Lemma 37.28.5. Let $f : X \to Y$ be a morphism of schemes. Let $n_{X/Y}$ be the function on $Y$ counting the numbers of geometrically connected components of fibres of $f$ introduced in Lemma 37.28.3. Assume $f$ of finite type. Let $y \in Y$ be a point. Then there exists a nonempty open $V \subset \overline{\{ y\} }$ such that $n_{X/Y}|_ V$ is constant.

**Proof.**
Let $Z$ be the reduced induced scheme structure on $\overline{\{ y\} }$. Let $f_ Z : X_ Z \to Z$ be the base change of $f$. Clearly it suffices to prove the lemma for $f_ Z$ and the generic point of $Z$. Hence we may assume that $Y$ is an integral scheme, see Properties, Lemma 28.3.4. Our goal in this case is to produce a nonempty open $V \subset Y$ such that $n_{X/Y}|_ V$ is constant.

We apply Lemma 37.24.8 to $f : X \to Y$ and we get $g : Y' \to V \subset Y$. As $g : Y' \to V$ is surjective finite étale, in particular open (see Morphisms, Lemma 29.36.13), it suffices to prove that there exists an open $V' \subset Y'$ such that $n_{X'/Y'}|_{V'}$ is constant, see Lemma 37.27.3. Thus we see that we may assume that all irreducible components of the generic fibre $X_\eta $ are geometrically irreducible over $\kappa (\eta )$. By Varieties, Lemma 33.8.16 this implies that also the connected components of $X_\eta $ are geometrically connected.

At this point suppose that $X_\eta = X_{1, \eta } \bigcup \ldots \bigcup X_{n, \eta }$ is the decomposition of the generic fibre into (geometrically) connected components. In particular $n_{X/Y}(\eta ) = n$. Let $X_ i$ be the closure of $X_{i, \eta }$ in $X$. After shrinking $Y$ we may assume that $X = \bigcup X_ i$, see Lemma 37.24.5. After shrinking $Y$ some more we see that each fibre of $f$ has at least $n$ connected components, see Lemma 37.28.1. Hence $n_{X/Y}(y) \geq n$ for all $y \in Y$. After shrinking $Y$ some more we obtain that $X_{i, y}$ is geometrically connected for each $i$ and all $y \in Y$, see Lemma 37.28.4. Since $X_ y = \bigcup X_{i, y}$ this shows that $n_{X/Y}(y) \leq n$ and finishes the proof. $\square$

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