The Stacks project

Separated and quasi-separated morphisms satisfy cancellation.

Lemma 26.21.13. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes. If $g \circ f$ is separated then so is $f$. If $g \circ f$ is quasi-separated then so is $f$.

Proof. Assume that $g \circ f$ is separated. Consider the factorization $X \to X \times _ Y X \to X \times _ Z X$ of the diagonal morphism of $g \circ f$. By Lemma 26.21.9 the last morphism is an immersion. By assumption the image of $X$ in $X \times _ Z X$ is closed. Hence it is also closed in $X \times _ Y X$. Thus we see that $X \to X \times _ Y X$ is a closed immersion by Lemma 26.10.4.

Assume that $g \circ f$ is quasi-separated. Let $V \subset Y$ be an affine open which maps into an affine open of $Z$. Let $U_1, U_2 \subset X$ be affine opens which map into $V$. Then $U_1 \cap U_2$ is a finite union of affine opens because $U_1, U_2$ map into a common affine open of $Z$. Since we may cover $Y$ by affine opens like $V$ we deduce the lemma from Lemma 26.21.6. $\square$


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Comment #822 by on

Suggested slogan: Separated and quasi-separated morphisms satisfy cancellation. (This is also a test of the system)

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