Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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The Jacobian sign of a regular C1 map is constant on a connected domain

Statement

Let n1, let URn be nonempty, open, and connected, and let f:URn be C1 with invertible derivative everywhere. Then either detDf(x)>0 for every xU or detDf(x)<0 for every xU. Thus f has one local orientation throughout U (Local orientation of a regular C1 Euclidean map).

Facts & Assumptions

Proof

technique · direct
1.1

The entries of Df are continuous, and [L1] expresses xdetDf(x) as a polynomial in them. Hence the Jacobian determinant is continuous on U.

L1given
2.1

By [L2], its image is a connected subset of R. Regularity excludes zero. If the image contained both a negative and a positive value, order-convexity would force it to contain zero, a contradiction. Nonemptiness therefore leaves exactly one sign throughout U.

step 1.1L2given

Depends on

Used by

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Dependency tree · two levels

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Sources