3
$\begingroup$

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.

$\endgroup$
1

1 Answer 1

7
$\begingroup$

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 $f(0,0)=f(0,2\pi)=(1,0)$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .