TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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Graphs of morphisms to a projective classical variety are closed

Statement

Let $f : X \to Y$ be a classical variety morphism for which a classical product $P = X \times Y$ has been constructed, and suppose $Y$ is projective. Define $Γ_{f}$ as the image of the product morphism $⟨ {id}_{X}, f ⟩ : X \to P$ . Then $Γ_{f}$ is closed in $P$ . No assertion is made for unconstructed products or arbitrary nonseparated prevarieties.

Facts & Assumptions

Given: $f : X \to Y$ and a constructed product $X \times Y$ , with $Y$ projective.

Proof

1.1given

The diagonal $Δ_{Y} \subseteq Y \times Y$ is closed: embed $Y$ projectively and use the Segre/projective coordinate model, where equality of two projective points is cut out on the standard charts by coordinate differences.

2.1step 1.1

Let $p_{X} : P \to X$ and $p_{Y} : P \to Y$ be the structure maps. The morphism $F = ⟨ f \circ p_{X}, p_{Y} ⟩ : P ⟶ Y \times Y$ is defined by the two product universal properties, without choosing a pair-set realization of $P$ . Its inverse image of $Δ_{Y}$ is exactly the image of $⟨ {id}_{X}, f ⟩$ .

3.1step 2.1∎

Since inverse images of closed algebraic sets under a morphism are closed, $F^{- 1} (Δ_{Y}) = Γ_{f}$ is closed. This is precisely Milne's projective-target graph criterion under the stated construction hypothesis.

Depends on

Used by

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Dependency tree · two levels

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Sources

J. S. Milne, Algebraic Geometry, Corollaries 5.27 and 5.28 (standard reference, not scraped)
MIT 18.725 Algebraic Geometry, Lecture 7, separatedness (standard reference, not scraped)