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Jul 8, 2015 at 18:47 vote accept Ron
Jun 30, 2015 at 18:32 comment added Allen Knutson I usually use "$f$ is equivariant".
Jun 30, 2015 at 16:44 answer added Jason Starr timeline score: 1
Jun 30, 2015 at 13:02 comment added Ron @Starr: Thank you very much for your answer. I am reading one of your articles "Families of rationally simply connected varieties over surface and torsors for semi-simple groups". In the first line of the proof of Lemma 15.7 you say that the considered evaluation map has isomorphic fibers because the target is homogeneous. I was trying to understand this statement.
Jun 30, 2015 at 12:59 comment added Jason Starr Also, there is another example: starting with $X' = Y\times Z$ and with $N$ everywhere disjoint graphs of morphisms from $Y$ to $Z$, let $X$ be the blowing up of these graphs. Because of these examples, usually people ask for the weaker conclusion that the fibers are birational. Moret-Bailly's pencils are counterexamples to "birational triviality".
Jun 30, 2015 at 12:14 comment added Jason Starr This question is similar to others that come up from time-to-time, and I will remind of the following (sorry to keep repeating this): Moret-Bailly constructed pencils of Abelian surfaces over $\mathbb{P}^1$ that are not isotrivial.
Jun 30, 2015 at 11:54 comment added Ron Sorry, typo. Corrected now.
Jun 30, 2015 at 11:54 history edited Ron CC BY-SA 3.0
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Jun 30, 2015 at 11:48 comment added Achim Krause how's $Z$ related to the situation?
Jun 30, 2015 at 10:32 history asked Ron CC BY-SA 3.0