Lemma 37.27.1. 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 37.23.1. After shrinking $Y$ some more we may assume that $X_ y = \bigcup X_{i, y}$ for $y \in Y$, see Lemma 37.23.5. 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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