Lemma 37.28.1. Let f : X \to Y be a morphism of schemes. Assume Y irreducible with generic point \eta and f of finite type. If X_\eta has n connected components, then there exists a nonempty open V \subset Y such that for all y \in V the fibre X_ y has at least n connected components.
Proof. As the question is purely topological we may replace X and Y by their reductions. In particular this implies that Y is integral, see Properties, Lemma 28.3.4. Let X_\eta = X_{1, \eta } \cup \ldots \cup X_{n, \eta } be the decomposition of X_\eta into connected components. Let X_ i \subset X be the reduced closed subscheme whose generic fibre is X_{i, \eta }. Note that Z_{i, j} = X_ i \cap X_ j is a closed subset of X whose generic fibre Z_{i, j, \eta } is empty. Hence after shrinking Y we may assume that Z_{i, j} = \emptyset , see Lemma 37.24.1. After shrinking Y some more we may assume that X_ y = \bigcup X_{i, y} for y \in Y, see Lemma 37.24.5. Moreover, after shrinking Y we may assume that each X_ i \to Y is flat and of finite presentation, see Morphisms, Proposition 29.27.1. The morphisms X_ i \to Y are open, see Morphisms, Lemma 29.25.10. Thus there exists an open neighbourhood V of \eta which is contained in f(X_ i) for each i. For each y \in V the schemes X_{i, y} are nonempty closed subsets of X_ y, we have X_ y = \bigcup X_{i, y} and the intersections Z_{i, j, y} = X_{i, y} \cap X_{j, y} are empty! Clearly this implies that X_ y has at least n connected components. \square
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