Deck transformation of $p : Y \to X : z \mapsto z^3 - 3z$ Let $X = \mathbb{C} \setminus \{ \pm 2 \}$ and $Y = \mathbb{C} \setminus \{ \pm 1, \pm 2 \}$. The map
$$
p : Y \to X : z \mapsto z^3 - 3z
$$
is a 3-branched covering.
Problem: Find $\operatorname{Deck}(Y/X)$, the group of Deck transformations of $Y$.
My try: My only idea is that $\operatorname{Deck}(Y/X) = \pi_1(Y)$ when $Y$ is the universal covering, but I don't think it is.
 A: The group of deck transformations $\operatorname{Deck}(Y \xrightarrow{p} X)$ is trivial, i.e. $\operatorname{Deck}(Y \xrightarrow{p} X)=\{\operatorname{id}_Y\}$. In particular, the covering $p\colon Y \rightarrow X$ is not Galois.
To show this, we will employ the following result. A proof can be found in Forster's Lectures on Riemann Surfaces (Theorem 8.5., page 52).

Theorem
Suppose A, B and C are Riemann surfaces and $\pi\colon B \rightarrow A, \tau \colon C\rightarrow A$ are proper holomorphic covering maps. Let $D \subseteq A$ be a closed discete subset and let $A'\colon =A\setminus D$. Set $B'\colon=\pi^{-1}(A')$ and $C'\colon =\tau^{-1}(A')$. Then every fiber-preserving biholomorphic mapping $\sigma'\colon B'\rightarrow C'$ can be extended to a fiber-preserving biholomorphic mapping $\sigma\colon B\rightarrow C$.

To show that $\operatorname{Deck}(Y \xrightarrow{p} X)=\{\operatorname{id}_Y\}$, let $f \in\operatorname{Deck}(Y \xrightarrow{p} X)$. Denote the Riemann sphere by $\mathbb{P}^1$. Define $A\colon =B\colon= C\colon=\mathbb{P}^1$.
Note that $p$ can be extended to a proper (branched) holomorphic covering map $\hat{p}:\mathbb{P}^1\rightarrow \mathbb{P}^1$ by setting $$\hat{p}(z)=\begin{cases} \infty & \text{if}\ z =\infty \\ z^3-3z &\text{else} \end{cases}$$
Next, observe that $D\colon=\mathbb{P}^1 \setminus X \subseteq \mathbb{P}^1$ is a closed discrete subset of $\mathbb{P}^1$.
Define $A'\colon=X$ and $B'\colon=C'\colon =Y$.
Being a lift along the covering map $p$, the homeomorphism $f$ is biholomorphic. We can thus apply above theorem by setting $\sigma'\colon=f$ and $\pi\colon=\tau\colon=\hat{p}$. In other words, the deck transformation $f$ of $p$ can be extended to a biholomorphic map $\hat{f}\colon \mathbb{P}^1 \rightarrow \mathbb{P}^1$ which is fiber-preserving with respect to $\hat{p}$.
Any automorphism of the Riemann sphere is a Möbius transformation. Therefore, we can find a matrix $\big(\begin{smallmatrix}
  a & b\\ 
  c & d
\end{smallmatrix}\big)\in \operatorname{GL}_2(\mathbb{C})$ such that $\hat{f}(z)=\frac{az+b}{cz+d}$ for all $z\in \mathbb{P}^1$.
We will now convince ourselves that $a=1, b=0, c=0$ and $d=1$.

*

*Since the fiber of $\hat{p}$ over $\infty$ is $\hat{p}^{-1}(\infty)=\{\infty\}$ and since $\hat{f}$ is fiber-preserving, we have $\hat{f}(\infty)=\infty$. We conclude $c=0$.


*Moreover, since $\hat{f}$ permutes the fiber $\hat{p}^{-1}(-2)=\{1,-2\}$, we have $\hat{f}(1)\in\{1,-2\}$. We make a case distinction.
If $\hat{f}(1)=1$, then since $\hat{f}$ is bijective, we have $\hat{f}(-2)=-2$. A Möbius transformation is determined by three function values. Thus, in the case that $\hat{f}(1)=1$, the mapping $\hat{f}$ is the identity on $\mathbb{P}^1$.
If on the other hand $\hat{f}(-2)=1$, then $\hat{f}(1)=-2$. By plugging these values into $\hat{f}(z)=\frac{az+b}{d}$, one immediately sees $a=b$. In particular, we have $\hat{f}(-1)=0$. This contradicts the fact that $\hat{f}$ permutes the fiber $\hat{p}^{-1}(2)=\{-1,2\}$. Thus, the second case $\hat{f}(-2)=1$ cannot occur.
In total, we have shown that the map $\hat{f}$ and thus its restriction $f$ is the identity.
A: Too long for a comment. This is an idea for a solution. It seems that the covering is not normal.
Let $a$, $b$, $c$, $d$ be loops around $-2,-1,1,2$ which generate $\pi_1(Y)=F(a,b,c,d)$ and $u,v$ be loops around $-2,2$ which generate $\pi_1(X)=F(u,v)$ then $p_{*}(a)=u^3$, $p_*(b)=uv^2$, $p_{*}(c)=u^2v$, $p_{*}(d)=v^3$ according to the multplicities of $-2,-1,1,2$. But, this claim needs a proof.
Let $f:Y\rightarrow Y$ be a deck transformation which is not (homeomorphic to) the identity deck transformation. Then it must interchange $a\&c$ or  $b\&d$. Assume that $f_*(a)=c$ then $u^3=p_*(a)=p_*f_*(a)=p_*(c)=u^2v$ with contradiction. Similarly $f_*(b)=d$ gives a contradiction. So, $f_*(a)=a$, $f_*(b)=b$, $f_*(c)=c$ and $f_*(d)=d$ and hence we have a contradiction with the asumption that $f$ is not the identity deck transformation.
