Doubt regarding fibre bundles under group action Let $X, Y$ be smooth projective varieties over $\mathbb{C} $, and let $\pi: Y \rightarrow X$ be a zariski-locally trivial fibre bundle map with smooth fibre $F$. Moreover, let a finite group $\Gamma$ act freely on both $X$ and $Y$, such that $\pi$ is $\Gamma$-equivariant. It is known that both the quotients $X':=X/\Gamma$ and $Y':=Y/\Gamma$ exist as smooth varieties.
Now, because of equivariance, $\pi$ will induce a map $\pi': Y'\rightarrow X'$. My question is:
Is $\pi'$ also a fibre bundle map with same fibre $F$?
I think it should be true; my idea is that if $U$ is a locally trivial neighborhood in $X$ such that $Y|_{U} \simeq U\times F$, then $\pi(U) $ should be locally trivial for $\pi'$. Is this correct?
Any help would be really appreciated.
 A: I think there exist many counterexamples such as follows:
Let $f:X\to Y$ be a Galois cover of curves whose Galois group is $G = \mathrm{Aut}(X/Y)$.
Assume that the genus $g(Y)$ of $Y$ is greater than $2$, then by Riemann-Hurwitz formula, it holds that $g(X) > g(Y)$.
Consider the trivial fiber bundle $p:X\times X \to X$ (second projection) and $G$-action, where $G$ acts on $X\times X$ by diagonal.
It is clear that $p$ is compatible with both $G$-action on $X\times X$ and $X$.
Hence we can consider the quotient $q: Q = (X\times X)/G \to Y$.
Then each fiber of $q$ is isomophic to $X$, but $q$ is not Zariski-trivial.
Let us prove that $q$ is not Zariski-trivial.
Assume that there exists a Zariski open subset $V\subset Y$ and local trivialization $X\times V \cong Q\times_Y V$.
Write $U := f^{-1}(V) \subset X$, $r:X\times X\to Q$ for the quotient morphism, and $i:U \to X\times X$ for the diagonal.
Then $i$ is compatible with $G$-action, so we obtain a morphism $j:V\to Q$ such that $j\circ f|_U = r\circ i$.
$$\require{AMScd}
\begin{CD}
U @>{i}>> X\times X @>{p}>> X \\
@V{f|_U}VV @V{r}VV  @VV{f}V \\
V @>{j}>> Q @>{q}>> Y. 
\end{CD}
$$
Note that the both squares in the above diagram are Cartesian.
Since $q$ is trivial over $V$, by composing $(j,q\circ j):V\to Q\times_Y V$, the isomorphism $Q\times_YV \cong X\times V$, and the first projection $X\times V \to X$, we obtain a morphism $g:V\to X$.
Moreover, $g$ is not constant because $i$ is the diagonal.
By extending $g: V \to X$ to $Y\to X$ and using Riemann-Hurwits formula, we conclude that $g(Y) > g(X)$, and this contradicts to the inequality $g(X) > g(Y)$.
