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Let $f:X \to Y$ be a morphism of varieties and let $Z \subset X$ be a closed subset. Assume that $f^{-1}(p) \cap Z$ is irreducible and of the same dimension for all $p \in Y.$ Show that $Z$ is irreducible.

Here is what I tried:

Assume the contrary and let $Z=Z_1 \cup Z_2$ where $Z_i$s are closed in $Z.$ By assumption, $f^{-1}(p) \cap Z=(f^{-1}(p) \cap Z_1) \cup (f^{-1}(p) \cap Z_2)$ is irreducible, so let $U=\{p \in Y|\; f^{-1}(p) \cap Z=f^{-1}(p) \cap Z_1 \}$ and $V=\{p \in Y|\; f^{-1}(p) \cap Z=f^{-1}(p) \cap Z_2 \}.$ My strategy is to prove that $U,V$ are closed in $Y$ and get the contradiction ( $Y$ is a variety, so irreducible). For instance, consider $U \subset Y.$ Then $Y$ has a finite open affine cover by $Y_i$s, so is enough to show that $U \cap Y_i$ is closed in $Y_i$ for all $i.$ I also know that there is an open subset in $W \subset Y$ s.t. all fibers $f^{-1}(x)$ have the same dimension for all $x \in W,$ but I got stuck and don’t know how to proceed and use the fact that $f^{-1}(p) \cap Z$ have the same dimension for all $p \in Y.$

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    I don't know what do you meaning by "varieties". At least $Y$ must be irreducible (obvious), and $Z$ must be connected (otherwise let $Y$ be the affine line and $Z=X$ be dijoint union of one point of $Y$ and of its complement).2012-11-22
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    Dear @QiL by variety I mean, an irreducible prevariety which is covered by finitely many open affines or a separated and irreducible scheme of finite type over an alg. closed field.2012-11-23
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    OK now I see in your post it is said $Y$ is irreducible. Sorry. I remember I did an exercise like this one in Harthorne, but how ? I can show that for any *irreducible component* $Z_i$ of $Z$, we have $Z_i=f^{-1}(f(Z_i))\cap Z$, but then I also got stuck. The images $f(Z_i)$ are constructible in $Y$, but this is not enough to conclude.2012-11-23
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    Note that to simply the settings, one can replace $X$ with $Z$ (a scheme of finite type over $Y$) and try to show $X$ is irreducible under the assumption that $f : X\to Y$ has irreducible fibers of the same dimension ($Y$ irreducible and $X$ connected).2012-11-23

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