TITLE: Finite etale covers of products of curves
QUESTION [5 upvotes]: Probably this question can be phrased in a much greater generality, but I will just state it in the generality I require. I work over $\mathbb{C}$.

Let $C_1, C_2 \subset \mathbb{P}^1$ be non-empty open subsets and $f: X \to C_1 \times C_2$ a non-trivial finite etale cover. Does there exist $i\in \{1,2\}$ such that the composition $X \to C_1 \times C_2 \to C_i$ has non-connected fibres?

REPLY [6 votes]: The question already has a beautiful answer, but here's a different point of view which you may find helpful.
Let $F_i = \pi_1(C_i, x_i)$, which is a free group on $\#(\mathbf{P}^1\setminus C_i) - 1$ generators (the etale fundamental group will be the profinite completion of this). Then $F = \pi_1(C_1\times C_2, x_1\times x_2) = F_1\times F_2$.
A finite etale cover of $C_i$ or $C_1\times C_2$ corresponds to a finite set (its fiber at the basepoint $x_i$ or $x_1\times x_2$) with an action of $F_i$ or $F$. The cover is connected if and only if the action on that set is transitive.
Let $\sigma_i \colon C_i\to C_1\times C_2$ be the section $\sigma_1(x) = (x, x_2)$, $\sigma_2(x) = (x_1, x)$. Then for a finite etale cover $X\to C_1\times C_2$, the composition $X\to C_1\times C_2\to C_i$ has connected fibres if and only if the pull-back of $X$ along $\sigma_{2-i}$ is connected.
So now the question is equivalent to: suppose that $S$ is a finite set with more than one element with an action of $F=F_1\times F_2$. Is it possible that $F_1$ and $F_2$ both act transitively on $S$?  It is very easy to construct such examples. 
The easiest one could be $S$ with two elements, with every generator of each $F_i$ acting by a nontrivial involution. If there are only two punctures on each curve, this coincides with Francesco Polizzi's construction, and we see that his example is in some sense minimal.