CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07

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.

A submanifold meeting the ambient boundary nonneatly

Statement refuted

In $M = [0, \infty) \times R$ , the embedded interval $S = {0} \times [- 1, 1]$ is not neat.

Facts & Assumptions

Given: The standard manifold with boundary $M = [0, \infty) \times R$ and the subset $S = {0} \times [- 1, 1]$ , supplied with the smooth structure transported from $[- 1, 1]$ by $s \mapsto (0, s)$ .

[L1]

An embedded submanifold with boundary is a subset carrying a manifold-with-boundary structure for which inclusion into the ambient manifold is a smooth embedding (Embedded smooth submanifolds with boundary).

[L2]

Neatness requires both $S \cap \partial M = \partial S$ and transversality to $\partial M$ (Neat submanifolds of a manifold with boundary).

Counterexample

technique · direct

1.1

The parametrization $[- 1, 1] \to S$ , $s \mapsto (0, s)$ , is a diffeomorphism onto $S$ with its subspace topology, and its derivative $v \mapsto (0, v)$ is injective. Thus the inclusion is a smooth embedding, so [L1] makes $S$ an embedded submanifold with boundary $\partial S = {(0, - 1), (0, 1)}$ .

givenL1constructalgebra

2.1

Since $\partial M = {0} \times R$ , one has $S \cap \partial M = S$ , which is not the two-point set $\partial S$ . The equality required by [L2] therefore fails, so $S$ is not neat (independently of the transversality condition).

givenL2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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

Ioan Mărcuț, Manifolds (2017 lecture notes), §§14.5, 15.1 (standard reference, not scraped)
Will Merry, Differential Geometry (2021), Lecture 24 (standard reference, not scraped)