TheoremStatement: 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.

A morphism of sheaves is an isomorphism exactly when it is an isomorphism on every stalk

Statement

Let $φ : F \to G$ be a morphism of sheaves of sets on a topological space $X$ . Then $φ$ is an isomorphism if and only if, for every $x \in X$ , the induced map on stalks $φ_{x} : F_{x} ⟶ G_{x}$ is a bijection.

Facts & Assumptions

Given: A morphism of sheaves $φ : F \to G$ .

[F1]

A morphism of presheaves is an isomorphism when it has a two-sided inverse whose components commute with restriction (Morphisms of presheaves).

[F2]

The stalk construction sends morphisms of presheaves to induced stalk maps, 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).

[L2]

A sheaf is glued from local sections that agree on overlaps (A sheaf on a topological space).

Proof

technique · direct

1.1

If $φ$ is an isomorphism with inverse $ψ$ , then for every $x$ the induced maps $φ_{x}$ and $ψ_{x}$ are inverse because taking stalks preserves composition and identities by [F2]. Hence each $φ_{x}$ is a bijection.

F1F2

1.2

Assume now that every $φ_{x}$ is a bijection. Fix an open set $U \subseteq X$ and a section $t \in G (U)$ . For each $x \in U$ , choose $a_{x} \in F_{x}$ with $φ_{x} (a_{x}) = t_{x}$ . Pick a representative $s_{x} \in F (V_{x})$ of $a_{x}$ on some open neighbourhood $V_{x} \subseteq U$ of $x$ . Then $(φ_{V_{x}} (s_{x}))_{x} = t_{x}$ , so [F3] lets us shrink $V_{x}$ and assume $φ_{V_{x}} (s_{x}) = t ∣_{V_{x}}$ .

F2F3givenchoose

1.3

If $s, s^{'} \in F (U)$ satisfy $φ_{U} (s) = φ_{U} (s^{'})$ , then for every $x \in U$ one has $φ_{x} (s_{x}) = φ_{x} (s_{x}^{'})$ by [F2]. Injectivity of $φ_{x}$ gives $s_{x} = s_{x}^{'}$ . By [F3], for each $x$ there exists an open neighbourhood $W_{x} \subseteq U$ with $s ∣_{W_{x}} = s^{'} ∣_{W_{x}}$ . The sets $W_{x}$ cover $U$ , so [L2] gives $s = s^{'}$ . Thus each $φ_{U}$ is injective.

F2F3L2

2.1

Let $x, y \in U$ . For any $z \in V_{x} \cap V_{y}$ , the two sections $s_{x} ∣_{V_{x} \cap V_{y}}$ and $s_{y} ∣_{V_{x} \cap V_{y}}$ have images under $φ$ whose germs at $z$ both equal $t_{z}$ . Since $φ_{z}$ is injective, the germs of $s_{x}$ and $s_{y}$ at $z$ are equal. By [F3], there is a neighbourhood $W_{z} \subseteq V_{x} \cap V_{y}$ of $z$ on which $s_{x} ∣_{W_{z}} = s_{y} ∣_{W_{z}}$ . The sets $W_{z}$ cover $V_{x} \cap V_{y}$ , so [L2] gives $s_{x} ∣_{V_{x} \cap V_{y}} = s_{y} ∣_{V_{x} \cap V_{y}} .$ Therefore the family $(s_{x})$ is compatible on the cover ${V_{x}}_{x \in U}$ of $U$ .

F2F3L2step 1.2

3.1

By [L2], the compatible family of step 2.1 glues to a unique section $ψ_{U} (t) \in F (U)$ satisfying $φ_{U} (ψ_{U} (t)) = t$ . Thus $φ_{U}$ is surjective for every $U$ .

L2step 2.1construct

4.1

Let $V \subseteq U$ and $t \in G (U)$ . Then $φ_{V} (ψ_{U} (t) ∣_{V}) = φ_{U} (ψ_{U} (t)) ∣_{V} = t ∣_{V} = φ_{V} (ψ_{V} (t ∣_{V})) .$ Step 1.3 gives injectivity of $φ_{V}$ , so $ψ_{U} (t) ∣_{V} = ψ_{V} (t ∣_{V})$ . Hence the maps $ψ_{U}$ commute with restriction and define a morphism of sheaves $ψ : G \to F$ . By steps 3.1 and 1.3, $ψ$ is a two-sided inverse to $φ$ , so [F1] shows that $φ$ is an isomorphism. Together with step 1.1 this proves both directions.

F1step 3.1step 1.3∎

Depends on

Used by

Nothing in the library uses this result yet.

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 4, Exercise 4.5 (standard reference, not scraped)
The Stacks Project, Sheaves on Spaces, Section 16 (standard reference, not scraped)