Lifting of automorphism of rational surface to that on abelian variety The paper I am referencing is "Normal Subgroups of the Cremona Group." https://arxiv.org/abs/1007.0895. In theorem 5.14, at the bottom of page 52, the author stated for the abelian surface $Y= \mathbb{C}/\mathbb{Z}[i] \times \mathbb{C}/\mathbb{Z}[i]$, we consider the quotient $X= Y/ \Gamma$ where $\Gamma$ is the group of order 4 generated by the map $(x,y)\mapsto (ix, iy)$ on $Y$. Moreover one can show that $X$ is rational.
In general the group of automorphism of $X$ cannot be lifted to that on $Y$ since there are more automorphisms on a rational surface than on an abelian one. However, the paper stated that in this case we can do so since the fundamental group $\Lambda$ of $X\backslash \text{Sing}(X)$ is the group $\mathbb{Z}/4\mathbb{Z} \ltimes (\mathbb{Z}[i]\times \mathbb{Z}[i])$. They further say that $\mathbb{Z}[i]\times \mathbb{Z}[i]$ is invariant under all automorphism of $\Lambda$ and used this to conclude that we can lift an aut on $X$ to that of $Y$. 
How did they arrive at this conclusion? Would be grateful if anyone can help to explain what that paragraph meant.
 A: Denote $X\backslash\text{Sing}(X)$ by $X_0$ and its preimage in $Y$ as $Y_0$. Note that $Y_0$ is the Galois cover of $X_0$ corresponding to the normal subgroup $\mathbb{Z}[i]\times\mathbb{Z}[i]$ inside $\Lambda.$ For any automorphism $f$ of $X_0$, the pullback of $Y_0$ along $f$ is the cover corresponding to the image of $\mathbb{Z}[i]\times\mathbb{Z}[i]$ under the automorphism of $\Lambda$ induced by $f$. By the reasoning you mentioned, this implies that the pullback of $Y_0$ along $f$ is again $Y_0$ and so we get an automorphism of $Y_0.$
It remains to show that any automorphism of $Y_0$ extends to an automorphism of $Y$. Here is one of many ways to show this. It is a theorem (see e.g. Theorem 3.2 in Milne's notes) that any rational map from a nonsingular variety to an abelian variety extends to a regular map. So we know that our automorphism of $Y_0$ comes from a map $Y\rightarrow Y$. This latter map is necessarily birational (because it restricts to $f$), quasi-finite, and proper, so by normality of $Y$ it must be an isomorphism.
