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.

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.
