Let $h(x)$ be a polynomial of degree $2g+2$ assume $h(x)$ has distinct roots.
- Form the smooth affine plane curve $X=\{(x,y)\in\mathbb{C×C} :F(x,y)=y^2-f(x)=0\}$. Let $U=\{(x,y)\in X: x\neq 0\}$; $U$ is an open subset of $X$.
Let $k(z)=z^{2g+2}h(1/z)$; $k(z)$ is also a polynomial in $z$ .
2)Form the smooth affine plane curve $Y=\{(z,w)\in\mathbb{C×C} :K(z,w)=w^2-k(z)=0\}$. Let $V=\{(z,w)\in Y: z\neq 0\}$; $V$ is an open subset of $Y$.
Show that ,$\phi:U\rightarrow V$ defined by $\phi(x,y)=(z,w)=(1/x,y/x^{g+1})$ is a holomorphic map between two Riemann surfaces.
Proof: As $F(x,y)$ and $K(z,w)$ smooth affine plane curves for any point of their domains at least one of the partial derivatives are non zero. Let $P=(x_0,y_0)\in U$ and $Q= \phi(P)=(z_0,w_0)$ . Wolg assume $ \frac{\partial F}{\partial y} \neq 0$ at $P$ and $ \frac{\partial K}{\partial w} \neq 0$ at $Q$ .
Using Implicit function theorem,there exists an open neighborhood $U'$ of $P$ in $X$ and a holomorphic function $g(x)$ such that on $U'$ ,$X$ locally looks like a graph i.e $X=\{(x,g(x)):x\in U'\}$, Also take a chart map $\pi_{P}(x,y)=x, {\pi_{P}}^{-1}(x)=(x,g(x))$ on $U'$ .
Similarly there exists an open neighborhood $V'$ of $Q$ in $Y$ and a holomorphic function $r(z)$ such that on $V'$ ,$Y$ locally looks like a graph i.e $X=\{(z,r(z)):z\in V'\}$, Also take a chart map $\pi_{Q}(z,w)=z, {\pi_{Q}}^{-1}(z)=(z,r(z))$.
Now if needed intersect $U'$ and $V'$ with $U$ and $V$ respectively. So let's compute the composition map $ \pi_Q\circ\phi \circ {\pi_{P}}^{-1}(x)=\pi_Q(\phi (x,g(x)))=\pi_Q(1/x,g(x)/x^{g+1})=1/x$ which is a holomorphic function from $\pi_P(U'\cap U)$ to $\pi_Q(V'\cap V)$. As $0\neq \pi_P(U'\cap U)$ there holomorphicity of the composition map is fine. Similarly we can show holomorphicity of $\phi$ for other charts also.