Lemma 76.29.3 (0E09)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 76: More on Morphisms of Spaces
Section 76.29: Reduced fibres
Lemma 76.29.3 (cite)

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Lemma 76.29.3. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y' \to Y$ be morphisms of algebraic spaces over $S$ . Denote $f' : X' \to Y'$ the base change of $f$ by $g$ . Then

$\begin{align*} \{ y' \in |Y'| : \text{the fibre of }f' : X' \to Y'\text{ at }y' \text{ is geometrically reduced}\} \\ = g^{-1}(\{ y \in |Y| : \text{the fibre of }f : X \to Y\text{ at }y \text{ is geometrically reduced}\} ). \end{align*}$

Proof. For $y' \in |Y'|$ choose a morphism $\mathop{\mathrm{Spec}}(k) \to Y'$ in the equivalence class of $y'$ . Then $g(y')$ is represented by the composition $\mathop{\mathrm{Spec}}(k) \to Y' \to Y$ . Hence $X' \times _{Y'} \mathop{\mathrm{Spec}}(k) = X \times _ Y \mathop{\mathrm{Spec}}(k)$ and the result follows from the definition. $\square$

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