Let $f:X \to Y$ be a proper surjective morphism of quasiprojective noetherian schemes over $\mathbb{C}$. Assume that $Y$ is irreducible and $X$ is reduced, connected with finitely many irreducible components. Suppose further that every closed fiber of $f$ is an integral scheme. Is there any known additional condition on $f$ (other than flatness) under which we can conclude that $X$ is irreducible?
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$\begingroup$ One possible condition would be that all closed fibers have the same dimension. $\endgroup$– Philipp HartwigCommented Mar 15, 2015 at 20:52

$\begingroup$ $f$ open is sufficient (proof immediate). $\endgroup$– abxCommented Mar 15, 2015 at 20:55
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This is false. Take for $X$ the union of the two axes in $\mathbf P^1\times\mathbf P^1$ and take for $f$ the first projection to $Y=\mathbf P^1$.