TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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A local inverse of a $C^{k}$ regular map is $C^{k}$

Statement

Let $k, n \geq 1$ , let $U \subseteq R^{n}$ be open, and let $f : U \to R^{n}$ be $C^{k}$ , with $D f (a)$ invertible. Then every local inverse supplied by the inverse function theorem at $a$ is $C^{k}$ . In its inverse neighbourhood it satisfies

$D g (y) = D f (g (y))^{- 1} .$

Facts & Assumptions

Given: The hypotheses in the Statement and the $C^{k}$ closure theorem $C^{k}$ Euclidean maps are closed under componentwise algebra and composition.

[L1]

The inverse function theorem supplies an inverse $g : W \to V$ that is $C^{1}$ and satisfies $D g (y) = D f (g (y))^{- 1}$ (The Euclidean inverse function theorem).

[L2]

If the entries of $A : U \to M_{n} (R)$ are $C^{r}$ and $\det A$ never vanishes, then the entries of $A^{- 1}$ are $C^{r}$ (Matrix inversion preserves $C^{k}$ regularity where the determinant is nonzero).

Proof

technique · direct

1.1

By [L1], the local inverse exists, is $C^{1}$ , and satisfies $D g = (D f \circ g)^{- 1}$ throughout its domain.

2.1

Suppose $1 \leq r < k$ and $g$ is $C^{r}$ . Because $f$ is $C^{k}$ , the entries of $D f$ are $C^{k - 1}$ , hence $C^{r}$ when $r < k$ ; closure under composition makes $D f \circ g$ $C^{r}$ , and [L2] makes $D g$ $C^{r}$ . Therefore $g$ is $C^{r + 1}$ . Starting from step 1.1 and repeating this finite bootstrap reaches $C^{k}$ .

step 1.1L2givenalgebra∎

Depends on

Used by

Dependency tree · two levels

26 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. Lebl, Basic Analysis II, Remark 8.5.8 (standard reference, not scraped)

A local inverse of a CkCk regular map is CkCk