Is the following statement true? A C¹ map with everywhere non-zero Jacobian is injective

No. f(x,y) = (eˣ cos y, eˣ sin y) on ℝ² The Jacobian determinant is e^{2x} ≠ 0, but f(x, y) = f(x, y + 2π). Invertibility is only local.

Counterexample

Is this true?

$A C^{1}$ map with everywhere non-zero Jacobian is injective

No — it is false.

The counterexample

f(x,y) = (eˣ cos y, eˣ sin y) on $\mathbb{R}^{2}$

Why it fails

The Jacobian determinant is $e^{2x}$ ≠ 0, but $f(x, y) = f(x, y + 2\pi)$ . Invertibility is only local.

Keep practicing

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.