LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A nonzero rank minor supplies the source coordinates for the constant-rank theorem

Statement

Let $k \geq 1$ , let $f : U \subseteq R^{m} \to R^{n}$ be $C^{k}$ , and suppose $rank D f (a) = r$ . After permuting source and target coordinates, if $r > 0$ the leading $r \times r$ minor of $D f (a)$ is nonzero and $Φ (x) = (f_{0} (x), \dots, f_{r - 1} (x), x_{r}, \dots, x_{m - 1})$ is a local $C^{k}$ diffeomorphism at $a$ . If $r = 0$ , the same conclusion holds with $Φ$ equal to the identity map. Empty coordinate blocks are omitted.

Facts & Assumptions

Given: The map $f$ , the point $a$ , and $r = rank D f (a)$ .

[L1]

Positive matrix rank is detected by a nonzero minor (A matrix has rank at least $r$ exactly when it has a nonzero $r$ -rowed minor, The rank of a derivative and constant-rank Euclidean maps).

[L2]

A real square matrix is invertible exactly when its determinant is nonzero; a $C^{1}$ map between equal-dimensional Euclidean open sets with invertible derivative at a point is a local $C^{1}$ diffeomorphism there, and a $C^{k}$ map has a $C^{k}$ local inverse (A finite square real matrix is invertible if and only if its determinant is nonzero, The Euclidean inverse function theorem, A local inverse of a $C^{k}$ regular map is $C^{k}$ , $C^{k}$ Euclidean maps and diffeomorphisms).

Proof

technique · direct

1.1

If $r = 0$ , take $Φ = {id}_{U}$ ; it is a $C^{k}$ diffeomorphism on every open neighbourhood of $a$ .

given

1.2

Suppose $r > 0$ . By [L1], choose a nonzero $r$ -rowed minor and permute coordinates so it is the leading minor. The derivative $D Φ (a)$ is block triangular with that $r \times r$ block and an identity block of size $m - r$ on its diagonal.

givenL1choose

2.1

Its determinant is the nonzero leading minor, including the full-rank case $r = m$ where the identity block is empty. Thus [L2] makes $D Φ (a)$ invertible and $Φ$ a local $C^{k}$ diffeomorphism at $a$ .

step 1.2L2algebra

3.1

Steps 1.1 and 2.1 cover every possible rank and give the asserted source coordinates.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

36 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

J. M. Lee, Introduction to Smooth Manifolds, proof of Theorem 7.13 (standard reference, not scraped)
L. W. Tu, An Introduction to Manifolds, proof of Theorem 11.1 (standard reference, not scraped)