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May 21, 2017 at 9:18 vote accept Heitor
May 20, 2017 at 9:41 vote accept Heitor
May 20, 2017 at 9:41
May 19, 2017 at 12:16 answer added Jason Starr timeline score: 1
May 19, 2017 at 10:40 comment added Jason Starr You can prove that by Noetherian induction. In the usual way, you reduce to the case that $X$ is integral. By generic smoothness, there exists a dense open subset $V$ of $X$ on which $\pi$ is smooth. If the closed complement $Z$ of $V$ does not dominate $Y$, then choose $U$ to be the open complement of the closure of $\pi(Z)$. If $Z$ does dominate $Y$, then by the induction hypothesis, there exists a dense open subset $U$ of $Y$ such that $Z\times_Y U \to U$ has surjective $d\pi$. Since also $d\pi$ is surjective on $V\times_Y U$, the result is proved by Noetherian induction.
May 19, 2017 at 9:58 comment added Francesco Polizzi If $\pi$ is generically unramified and $Y$ is smooth, there is an exact sequence $$ 0 \longrightarrow T_X \stackrel{d \pi}{\longrightarrow}\pi^*T_Y \to N_{\pi} \longrightarrow 0,$$ where $N_{\pi}=T_{X/Y}$ if $X$ is also smooth. See [Sernesi, Deformations of algebraic schemes, p. 162].
May 19, 2017 at 9:43 history asked Heitor CC BY-SA 3.0