Let $S_{1},S_{2}$ be the hyperelliptic curves given by \begin{align*}S_{1}&: y^{2} = x^{8} - 1,\\S_{2}&: y^{2} = x^{5} - x,\end{align*} respectively. Let $f: S_{1} \rightarrow S_{2}$ be the morphism given by \begin{align*} (x,y) &\mapsto (x^{2}, xy)\\ \infty_{1} &\mapsto \infty\\ \infty_{2} &\mapsto \infty, \end{align*} where $\infty_{i}$ are the points at infinity on $S_{1}$ and $\infty$ is the point at infinity on $S_{2}$.
Show that $f$ is unramified.
The only tool available at this point in the book is the definition of unramified ($1+\mathrm{ord}_{P}(\psi\circ f)' < 2$ where $\psi$ is a chart on $S_{2}$ or $\mathrm{ord}_{P}(\psi\circ f) < 2$ if $\psi$ is centered at $P$).
I was able to show it is unramified at points which are not zeros of the equation for $S_{1}$, but I struggled for the other points. Let the zeros of $S_{1}$ be $a_{j}$, $\varphi_{j}$ be the chart determined by $$\varphi_{j}^{-1}(z) = (z^{2} + a_{j}, z\sqrt{\prod_{k}(z^{2} + a_{j} - a_{k})}),$$ the zeros of $S_{2}$ be $b_{j}$, and $\psi_{j}$ be the chart determined by $$\psi_{j}^{-1}(z) = (z^{2} + b_{j}, z\sqrt{\prod_{k}(z^{2} + b_{j} - b_{k})}).$$ Then (for $b_{j} = a_{j}^{2}$) I got (using the fact that $-a_{k}$ is a zero of $S_{1}$ and $a_{k}(-a_{k}) = b_{k}$) \begin{align*}\psi_{j}f \varphi_{j}^{-1}(z) &= \psi_{j}f\Big(z^{2} + a_{j}, z\sqrt{\prod(z^{2} + a_{j} - a_{k})}\Big)\\ &= \psi_{j}\Big((z^{2} + a_{j})^{2}, (z^{2} + a_{j})z\sqrt{(z^{2} + a_{j} + a_{j})\prod\big((z^{2} + a_{j})^{2} - b_{k}\big)}\Big)\\ ``\!\!&=" \psi_{j}\Big((z^{2} + a_{j})^{2}, (z^{2} + a_{j})\sqrt{(z^{2} + a_{j} - a_{j})(z^{2} + a_{j} + a_{j})\prod\big((z^{2} + a_{j})^{2} - b_{k}\big)}\Big)\\ &= \psi_{j}\Big((z^{2} + a_{j})^{2}, (z^{2} + a_{j})\sqrt{\prod\big((z^{2} + a_{j})^{2} - b_{k}\big)}\Big). \end{align*} But then I'm not really sure what to do from here. I guess my problem comes down to when I try to apply $\psi_{j}$, I feel like the $a_{j}^{2}$ should come out of $(z^{2} + a_{j})^{2}$ because the map should be centered at $b_{j}$, but then there isn't a square anymore.