LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 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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Morphisms of sheaves are determined by their maps on stalks

Statement

Let $φ, ψ : F \to G$ be morphisms of sheaves of sets on a topological space $X$ . If the induced maps on stalks $φ_{x}, ψ_{x} : F_{x} ⟶ G_{x}$ are equal for every $x \in X$ , then $φ = ψ$ .

Facts & Assumptions

Given: Two morphisms of sheaves $φ, ψ : F \to G$ .

[F1]

A morphism of presheaves is given by component maps compatible with restriction (Morphisms of presheaves).

[F2]

The stalk construction sends a morphism of presheaves to induced maps on stalks, and for every $s \in F (U)$ and $x \in U$ one has $(φ_{U} (s))_{x} = φ_{x} (s_{x})$ (Morphisms of presheaves, The stalk of a presheaf at a point, Germs of sections).

[F3]

Two germs at $x$ are equal exactly when their representing sections agree on some smaller neighbourhood of $x$ (The stalk of a presheaf at a point).

[L1]

Sections of a sheaf are equal once they agree on an open cover (A sheaf on a topological space).

Proof

technique · direct

1.1F2given

Fix an open set $U \subseteq X$ and a section $s \in F (U)$ . For every $x \in U$ , the hypothesis $φ_{x} = ψ_{x}$ and [F2] give $(φ_{U} (s))_{x} = φ_{x} (s_{x}) = ψ_{x} (s_{x}) = (ψ_{U} (s))_{x} .$

2.1F3step 1.1

By [F3], equality of the germs in step 1.1 means that for each $x \in U$ there exists an open neighbourhood $V_{x} \subseteq U$ such that $φ_{U} (s) ∣_{V_{x}} = ψ_{U} (s) ∣_{V_{x}} .$ The sets $V_{x}$ cover $U$ .

3.1F1L1step 2.1∎

By [L1], the two sections $φ_{U} (s)$ and $ψ_{U} (s)$ are equal on $U$ . Since $U$ and $s$ were arbitrary, the component maps $φ_{U}$ and $ψ_{U}$ agree for every open $U$ . Therefore $φ = ψ$ by [F1].

Depends on

Used by

Dependency tree · two levels

9 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 3, Exercise 4.4 (standard reference, not scraped)