CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-31

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 figure-eight curve is an immersed image but not an embedded submanifold

Statement refuted

Every immersed image is an embedded submanifold.

Facts & Assumptions

Given: The map $γ (t) = (\sin t, \sin 2 t)$ from $R$ to $R^{2}$ .

[F1]

The displayed claim is the false statement under discussion (The image of every immersion need not be an embedded submanifold).

[L1]

$(\sin t)^{'} = \cos t$ (The derivatives of sine and cosine are cosine and minus sine), and the chain rule gives $(\sin 2 t)^{'} = 2 \cos 2 t$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

Counterexample

technique · direct

1.1

By [L1], $γ^{'} (t) = (\cos t, 2 \cos 2 t)$ , and as in the false-statement proof this never vanishes. Hence $γ$ is an immersion.

L1given

2.1

The image passes through $(0, 0)$ at $t = 0$ and $t = π$ , and in fact at every integer multiple of $π$ . The two branches coming from $t = 0$ and $t = π$ already have distinct tangent directions, so the image has a transverse self-crossing and is not locally homeomorphic to an interval at the origin.

step 1.1

3.1

Thus $γ (R)$ is an immersed image that is not an embedded submanifold, refuting [F1].

F1step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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, Immersed Submanifolds (standard reference, not scraped)