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If we have a map $f:X\to Y$ with $Y$ and every fiber $f^{-1}(y)$ (where $y\in Y$) is irreducible. Can we say $X$ itself is irreducible?

If it's right, how to prove?

Akatsuki
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  • @AndrewMiloradovsky I mean they are irreducible varieties. – Akatsuki Feb 27 '16 at 08:52
  • Do you want to assume that $Y$ is irreducible? Else you could take $X=Y$ some non-irreducible space, and $f$ to be the identity map. – Alex Youcis Feb 27 '16 at 09:12
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    It also is not true if $Y$ is irreducible. See here: http://math.stackexchange.com/questions/242360/irreducible-fibers-of-a-closed-subset-implies-irreducibility. If $f$ is proper though, this is a classic exercise (assuming $Y$ is irreducible). – Alex Youcis Feb 27 '16 at 09:14

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