LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Separated morphisms compose

Statement

Let $X \overset{g}{\to} Y \overset{f}{\to} S$ be morphisms of schemes. If $g$ and $f$ are separated, then $f \circ g$ is separated.

Facts & Assumptions

Given: Morphisms of schemes $g : X \to Y$ and $f : Y \to S$ with diagonals $Δ_{X / Y}$ , $Δ_{X / S}$ , $Δ_{Y / S}$ , and the canonical morphism $u : X \times_{Y} X \to X \times_{S} X$ induced by $g$ .

[F1]

A morphism is separated when its diagonal is a closed immersion. (Separated morphism of schemes)

[F2]

The diagonal morphism $Δ_{X / S}$ is the unique morphism to $X \times_{S} X$ whose two composites with the projections are the identity; it exists by Existence of all scheme fibre products. (The diagonal morphism)

[F3]

A morphism $i : Z \to T$ is a closed immersion when its underlying map is a homeomorphism onto a closed subset and $O_{T} \to i_{*} O_{Z}$ is surjective. (Closed immersions of schemes)

[F4]

Closed immersions remain closed immersions after arbitrary base change. (Base change of immersions)

Proof

technique · direct

1.1

For closed immersions $i : Z \to Y$ and $j : Y \to T$ the composite $j i$ is a closed immersion: its underlying map is a composite of homeomorphisms onto closed subsets, hence a homeomorphism onto a closed subset of $T$ , and $O_{T} \to j_{*} (i_{*} O_{Z}) = (j i)_{*} O_{Z}$ is a composite of the surjections $O_{T} \to j_{*} O_{Y}$ and $j_{*} O_{Y} \to j_{*} (i_{*} O_{Z})$ of [F3], hence surjective.

F3given

1.2

By [F2] the diagonal $Δ_{X / S}$ is determined by ${pr}_{i} Δ_{X / S} = {id}_{X}$ for $i = 1, 2$ ; the composite $u \circ Δ_{X / Y}$ of the two canonical maps into $X \times_{S} X$ has the same two composites with the projections of $X \times_{S} X$ as does $Δ_{X / S}$ , because the projections of $X \times_{Y} X$ restrict to the projections of $X \times_{S} X$ along $u$ . Hence $Δ_{X / S} = u \circ Δ_{X / Y}$ by uniqueness in [F2].

F2given

1.3

Consider $g \times_{S} g : X \times_{S} X \to Y \times_{S} Y$ . The pullback of $Δ_{Y / S} : Y \to Y \times_{S} Y$ along this morphism is canonically $X \times_{Y} X$ : a map $T \to X \times_{S} X$ factors through that pullback exactly when its two composites $T \to X \overset{g}{\to} Y$ agree, which is the defining universal property of $X \times_{Y} X$ . Thus $u : X \times_{Y} X \to X \times_{S} X$ is this base change of $Δ_{Y / S}$ . Since $f$ is separated, [F1] makes $Δ_{Y / S}$ a closed immersion, and [F4] makes $u$ a closed immersion.

F1F4given

1.4

Since $g$ is separated, [F1] makes $Δ_{X / Y}$ a closed immersion.

F1given

2.1

By step 1.1 the composite $u \circ Δ_{X / Y}$ of the closed immersions $u$ of step 1.3 and $Δ_{X / Y}$ of step 1.4 is a closed immersion; by step 1.2 this composite is $Δ_{X / S}$ .

step 1.1step 1.2step 1.3step 1.4

3.1

Since its diagonal $Δ_{X / S}$ is a closed immersion, the composite morphism $f \circ g : X \to S$ is separated by [F1].

F1step 2.1∎

Depends on

Used by

Open and closed immersions are separated Lemma

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Schemes, Lemmas 26.21.9 and 26.21.12, printed p.41 (standard reference, not scraped)
Vakil, The Rising Sea, Section 11.3.3, printed p.308 (standard reference, not scraped)