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Jul 2, 2014 at 10:10 vote accept Alex Gavrilov
Jul 2, 2014 at 8:51 comment added abx Even if it is only birationally equivalent, its $\pi_1$ is isomorphic to that of $S$, hence is arbitrary.
Jul 2, 2014 at 6:52 comment added Alex Gavrilov You are probably right. But I am not sure I understand why the strict transform is isomorphic to $S$, not only birationally equivalent.
Jul 2, 2014 at 6:35 history answered abx CC BY-SA 3.0