Fdtxjr

I am reading the book, Birational Geometry of Algebraic Varieties, by J$acutemathrm a$nos Koll$acutemathrm a$r et al..

(Rigidity Lemma) Let $Y$ be an irreducible variety and $f:Yto Z$ a proper surjective morphism. Assume that every fiber of $f$ is connected and of dimension $n$. Let $g:Yto X$ be a morphism such that $g(f^-1(z_0))$ is a point for some $z_0in Z$. Then $g(f^-1(z))$ is a point for any $zin Z$.

Proof. Set $W=im(ftimes g)subset Ztimes X$. We obtain proper morphisms $f:Yto Wto Z$, where $h:Yto W$ and $p:Wto Z$. $p^-1(z)=h(f^-1(z))$ and $dim p^-1(z_0)=0$. By the upper semi-continuity of fiber dimensions, there is an open subset $z_0in Usubset Z$ such that $dim p^-1(z)=0$ for any $zin U$. Thus $h$ has fiber dimension $n$ over $p^-1(U)$, hence $h$ has fiber dimension at least $n$ everywhere. For any $win W$, $h^-1(w)subset f^-1(p(w))$, $dim h^-1(w)geqslant n$ and $dim f^-1(p(w))=n$. Therefore $h^-1(w)$ is a union of irreducible components of $f^-1(p(w))$, and so $h(f^-1(p(w)))=p^-1(p(w))$ is finite. It is a single point since $f^-1(p(w))$ is connected. Q.E.D.

I have no idea on the claim that $h(f^-1(p(w)))=p^-1(p(w))$ is a finite set. Maybe I should use the density of $p^-1(U)$ in $W$ ?

My try:

If $p^-1(p(w))$ is infinite, since $p^-1(U)$ is dense in $W$, one has $p^-1(p(w))cap p^-1(U)neqemptyset$, equivalently , $p(w)cap Uneqemptyset$. Hence there exists some $zin U$ such that $p(w)=z$. Note that $dim p^-1(z)=0$ for any $zin U$, impossible (since $w$ is arbitrary).

So could anyone give me some hints? Thanks!

I will assume that you are fine with the fact that $h$ has fiber dimension at least $n$ everywhere. Then, set theoretically, we can check $h^-1(w) subset f^-1(p(w))$ for any $w in W$. By assumption, $f$ has fiber dimension $n$ everywhere; thus, by the inclusion, we deduce that $h$ has fiber dimension $n$ everywhere. Now, as $f$ is proper, then so is $h$. Therefore, $h^-1(w)$ is a proper subvariety of $f^-1(p(w))$ of the same dimension. It follows that it is the union of some of the irreducible components of $f^-1(p(w))$.

Now, notice that $f^-1(p(w))=h^-1(p^-1(p(w))$. Then, $p^-1(p(w))$ has dimension 0. Indeed, if it had bigger dimension, $f^-1(p(w))$ would have dimension at least $n+1$, as $h$ has fiber dimension $n$. Since $f$ is a proper morphism, $f^-1(p(w))$ is a proper variety over the residue field of the point $p(w)$. In particular, $f^-1(p(w))$ is a variety of finite type. This implies that it has finitely many irreducible components.

Now, we know that $p^-1(p(w))$ is a collection of points. Since $h$ is proper and has fiber dimension $n$, for every point $w_i$ in $p^-1(p(w))$, the fiber $h^-1(w_i)$ provides a positive finite number of irreducible components of $f^-1(p(w))$ (they are finite since $h$ is proper, we have at least 1 since we are working in the image of the morphism, but they may be more). Thus, since $f^-1(p(w))$ has finitely many irreducible components, $p^-1(p(w))$ needs to be finite.

From the identity $f^-1(p(w))=h^-1(p^-1(p(w))$, then follows the identity $h(f^-1(p(w)))=p^-1(p(w))$.