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.

Projecting a space curve can create a double point

Example

On $S^{1} = R / 2 π Z$ , consider the embedded space curve $γ ([t]) : = (\cos t, \sin t, \cos 2 t) \subseteq R^{3} .$ Projecting along the $y$ -axis gives $π (γ ([t])) = (\cos t, \cos 2 t) \in R^{2} .$ The distinct parameters $t$ and $- t$ have the same projected image, so this projection creates double points.

Facts & Assumptions

Given: The space curve $γ : S^{1} \to R^{3}$ above and the $y$ -coordinate projection $π$ .

Verification

technique · direct

1.1givenalgebra

The curve $γ$ is smooth and embedded: the first two coordinates already parametrize the standard unit circle, so the third coordinate only lifts that circle into space.

2.1step 1.1algebra

The projection $π$ forgets the $y$ -coordinate. Since $\cos (- t) = \cos t$ and $\cos (- 2 t) = \cos 2 t$ , one has $π (γ ([- t])) = π (γ ([t]))$ for every $t$ , while $γ ([- t]) \neq γ ([t])$ away from $[0]$ and $[π]$ . Thus the projected curve has double points.

3.1step 2.1∎

This is a concrete bad direction of the type excluded by A generic linear projection preserves injectivity and immersion, so it witnesses the false claim FALSE: an arbitrary linear projection of an embedding is an embedding.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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, 2nd ed., Lemma 6.13 (standard reference, not scraped)