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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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A local inverse of a Ck regular map is Ck

Statement

Let k,n≥1, let U⊆Rn be open, and let f:U→Rn be Ck, with Df(a) invertible. Then every local inverse supplied by the inverse function theorem at a is Ck. In its inverse neighbourhood it satisfies

Dg(y)=Df(g(y))−1.

Facts & Assumptions

Given: The hypotheses in the Statement and the Ck closure theorem Ck Euclidean maps are closed under componentwise algebra and composition.

[L1]

The inverse function theorem supplies an inverse g:W→V that is C1 and satisfies Dg(y)=Df(g(y))−1 (The Euclidean inverse function theorem).

[L2]

If the entries of A:U→Mn(R) are Cr and det⁡A never vanishes, then the entries of A−1 are Cr (Matrix inversion preserves Ck regularity where the determinant is nonzero).

Proof

technique · direct
1.1L1

By [L1], the local inverse exists, is C1, and satisfies Dg=(Df∘g)−1 throughout its domain.

2.1step 1.1L2givenalgebra∎

Suppose 1≤r<k and g is Cr. Because f is Ck, the entries of Df are Ck−1, hence Cr when r<k; closure under composition makes Df∘g Cr, and [L2] makes Dg Cr. Therefore g is Cr+1. Starting from step 1.1 and repeating this finite bootstrap reaches Ck.

Depends on

Dependency tree · 2 levels

Sources