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Tony Pantev
  • Member for 15 years, 2 months
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Elements of order two in a Prym variety
So, yes, you are absolutely right - the kernel K is connected and the snake lemma argument works.
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Elements of order two in a Prym variety
Ah, sorry - my mistake! The statement I was thinking about is that for an \'{e}tale double cover the kernel of $1 + \sigma$ is disconnected and consists of $P$ and a translate of $P$ by a point of order $2$.
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Elements of order two in a Prym variety
The kernel of $1 - \sigma$ is not connected for an \'{e}tale cover it is only connected when $\sigma$ has fixed points. For $\sigma$ with no fixed points the kernel has two connected components: the image of $J_{0} \to J_{1}$ and a translate of that image by a point of order $2$ on $J_{1}$.
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Analytic vs Zariski neighbourhood of a fibre
Yes, this what I had in mind. And the point is that for a Zariski open set $U$, the cover $\widetilde{U}$ is connected while for a small disk $\Delta$, the cover $\widetilde{\Delta}$ has two components.
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