LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04‡ rests on unproved material (inherited)

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

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

1.1

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 $(φ_{U} (s))_{x} = φ_{x} (s_{x}) = ψ_{x} (s_{x}) = (ψ_{U} (s))_{x} .$

F2given

2.1

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 $U$ .

F3step 1.1

3.1

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].

F1L1step 2.1∎

Depends on

Used by

Distinct continuous functions can share one germ, but equal germs everywhere force equality Example

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)