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False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-31
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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→R2, γ(t)=(sin⁡t,sin⁡2t).

[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⁡2t)′=2cos⁡2t (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

Refutation

technique · direct
1.1L1given

By [L1], γ′(t)=(cos⁡t,2cos⁡2t). If cos⁡t=0, then t=π/2+kπ, so cos⁡2t=−1 and the second component of γ′(t) is not zero. Thus γ′(t)≠0 for all t, and γ is an immersion.

2.1F1step 1.1

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.

3.1F2step 1.1step 2.1∎

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

Depends on

Used by

Dependency tree · two levels

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Sources