Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21
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An injective regular C1 map is a diffeomorphism onto its image

Statement

Let n≥1, let U⊆Rn be open, and let f:U→Rn be injective and C1. Suppose Df(x) is invertible for every x∈U, equivalently U=Reg⁡(f) (The regular locus of a square-dimensional C1 map). Then f[U] is open and f:U→f[U] is a C1 diffeomorphism. Its inverse g satisfies

Dg(f(x))=Df(x)−1(x∈U).

Facts & Assumptions

[L1]

Let f:U→Rn be C1 and suppose Df(x) is invertible for every x∈U. Then f maps every open subset of U to an open subset of Rn (A C1 map with everywhere-invertible derivative is open).

[L2]

For the local inverse g supplied by the inverse function theorem, Dg(y)=Df(g(y))−1(y∈W). (The Euclidean inverse function theorem)

Proof

technique · direct
1.1L1given

By [L1], f[U] is open and the continuous bijection f:U→f[U] is open. Therefore its inverse g:f[U]→U is continuous.

2.1step 1.1L2given∎

Fix y∈f[U]. The unique global inverse agrees near y, by injectivity, with the local inverse from [L2]. Hence g is C1 near every image point and satisfies Dg(y)=Df(g(y))−1 there. This proves the stated global C1 diffeomorphism and derivative formula.

Depends on

Used by

Dependency tree · two levels

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Sources