PomeTheorem 9.10.5 (Zariski's main theorem).E707851D5DBA Let $\pi \colon Y\rightarrow X$ be a proper bimeromorphic morphism between complex spaces. Assume that $X$ is unibranch. Then $\pi $ has connected fibers.
Theorem 9.10.5 (Zariski's main theorem)E707851D5DBA
Proof. Let $Z\subseteq X$ be the smallest nowhere dense analytic subset such that $\pi $ is an isomorphism over $X\setminus Z$. Assume that the fiber $F=\pi ^{-1}(x)$ over some $x\in X$ is disconnected. Since $\pi $ is proper, $F$ is compact. Choose a decomposition $F=F_1\sqcup F_2$ into two non-empty disjoint compact open subsets of $F$, and choose disjoint open neighborhoods $W_1,W_2$ of $F_1,F_2$ in $Y$. Since proper maps are closed, after shrinking an open neighborhood $U$ of $x$, we may assume that
\[ \pi ^{-1}(U)\subseteq W_1\cup W_2. \]
Thus $\pi ^{-1}(U)$ is disconnected.
Since $X$ is unibranch, we may shrink $U$ further so that $U$ is irreducible. Then $U\setminus Z$ is connected (cf. [Chapter 9, §1.2, CAS]). Since $\pi $ is bimeromorphic, every irreducible component of $Y$ maps onto an irreducible component of $X$. Hence $\pi ^{-1}(Z)$ contains no irreducible component of $Y$ and is nowhere dense. On the other hand, write
\[ \pi ^{-1}(U)=V_1\sqcup V_2 \]
as a union of two non-empty disjoint open subsets. Each $V_ i$ meets $\pi ^{-1}(U\setminus Z)$: otherwise $V_ i$ would be a non-empty open subset of $Y$ contained in the nowhere dense analytic subset $\pi ^{-1}(Z)$, which is impossible. Hence
\[ \pi ^{-1}(U\setminus Z) = \bigl (V_1\cap \pi ^{-1}(U\setminus Z)\bigr ) \sqcup \bigl (V_2\cap \pi ^{-1}(U\setminus Z)\bigr ) \]
is disconnected. This contradicts the fact that $\pi \colon \pi ^{-1}(U\setminus Z)\rightarrow U\setminus Z$ is an isomorphism and $U\setminus Z$ is connected. $\square$
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