LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A section of a sheaf of groups is zero exactly when all of its germs are zero

Statement

Let $F$ be a sheaf of groups on a topological space $X$ , let $U \subseteq X$ be open, and let $s \in F (U)$ . Then $s = 0 in F (U) ⟺ s_{x} = 0 in F_{x} for every x \in U .$

Facts & Assumptions

Given: A sheaf of groups $F$ , an open set $U$ , and a section $s \in F (U)$ .

[F1]

A sheaf of groups has an identity section on every open set; when the group law is written additively, this section is denoted by $0$ , and all restrictions preserve it (Presheaves and sheaves of groups, rings, and modules).

[L1]

A sheaf is determined by local agreement of sections on an open cover (A sheaf on a topological space).

[F2]

The germ $s_{x}$ is the class of $s$ in the stalk at $x$ (Germs of sections). Here $0_{x}$ (also written $0$ in $F_{x}$ ) denotes the germ of an identity section. This is independent of the neighbourhood used: any two identity sections restrict to the identity on their intersection by [F1], and so have the same germ. No group operation on the stalk is needed for this notation.

Proof

technique · direct

1.1F1F2

If $s = 0$ in $F (U)$ , then for every $x \in U$ the induced germ is $s_{x} = 0_{x}$ by the definition of $0_{x}$ in [F2].

1.2F2given

Assume $s_{x} = 0$ for every $x \in U$ . Fix $x \in U$ . By lem-germ-equivalence-relation, equality of the two germ classes means that their representatives agree on some open neighbourhood $V_{x} \subseteq U$ of $x$ , so $s ∣_{V_{x}} = 0 ∣_{V_{x}}$ . The sets $V_{x}$ cover $U$ .

2.1L1step 1.1step 1.2∎

On the open cover ${V_{x}}_{x \in U}$ , the sections $s$ and $0$ have the same restriction by step 1.2. Locality in [L1] therefore gives $s = 0$ in $F (U)$ . For $U = \emptyset$ , this uses the empty cover, which is included in [L1]. Together with step 1.1, this proves both directions.

Depends on

Used by

Global functions on geometrically connected and geometrically reduced proper schemes Corollary
Godement resolution of an abelian sheaf Definition
Support of a section of a sheaf Definition
Godement terms are flasque and compute cohomology Theorem

Dependency tree · two levels

11 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

Ravi Vakil, Foundations of Algebraic Geometry, Class 4 (standard reference, not scraped)
The Stacks Project, Sheaves on Spaces, Section 15 (standard reference, not scraped)