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Jul 12, 2020 at 15:41 comment added Tony Pantev 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.
Jul 12, 2020 at 3:23 comment added bog Just to be sure that I have understood completely your answer: you are using the fact that the analytic Picard group of $f^{-1}(U)$ is the invariant Picard group in $E\times \tilde U$ by the involution, right?
Jul 12, 2020 at 2:37 comment added bog Thank you very much!
Jul 12, 2020 at 2:36 vote accept bog
Jul 10, 2020 at 17:55 history answered Tony Pantev CC BY-SA 4.0