False statementConstruction: 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.

The image of every immersion need not be an embedded submanifold

Statement

False claim: the image of every immersion is an embedded submanifold.

Facts & Assumptions

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

[F1]

Embedded submanifolds are locally modeled on coordinate slices, hence on a one-manifold they cannot have a self-crossing neighbourhood (Embedded submanifolds and slice charts).

[F2]

An immersed submanifold need only come from an injective immersion on its own manifold (Immersed submanifolds).

[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: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

Refutation

technique · direct

1.1

By [L1], $γ^{'} (t) = (\cos t, 2 \cos 2 t)$ . If $\cos t = 0$ , then $t = π / 2 + k π$ , so $\cos 2 t = - 1$ and the second component of $γ^{'} (t)$ is not zero. Thus $γ^{'} (t) \neq 0$ for all $t$ , and $γ$ is an immersion.

L1given

2.1

The points $t = 0$ and $t = π$ both map to $(0, 0)$ , and the two tangent directions there are $(1, 2)$ and $(- 1, 2)$ . Hence the image has a transverse self-crossing at the origin. A neighbourhood of that crossing is not homeomorphic to an interval, so by [F1] the image is not an embedded one-submanifold.

F1step 1.1

3.1

Therefore the image of the immersion $γ$ fails to be embedded, refuting the claim.

F2step 1.1step 2.1∎

Depends on

Used by

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

Dependency tree · two levels

14 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)