ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-01

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.

Generic affine hyperplanes meet an embedded submanifold transversely

Example

For the unit circle $S^{1} \subseteq R^{2}$ , the vertical line $H_{a} = {x = a}$ meets $S^{1}$ transversely for every $a \in R ∖ {- 1, 1}$ . For $∣ a ∣ > 1$ the intersection is empty, and for $∣ a ∣ < 1$ it consists of two transverse points.

Facts & Assumptions

Given: The height map $h : S^{1} \to R$ , $h (x, y) = x$ , and a real parameter $a$ .

[L1]

Outside a null subset of parameters, translating a Euclidean-valued map makes a chosen point a regular value (Outside a null set every translation makes a chosen value a transverse zero).

Verification

technique · direct

1.1

The fibre $h^{- 1} (a)$ is $S^{1} \cap H_{a}$ . When $∣ a ∣ < 1$ , the equation $x = a$ on $x^{2} + y^{2} = 1$ gives the two points $(a, \pm \sqrt{1 - a^{2}})$ . At either point, $d h$ is nonzero on the circle tangent line exactly when $a \neq \pm 1$ .

givenalgebra

2.1

Therefore for every $a \neq \pm 1$ , the line $H_{a}$ either misses the circle or meets it transversely. The exceptional set ${- 1, 1}$ is finite, hence null, exactly as [L1] predicts in this one-parameter family.

L1step 1.1

3.1

Thus generic affine hyperplanes meet the embedded circle transversely.

step 2.1∎

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

Marco Gualtieri, Topology I: Smooth Manifolds, cumulative notes (standard reference, not scraped)