Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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An injective regular C1 map is a diffeomorphism onto its image

Statement

Let n1, let URn be open, and let f:URn be injective and C1. Suppose Df(x) is invertible for every xU, equivalently U=Reg(f) (The regular locus of a square-dimensional C1 map). Then f[U] is open and f:Uf[U] is a C1 diffeomorphism. Its inverse g satisfies

Dg(f(x))=Df(x)1(xU).

Facts & Assumptions

[L1]

Let f:URn be C1 and suppose Df(x) is invertible for every xU. 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(yW). (The Euclidean inverse function theorem)

Proof

technique · direct
1.1

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

L1given
2.1

Fix yf[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.

step 1.1L2given

Depends on

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