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Jun 15, 2020 at 7:27 history edited CommunityBot
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Nov 16, 2014 at 22:38 vote accept Ginevra Carbone
Oct 29, 2014 at 16:26 answer added Pyramid timeline score: 5
Oct 29, 2014 at 9:21 comment added Francesco Polizzi In other words, since the fibration is stable, it provides a map to the stable moduli space of curves $\overline{\mathcal{M}}_g$. The image of this map meets both the boundary of the moduli space (because of the presence of the nodal, irreducible fibres) and its interior $\mathcal{M}_g$, hence it cannot be constant. This is precisely your argument, of course.
Oct 29, 2014 at 9:15 comment added Francesco Polizzi I think your argument works. The point is that Roberto's example is not stable (the rational tail meets the other component of the singular fibre in just one point), hence it does not provide directly a map to the moduli space (this is precisely the reason why (semi)-stable reduction is introduced).
Oct 29, 2014 at 9:02 comment added Felipe Voloch My argument above doesn't work for genus zero, where $M$ is a point. I only saw this after reading Roberto's answer that has a genus zero example.
Oct 29, 2014 at 7:21 answer added Roberto Pignatelli timeline score: 4
Oct 28, 2014 at 20:18 comment added Felipe Voloch A quick proof can be given if you take the existence of moduli spaces for granted. The fibration gives a map $\mathbb{P}^1 \to M, x \mapsto [f^{-1}(x)]$ where $M$ is the moduli space and $[C]$ is the class of a curve $C$. This map is non-constant since singular fibers cannot map to the same point as a non-singular fiber and your fibration has both. OTOH the map would be constant if the fibration were isotrivial. But I am sure this statement has to be proved on the process of constructing moduli spaces, so I am cheating a bit.
Oct 28, 2014 at 20:00 review First posts
Oct 28, 2014 at 20:22
Oct 28, 2014 at 19:58 history asked Ginevra Carbone CC BY-SA 3.0