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Let $f:X\rightarrow Y$ be a morphism of scheme over $\mathbb{C}$. Assume that $Y$ and the the general fiber $F_y = f^{-1}(y)$ of $f$ are irreducible.

Does there exists an irreducible component $X'$ of $X$ such that $f' := f_{|X'}:X'\rightarrow Y$ satisfies $(f')^{-1}(y) = f^{-1}(y)$ for $y\in Y$ a general point?

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1 Answer 1

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If you assume $f$ is of finite type and $X$ has finitely many irreducible components, then the answer is yes. There's a nonempty open neighborhood $V$ of $\eta_Y$ so that for $y\in V$, we have that $\dim X_y=\dim X_{\eta_y}$ (ref), so there's a nonempty open subset $U\subset Y$ where $X\times_Y U\to U$ has irreducible fibers all of the same dimension. This implies that $X\times_Y U$ is irreducible (ref), so the closure of $X\times_Y U$ in $X$ is the irreducible component you seek.

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