Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21
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The Jacobian sign of a regular C1 map is constant on a connected domain

Statement

Let n≥1, let U⊆Rn be nonempty, open, and connected, and let f:U→Rn be C1 with invertible derivative everywhere. Then either det⁡Df(x)>0 for every x∈U or det⁡Df(x)<0 for every x∈U. Thus f has one local orientation throughout U (Local orientation of a regular C1 Euclidean map).

Facts & Assumptions

Proof

technique · direct
1.1L1given

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

2.1step 1.1L2given∎

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.

Depends on

Used by

Dependency tree · two levels

36 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