Let $X$ be a hyperelliptic curve, and let $f:X\to \Bbb P^1$ be the degree 2 map induced by hyperelliptic involution. By elementary properties of holomorphic maps, $f$ is surjective, continuous, and open, so it is a quotient map. Suppose $g$ be a nonconstant meromorphic function on $X$, viewed as a holomorphic map $X\to \Bbb P^1$, and that $f(x)=f(y)$ implies $g(x)=g(y)$. Then $f$ induces a unique continuous map $h:\Bbb P^1\to \Bbb P^1$ such that $g=hf$. How can we show that $h$ is holomorphic?
If $x\in X$ is not a ramification point of $f$, then $f$ is a local biholomorphism at $x$, so $h$ is holomorphic at $f(x)$. But I can't show that $h$ is holomorphic at the $2g+2$ branch points of $f$ (where $g$ is the genus of $X$).
Edit: I wrote more details below.
Suppose $X$ is given by $\{(x,y)\in \Bbb C^2: y^2=k(x)\}$ locally. Then $f:X\to \Bbb C$ is locally given by $f(x,y)=x$. If $(x_0,y_0)$ is a ramification point of $f$, we must have $k(x_0)=0$ and $y_0=0$. At this point $(x_0,0)$, the coordinate $y$ can be served as a local coordinate. After choosing a chart near $g(x_0,0)$, we may assume $g$ is locally given by $y\mapsto g(y)\in \Bbb C$.
In summary, we are in the following situation: Suppose $g:\Bbb C\to \Bbb C$ is an even function, and $x=x(y):\Bbb C\to \Bbb C$ is a degree 2 map ramified at $0$. And there is a map $h:\Bbb C\to \Bbb C$ such that $h(x(y))=g(y)$. How can we show that $h$ is holomorphic at $0$?