Theorem 9.8.1176C7A4C3D95

Proof. Choose a nowhere dense analytic subset $A\subset X$ such that $f^{-1}(A)$ is nowhere dense and $f$ is an isomorphism over $X\setminus A$. Suppose that the compact fiber over $x$ is disconnected, and write it as the union of two disjoint nonempty compact sets. Since $Y$ is Hausdorff, they have disjoint open neighborhoods $V_1$ and $V_2$. Since $f$ is proper, hence closed, $U=X\setminus f(Y\setminus (V_1\cup V_2))$ is an open neighborhood of $x$ with $f^{-1}(U)\subset V_1\cup V_2$. As $(X,x)$ is irreducible, corollary 9.4.9 allows us to shrink $U$ so that $U\setminus A$ is connected. Each set $V_ i\cap f^{-1}(U)$ is open and nonempty, so it meets the dense set $Y\setminus f^{-1}(A)$. Therefore $V_1$ and $V_2$ split $f^{-1}(U\setminus A)$ into two disjoint nonempty open sets. This contradicts the connectedness of $U\setminus A$, which is isomorphic to $f^{-1}(U\setminus A)$. $\square$