Showing that a given covering is not normal This is a question regarding exercise 5.6 of Forster's Lectures on Riemann surfaces.
We have $X=\mathbb{C}\setminus\{0,1\}$, $Y=\mathbb{C}\setminus\{0,\pm i,\pm i\sqrt{2}\}$, $p\colon Y\to X$ given by $p(z)=(z^2+1)^2$.
It is easy to see that this defines an (unbranched) 4-sheeted covering map, and that $\varphi\colon z\mapsto -z$ is a deck transformation.
(1) I want to proof that apart from $\varphi$ and the identity, there are no other deck transformations for $p$.
One way to see this is as follows: We can extend $p$ to a branched holomorphic covering map $\overline{p}\colon \hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}\to\hat{\mathbb{C}}$, since $p$ is a meromorphic function. Then one can show that any deck transformation of $p$ extends to a deck transformation of $\overline{p}$ by Riemann's theorem on removable singularities. Now one notes that a deck transformation of a branched covering map must preserve the ramification index to see that our list of deck transformations was exhaustive.
My question is:
Is there a way to prove statement (1) without going to the extension of $p$ to an unbranched covering map?**
 A: We can do this by getting a concrete understanding of how explicit generators of $\pi_1(X)$ lift under $p$, but it's more involved than the quick approach in your solution.
Fix some base point $x_0 \in (0,1)$. Representatives of generators of $\pi_1(X)$ are given by loops starting at $x_0$ and going around $0$, $1$. Call these $\alpha, \beta$ respectively. Notice that $x_0$ lifts to four points on the imaginary axis, one in each interval $(-i, -i/2), (-i/2, 0), (0, i/2),$ and $ (i/2, i)$.
One can use elementary reasoning to see that $\alpha$ and $\beta$ have lifts under $p$ homotopic to those pictured below.
The point is that there are lifts $\widetilde{\beta}$ and $\widetilde{\beta}'$ of $\beta$ starting at basepoints $\tilde x_0, \tilde x_0'$ respectively so that $\tilde{\beta}$ is a nontrivial loop in $Y$ and $\tilde{\beta}'$ is not a loop (and hence represents $e$ in $\pi_1(Y)$). Now $p_*$ is injective as $p$ is a covering, so $p_*([\tilde \beta]) \neq e$. Then if $\varphi$ is a Deck transformation so $\varphi(\tilde{x_0}) = \tilde{x}_0'$, $$e = p_* ([\tilde \beta']) = p_* \circ \varphi_*([\tilde \beta]) \neq e.$$ This shows that at most $3$ deck transformations can exist. Since the order of the Deck group divides $\text{deg}(p)$, there must be at most $2$ Deck transformations, as desired.
