Hyperelliptic Riemann Surface, extension of a holomorphic map:

Question

Let $X,Y$ be compact compact hyper elliptic curve . $X=\{ (t,x):x^2=3+10t^4+3t^8\}$ and $Y=\{ (z,w):w^2=z^6-1\}$ . Let $U,V$ be corresponding affine plane curves respectively .Show that the function $F:U\rightarrow V$ defined by $z=\frac{(1+t^2)}{(1-t^2)}$ and $w=\frac{2tx}{(1-t^2)^3}$ extend to a holomorphic map from $X$ to $Y$ of degree $2$ which is nowhere ramified . This is one exercise from Rick miranda's Algebraic curve and Riemnan Surfaces , page $65$ , exercise $iii.1 . F. $ How to extend such maps ?

Answer 1

Exercise II.4.K gets you most of the way there. Then we need to worry about the points of $X$ at infinity as well as $t= \pm 1$. For the points of $X$ at infinity we can just examine the limit as $t$ approaches infinity of $z$ and $w$. If I'm not mistaken, these are sent to the point $z=-1$, $w=0$.

For $t= \pm 1$, Section 2 of Chapter III explains the idea. Examine the chart $\phi(w, z) = \frac{w}{z^3} = \frac{2tx}{(1+t^2)^3}$ when $t = \pm 1$. The bottom of page 66 explains that as $z$ approaches $\infty$, $\frac{w}{z^3}$ approaches one of the two square roots of $\alpha = \lim_{z \rightarrow \infty} \frac{z^6-1}{z^6} = 1$. These are the two points of infinity for $Y$. For example, when $t=1$, $w=4$, then $\phi(1, 4)$ is sent to the point of infinity of $Y$ corresponding to 1, while for $t=1$, $w=-4$, $\phi(1, -4)$ is sent to the point of infinity of $Y$ corresponding to -1.