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Is the following generalization of the Inverse Function Theorem true:

Let $f:\Bbb{R^2}\to\Bbb{R^2}$ be a smooth function. If the determinant of the derivative matrix is non-zero everywhere, then the function is globally one-to-one.

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No. Take, for instance, $f(x,y)=\bigl(e^x\cos(y),e^x\sin(y)\bigr)$. Its derivative has non-zero determinant everywhere, but $f(0,0)=f(2\pi,0)=(1,0)$.

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  • $\begingroup$ Sure. I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Feb 9 at 16:17

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