TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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A regular level set is an embedded submanifold

Statement

Let $F : M^{m} \to N^{n}$ be smooth, let $q \in N$ be a regular value, and assume $F^{- 1} (q)$ is nonempty. Then $F^{- 1} (q)$ is an embedded submanifold of codimension $n$ . Equivalently, it has dimension $m - n$ .

Facts & Assumptions

Given: A smooth map $F : M^{m} \to N^{n}$ and a regular value $q \in N$ with nonempty fibre.

[F1]

A regular value is one whose fibre points are all submersion points; the fibre may be empty (Regular and critical points and values).

[F2]

Codimension means ambient dimension minus submanifold dimension, and embedded submanifolds are defined by slice charts (Codimension and hypersurfaces, Embedded submanifolds and slice charts).

[L1]

Near any submersion point, suitable coordinates put $F$ into the form $(u, v) \mapsto u$ (Local normal form for submersions).

Proof

technique · direct

1.1F1given

Let $p \in F^{- 1} (q)$ be arbitrary. By [F1], $F$ is a submersion at $p$ .

2.1step 1.1L1algebra

Apply [L1] at $p$ . In suitable charts around $p$ and $q$ , the map becomes $(u, v) \mapsto u$ on $R^{n} \times R^{m - n}$ . After centering $q$ , the fibre is ${0} \times R^{m - n}$ . Permuting the two source-coordinate blocks sends it to the standard slice $R^{m - n} \times {0}$ . Thus $F^{- 1} (q)$ is locally an embedded submanifold.

3.1F2step 2.1∎

Since every point of the fibre has such a slice neighbourhood, $F^{- 1} (q)$ is an embedded submanifold. Its local model has dimension $m - n$ , so by [F2] the codimension is $n$ .

A regular level set is an embedded submanifold

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources