PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-09-01

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Two tubular neighbourhood germs are isomorphic near the zero section

Statement

Let $Φ_{1} : Ω_{1} \to M$ and $Φ_{2} : Ω_{2} \to M$ be two tubular neighbourhoods of the same closed embedded submanifold $S \subseteq M$ built on the same normal bundle $ν (S)$ . Then, after shrinking $Ω_{1}$ and $Ω_{2}$ around the zero section, there is a diffeomorphism $Ψ : Ω_{1}^{'} \to Ω_{2}^{'}$ such that $Φ_{2} \circ Ψ = Φ_{1}$ and $Ψ$ restricts to the identity on the zero section.

Facts & Assumptions

Given: Two tubular neighbourhood charts $Φ_{1} : Ω_{1} \to M$ and $Φ_{2} : Ω_{2} \to M$ for the same closed embedded submanifold $S \subseteq M$ .

[F1]

A tubular neighbourhood chart is a diffeomorphism from an open normal-bundle neighbourhood of the zero section onto an ambient open neighbourhood of $S$ (Tubular neighbourhoods of embedded submanifolds).

[L1]

Tubular neighbourhoods exist in smooth ambient manifolds (The tubular neighbourhood theorem in a smooth ambient manifold).

Proof

technique · direct

1.1F1givenconstruct

By [F1], both $Φ_{1}$ and $Φ_{2}$ are diffeomorphisms onto open neighbourhoods of $S$ . Shrink the domains so that their images lie in the common overlap. Then $Ψ : = Φ_{2}^{- 1} \circ Φ_{1}$ is a diffeomorphism between the shrunken domains.

2.1F1step 1.1

On the zero section both tubular charts agree with the inclusion of $S$ into $M$ , so $Ψ (p, 0) = (p, 0)$ for every $p \in S$ . Thus $Ψ$ restricts to the identity on the zero section.

3.1L1step 1.1step 2.1∎

The relation $Φ_{2} \circ Ψ = Φ_{1}$ is built into the definition of $Ψ$ , and [L1] guarantees that these tubular charts are honest smooth objects rather than formal placeholders. Hence the two tubular neighbourhoods define the same germ near the zero section.

Depends on

Dependency tree · 2 levels

Sources

John M. Lee, Introduction to Smooth Manifolds, 2nd ed., Tubular Neighborhoods