LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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A generic linear projection preserves injectivity and immersion

Statement

Assume the Axiom of Countable Choice. Let $f : M^{n} \to R^{N}$ be a smooth embedding with $N > 2 n + 1$ . Then the set of unit vectors $u \in S^{N - 1}$ for which the orthogonal projection $P_{u} : R^{N} \to u^{⊥}$ makes $P_{u} \circ f$ an injective immersion is dense in $S^{N - 1}$ .

Facts & Assumptions

Given: Countable Choice and a smooth embedding $f : M^{n} \to R^{N}$ with $N > 2 n + 1$ .

[F1]

The secant-direction map $σ_{f}$ is defined on $(M \times M) ∖ Δ_{M}$ , and the tangent-direction map $τ_{f}$ is defined on $T M ∖ 0_{M}$ (Secant and tangent direction maps of a Euclidean embedding).

[L1]

The image of a $C^{1}$ manifold of dimension strictly smaller than the target-manifold dimension is a null set (The image of a lower-dimensional $C^{1}$ manifold is null).

[L2]

A null subset of a positive-dimensional manifold has dense complement (A null set has dense complement in a positive-dimensional manifold).

[L3]

A finite union of manifold null sets is null under Countable Choice (Countable unions and subsets of manifold null sets are null).

Proof

technique · direct

1.1F1L1given

The manifold $(M \times M) ∖ Δ_{M}$ has dimension $2 n$ , and $T M ∖ 0_{M}$ also has dimension $2 n$ . Since $S^{N - 1}$ has dimension $N - 1 > 2 n$ , [L1] under the stated Countable Choice shows that both images $σ_{f} ((M \times M) ∖ Δ_{M})$ and $τ_{f} (T M ∖ 0_{M})$ are null subsets of $S^{N - 1}$ .

2.1L2L3step 1.1choose

By [L3], the union of the two bad sets is null. By [L2], its complement is dense in $S^{N - 1}$ . Fix $u$ in that complement.

3.1F1step 2.1algebra

If $P_{u} (f (p)) = P_{u} (f (q))$ , then $f (q) - f (p)$ is parallel to $u$ . Because $f$ is injective, either $p = q$ or $u = \pm σ_{f} (p, q)$ . The second alternative is impossible by step 2.1, so $p = q$ . Thus $P_{u} \circ f$ is injective.

3.2F1step 2.1algebra

If $d (P_{u} \circ f)_{p} (v) = 0$ for some $v \neq 0$ , then $d f_{p} (v)$ is parallel to $u$ , so $u = \pm τ_{f} (p, v)$ . This again contradicts step 2.1. Hence $d (P_{u} \circ f)_{p}$ is injective for every $p$ , and $P_{u} \circ f$ is an immersion.

4.1step 2.1step 3.1step 3.2∎

Therefore every $u$ outside the secant and tangent images gives an injective immersion after projection, and such $u$ form a dense set.

Depends on

Used by

Projecting a space curve can create a double point Example
FALSE: an arbitrary linear projection of an embedding is an embedding False statement
The weak Whitney proper embedding theorem Theorem

Dependency tree · two levels

21 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)
Marco Gualtieri, Topology I: Smooth Manifolds, Part 11 (standard reference, not scraped)