Lemma 72.12.3 (0A11)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 72: Algebraic Spaces over Fields
Section 72.12: Geometrically connected algebraic spaces
Lemma 72.12.3 (cite)

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Lemma 72.12.3. Let $k$ be a field. Let $X$, $Y$ be algebraic spaces over $k$. Assume $X$ is geometrically connected over $k$. Then the projection morphism

\[ p : X \times _ k Y \longrightarrow Y \]

induces a bijection between connected components.

Proof. Let $y \in |Y|$ be represented by a morphism $\mathop{\mathrm{Spec}}(K) \to Y$ where $K$ is a field. The fibre of $|X \times _ k Y| \to |Y|$ over $y$ is the image of $|X_ K| \to |X \times _ k Y|$ by Properties of Spaces, Lemma 66.4.3. Thus these fibres are connected by our assumption that $X$ is geometrically connected. By Morphisms of Spaces, Lemma 67.6.6 the map $|p|$ is open. Thus we may apply Topology, Lemma 5.7.6 to conclude. $\square$

Comments (3)

Comment #5941 by Dario Weißmann on February 19, 2021 at 07:25

typos: Y_K should be X_K when showing that the fibres are connected. Also in the next sentence "Y is geometrically connected" should be "X is geometrically connected".

Comment #5942 by Laurent Moret-Bailly on February 19, 2021 at 09:05

First line of proof: delete "be a morphism".

Comment #6130 by Johan on April 30, 2021 at 12:58

Thanks to both of you. The fix is in this commit.

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