LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27

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Closed immersions are local on the target

Statement

Let $i : Z \to X$ be a morphism of schemes and let $X = ⋃_{j} V_{j}$ be an open cover. Then $i$ is a closed immersion if and only if the restriction $i^{- 1} (V_{j}) \to V_{j}$ is a closed immersion for every $j$ .

Facts & Assumptions

Given: A morphism of schemes $i : Z \to X$ and an open cover $X = ⋃_{j} V_{j}$ , the restrictions carrying the restricted structure sheaves.

[F1]

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

Proof technique: direct.

Proof

1.1

Assume $i$ is a closed immersion and fix $j$ . The map $i$ restricts to a homeomorphism of $i^{- 1} (V_{j})$ onto $i (Z) \cap V_{j}$ , which is closed in $V_{j}$ . For $x \in V_{j}$ with $i (z) = x$ one has $(i_{*} O_{Z})_{x} = O_{Z, z}$ and $(i ∣_{i^{- 1} (V_{j})})_{*} O_{i^{- 1} (V_{j})}$ has the same stalk at $x$ , namely $O_{Z, z}$ ; the stalk map is the surjection $O_{X, x} \to O_{Z, z}$ of [F1]. Hence $O_{V_{j}} \to (i ∣_{i^{- 1} (V_{j})})_{*} O_{i^{- 1} (V_{j})}$ is surjective and the restriction is a closed immersion by [F1].

F1given

1.2

Conversely assume that every restriction $i^{- 1} (V_{j}) \to V_{j}$ is a closed immersion, hence injective on points. If $i (z) = i (z^{'}) = x$ , choose $j$ with $x \in V_{j}$ ; then $z, z^{'} \in i^{- 1} (V_{j})$ and injectivity gives $z = z^{'}$ . So $i$ is injective.

given

2.1

For each $j$ the set $i (Z) \cap V_{j} = i (i^{- 1} (V_{j}))$ is closed in $V_{j}$ , since it is the image of the restriction $i^{- 1} (V_{j}) \to V_{j}$ , a homeomorphism onto a closed subset. A subset of $X$ whose intersection with every member of an open cover is closed is closed, so $i (Z)$ is closed in $X$ .

givenstep 1.2

2.2

Let $C \subseteq Z$ be closed. Then $C \cap i^{- 1} (V_{j})$ is closed in $i^{- 1} (V_{j})$ , so its image $i (C) \cap V_{j} = i (C \cap i^{- 1} (V_{j}))$ is closed in $i (Z) \cap V_{j}$ for every $j$ , using the first direction of [F1] for each restriction. Hence $i (C)$ is closed in the closed subset $i (Z)$ , and $i : Z \to i (Z)$ is a homeomorphism.

F1givenstep 1.2

2.3

Finally $O_{X} \to i_{*} O_{Z}$ is surjective, because surjectivity of a morphism of sheaves is checked on stalks and at $x \in V_{j}$ the stalk map is the stalk at $x$ of the surjective map $O_{V_{j}} \to (i ∣_{i^{- 1} (V_{j})})_{*} O_{i^{- 1} (V_{j})}$ of step 1.2.

F1step 1.2

3.1

Steps 1.2, 2.1, 2.2 and 2.3 exhibit $i$ as a homeomorphism onto the closed subset $i (Z)$ with surjective structure map, so $i$ is a closed immersion by [F1]; the forward direction is step 1.1. ∎

F1step 1.1step 1.2step 2.1step 2.2step 2.3

Depends on

Closed immersions of schemes

Used by

The diagonal of the affine line Example
Affine morphisms are separated Lemma
Every scheme diagonal is an immersion Lemma
Separatedness is local on the base Lemma
Affine-overlap criterion for separatedness Theorem

Dependency tree · two levels

3 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, Lemma 26.4.2 (tag 01HL), printed p.5 (standard reference, not scraped)