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

1.1givenalgebra

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.

1.2givenalgebra

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.

2.1L1step 1.1step 1.2∎

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.

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

6 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)