linear system $|D - K_X|$ consists of a unique divisor, hence $P$ and $Q$ are canonically defined up to permutation.
A: Given the answers above, I think I can show that this is a surjection by considering the regularization $\mu:\overline{Y}\rightarrow Y$ , where $Y$ is the quartic curve. We claim that $Y$ is irreducible. (Though the regularization is defined for irreducible curves in the book, the process can still be carried out for a general curve, and if it is reducible, $\overline{Y}$ would be disjoint union of regularizations of each factor.) Now $\mu^{-1}\circ\sigma$ extends to be a isomorphism between $X$ and a connected component of $\overline{Y}$. If $Y$ is reducible, $X$ would be the regularization of some curve with lower degree, which is impossible by checking the genus. Then we must have $X$ as the regularization of $Y$, which implies surjectivity.
(I still don't konw why $\sigma(X)$ lies in a quartic curve actually. I wish you could explain it to me.)
