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?

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    $\begingroup$ I do not think so, at least in general. Think of a double cover $\bar{X}$ of $\mathbb{P}^1 \times \mathbb{P}^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=\mathbb{P}^1$ minus two points). The general fibres of the composition $\bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$ are smooth double cover of $\mathbb{P}^1$ branched at two points, hence they are again isomorphic to $\mathbb{P}^1$, and so the fibres of your composition are isomorphic to $\mathbb{P}^1$ minus the ramification, i.e. $\mathbb{P}^1$ minus two points. $\endgroup$ – Francesco Polizzi Apr 10 '19 at 9:01
  • $\begingroup$ If you take a branch curve of type $L_1+L_2+L_3+L_4+M_1+M_2+M_3+M_4$, then the fibres of your compositions will be elliptic curves minus four points, and so on... $\endgroup$ – Francesco Polizzi Apr 10 '19 at 9:08
  • $\begingroup$ @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example? $\endgroup$ – Daniel Loughran Apr 10 '19 at 9:58
  • $\begingroup$ @DanielLoughran: I will try $\endgroup$ – Francesco Polizzi Apr 10 '19 at 10:44

The answer is no, at least in general, as shown by the following counterexample.

Take a double cover $\bar{f} \colon \bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1$, branched over a reducible curve of the form $B=L_1 + L_2 + M_1 + M_2$ (here $|L|$ and $|M|$ are the two pencil of lines on the quadric).

Such a cover exists because $B$ is $2$-divisible in $\mathrm{Pic}(\mathbb{P}^1 \times \mathbb{P}^1)$, and corresponds to an étale cover $f \colon X \to C_1 \times C_2$, where each $C_i$ is $\mathbb{P}^1$ - {two points}.

If these points are (say) $0$ and $1$ in both factors, then the equation for $X \subset \mathbb{C} \times (\mathbb{C}-\{0, \, 1\})^2$ is
$$z^2 = xy(x-1)(y-1), \quad f(z, (x, \,y)) = (x,\, y).$$

It is clear that the general line in $|L|$ and $|M|$ intersects the branch locus $B$ transversally at two points, hence both compositions $$\bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$$ have connected fibres, the general one being isomorphic to $\mathbb{P}^1$ (double cover of $\mathbb{P}^1$ branched at two points).

Then both compositions $$X \to C_1 \times C_2 \to C_i$$ have connected fibres, the general one being isomorphic to the $\mathbb{P}^1$ above minus the ramification, i.e. $\mathbb{P}^1$ minus two points, that is clearly connected.

In the same vein, choosing as $B \subset \mathbb{P}^1 \times \mathbb{P}^1$ a divisor of type $$B = \sum_{i=1}^{2g+2} L_i + \sum_{i=1}^{2g+2} M_i,$$ both compositions $$\bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$$ have connected fibres, the general one being isomorphic to a hyperelliptic curve $\Sigma_g$ of genus $g$, and so both compositions $$X \to C_1 \times C_2 \to C_i$$ (here each $C_i$ is $\mathbb{P}^1$ minus $2g+2$ points) have connected fibres, the general one being isomorphic to $\Sigma_g$ minus $2g+2$ distinct points.

  • $\begingroup$ Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $\mathbb{P}^1$ are conic bundles on the surface. $\endgroup$ – Daniel Loughran Apr 10 '19 at 15:55
  • $\begingroup$ @DanielLoughran: You are welcome. Note that $\bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$. $\endgroup$ – Francesco Polizzi Apr 10 '19 at 16:02

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.


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