CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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An injective regular $C^{k}$ map is a $C^{k}$ diffeomorphism onto its image

Statement

Let $k, n \geq 1$ , let $U \subseteq R^{n}$ be open, and let $f : U \to R^{n}$ be an injective $C^{k}$ map whose derivative is invertible everywhere. Then $f [U]$ is open and the corestriction $f : U \to f [U]$ is a $C^{k}$ diffeomorphism.

Facts & Assumptions

Given: The hypotheses in the Statement and the $C^{k}$ diffeomorphism convention of $C^{k}$ Euclidean maps and diffeomorphisms.

[L1]

Under the corresponding $C^{1}$ hypotheses, $f [U]$ is open and $f : U \to f [U]$ is a $C^{1}$ diffeomorphism (An injective regular $C^{1}$ map is a diffeomorphism onto its image).

[L2]

If $f$ is $C^{k}$ for $k \geq 1$ , then every local inverse supplied by the inverse function theorem is $C^{k}$ (A local inverse of a $C^{k}$ regular map is $C^{k}$ ).

Proof

technique · direct

1.1

Since a $C^{k}$ map with $k \geq 1$ is $C^{1}$ , [L1] supplies the open image and the unique global $C^{1}$ inverse $g : f [U] \to U$ .

2.1

Around each $y \in f [U]$ , the inverse $g$ agrees with the unique local inverse of $f$ . By [L2] that restriction is $C^{k}$ . Thus $g$ is locally, and hence globally, $C^{k}$ , so the corestriction is a $C^{k}$ diffeomorphism.

step 1.1L2∎

Depends on

Used by

The map (x,y)↦(x,xy) is a diffeomorphism off the vertical axis Example

Dependency tree · two levels

13 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)
University of Toronto MAT237, §3.3 (standard reference, not scraped)

An injective regular CkCk map is a CkCk diffeomorphism onto its image