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

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Every closed embedded submanifold has a smooth neighborhood retraction

Statement

Assume countable choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $i : S ↪ M$ be a closed smooth embedded submanifold. Then $S$ has an open neighbourhood in $M$ that retracts smoothly onto $S$ .

Facts & Assumptions

Given: Countable choice and a closed smooth embedded submanifold $i : S ↪ M$ .

[L1]

Under countable choice, the ambient manifold tubular neighbourhood theorem provides a diffeomorphism $Φ : Ω \to U$ from a normal-bundle neighbourhood of the zero section onto an open neighbourhood $U$ of $S$ (The tubular neighbourhood theorem in a smooth ambient manifold).

Proof

technique · direct

1.1L1givenconstruct

Let $π : Ω \to S$ be the bundle projection, and define $r : = π \circ Φ^{- 1} : U \to S .$ This map is smooth because $π$ and $Φ^{- 1}$ are smooth.

2.1step 1.1algebra∎

For $p \in S$ , one has $Φ^{- 1} (p) = (p, 0)$ , so $r (p) = p$ . Therefore $r$ is a smooth retraction of $U$ onto $S$ .

Depends on

Used by

Dependency tree · two levels

22 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., Tubular Neighborhoods (standard reference, not scraped)