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?



Irreducibility of the base and of the general fiber
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](https://stacks.math.columbia.edu/tag/05F7)), 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](https://math.stackexchange.com/questions/242360/irreducible-fibers-of-a-closed-subset-implies-irreducibility)), so the closure of $X\times\_Y U$ in $X$ is the irreducible component you seek.


