Lemma 67.16.1 (082X)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.16: Scheme theoretic image
Lemma 67.16.1 (cite)

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The scheme-theoretic image of a morphism of algebraic spaces exists.

Lemma 67.16.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ . There exists a closed subspace $Z \subset Y$ such that $f$ factors through $Z$ and such that for any other closed subspace $Z' \subset Y$ such that $f$ factors through $Z'$ we have $Z \subset Z'$ .

Proof. Let $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$ . If $\mathcal{I}$ is quasi-coherent then we just take $Z$ to be the closed subscheme determined by $\mathcal{I}$ , see Lemma 67.13.1. In general the lemma requires us to show that there exists a largest quasi-coherent sheaf of ideals $\mathcal{I}'$ contained in $\mathcal{I}$ . This follows from Lemma 67.14.2. $\square$

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Comment #1285 by Johan Commelin on January 29, 2015 at 12:13

Suggested slogan: The scheme-theoretic image of a morphism of algebraic spaces exists.

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