Lemma 37.76.3. The base change of a completely decomposed morphism of schemes is completely decomposed. If $\{ f_ i : X_ i \to Y\} _{i \in I}$ is completely decomposed and $Y' \to Y$ is a morphism of schemes, then $\{ X_ i \times _ Y Y' \to Y'\} _{i \in I}$ is completely decomposed.

Proof. Let $f : X \to Y$ and $g : Y' \to Y$ be morphisms of schemes. Let $y' \in Y'$ be a point with image $y = g(y')$ in $Y$. If $x \in X$ is a point such that $f(x) = y$ and $\kappa (x) = \kappa (y)$, then there exists a unique point $x' \in X' = X \times _ Y Y'$ which maps to $y'$ in $Y'$ and to $x$ in $X$ and moreover $\kappa (x') = \kappa (y')$, see Schemes, Lemma 26.17.5. From this fact the lemma follows easily; we omit the details. $\square$

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