Let S be a singular complex algebraic surface. Let $\psi: \chi \rightarrow \Delta$ be a QQ-Gorenstein smoothing of S (central fibre is isomorphic to S). Let G be a finite group acting on the fibres of above deformation with no fixed points. What can we say about the new deformation family (where fibres are quotients of fibres of $\psi: \chi \rightarrow \Delta$). Is it QQ-Gorenstein smoothing of S/G? Also provide some reference related to this.



QQ-Gorenstein smoothing of a surface and a finite group action
I assume that by a "$\mathbb Q$-Gorenstein smoothing" you mean that $\chi$ is $\mathbb Q$-Gorenstein and probably $\Delta$ is the unit disk. 


I also assume that by "$\mathbb Q$-Gorenstein" you probably mean something like


1. $\chi$ is normal
2. $K\_\chi$ is $\mathbb Q$-Cartier,


but you're probably not requiring that $\chi$ be Cohen-Macaulay.


Anyway, I think that the answer is "yes" and the proof is very simple: 


Let $\pi:\chi\to \chi'=\chi/G$ be the quotient map. Since $G$ is finite and acts without fixed points, $\pi$ is étale and hence $\chi'$ is locally analytically isomorphic to $\chi$. More precisely for all $x\in\chi'$, and for all $y\in\pi^{-1}(x)\subset \chi$, the analytical or étale local schemes $(\chi',x)$ and $(\chi, y)$ are isomorphic. Hence $\omega\_{\chi'}^{[m]}$ (the reflexive hull of the $m^{\text{th}}$ tensor power) is locally free in the étale topology for some $m\in \mathbb N$ and hence the stalk $(\omega\_{\chi'}^{[m]})\_x$ has rank $1$. This is independent of the topology and so it follows that $\omega\_{\chi'}^{[m]}$ is a line bundle and so $\chi'$ is $\mathbb Q$-Gorenstein. 


You might want to replace "étale" with analytic if you prefer that language, but I think the proof works the same way. 


