ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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.

Homotopy equivalent annulus and circle have isomorphic de rham rings

Example

Under countable choice, the annulus $A = {x \in R^{2} : a < ∣ x ∣ < b}$ , where $0 < a < 1 < b$ , and $S^{1}$ have isomorphic de Rham graded algebras.

Facts & Assumptions

Given: Assume countable choice. Inclusion $i : S^{1} \to A$ and $r : A \to S^{1}$ , $r (x) = x / ∣ x ∣$ .

[F1]

De rham cohomology is smooth homotopy invariant: A smooth homotopy equivalence induces an isomorphism of de Rham graded real algebras.

[F2]

Pullback is a homomorphism of de rham cohomology algebras: Smooth pullback induces a unital graded real algebra homomorphism $H_{d R}^{∙} (N) \to H_{d R}^{∙} (M)$ .

[F3]

De rham cohomology of spheres: Assume countable choice. For $n \geq 1$ , $H_{d R}^{k} (S^{n})$ is $R$ in degrees $0, n$ and zero otherwise. For $S^{0}$ it is $R^{2}$ in degree zero and zero otherwise.

Verification

technique · direct

1.1

The homotopy $F (x, t) = ((1 - t) + t / ∣ x ∣) x$ has radius $(1 - t) ∣ x ∣ + t$ , between $∣ x ∣$ and $1$ , hence strictly between $a$ and $b$ . It is smooth, fixes the unit circle, and connects the identity to $i r$ ; also $r i = id$ .

givenalgebra

2.1

Thus $i^{*}, r^{*}$ are inverse graded algebra homomorphisms by smooth homotopy invariance. The sphere computation gives one generator $u$ in degree one and the unit in degree zero; $u^{2} = 0$ because $H^{2} (S^{1}) = 0$ . The annulus has the same multiplication, so its ring is the exterior algebra on one degree-one generator.

F1F2F3step 1.1∎

Source locator

Lee, Proposition 17.10 and Theorem 17.11, pp.445–446; the annulus retraction and its radial homotopy are explicit.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)