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.



Lifting of automorphism of rational surface to that on abelian variety
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](https://www.jmilne.org/math/CourseNotes/AV.pdf)) 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.


