False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-09-01

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FALSE: an arbitrary linear projection of an embedding is an embedding

Statement

False claim: every linear projection of an embedded submanifold is again an embedding.

Facts & Assumptions

Given: The standard unit circle

S^{1} = {(x, y) \in R^{2} : x^{2} + y^{2} = 1}

and the projection $π (x, y) = x$ onto the first coordinate.

[L1]

Only generic projection directions preserve injectivity and immersion (A generic linear projection preserves injectivity and immersion).

Refutation

technique · direct

1.1

The restriction $π ∣_{S^{1}} : S^{1} \to R$ identifies the antipodal pairs $(x, y)$ and $(x, - y)$ whenever $y \neq 0$ . Thus it is not injective.

givenalgebra

1.2

At the left and right points $(\pm 1, 0)$ , the tangent line to $S^{1}$ is vertical, so $d (π ∣_{S^{1}})$ vanishes there. Hence the projection is not even an immersion.

givenalgebra

2.1

Step 1.1 shows that one bad secant direction destroys injectivity, and step 1.2 shows that one bad tangent direction destroys immersion. So [L1] cannot be weakened to "every projection," and the claim is false.

L1step 1.1step 1.2∎

Depends on

A generic linear projection preserves injectivity and immersion

Used by

Projecting a space curve can create a double point Example

Dependency tree · two levels

4 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)