Prove $ \det \nabla f > 0$ for all $x \in G$, but f is not injective with $f(x,y):= (e^x \cos y, e^x \sin y) $

Question

Let $G \subset \mathbb{R^2}$ given by

$$f(x,y):= \begin{pmatrix} e^x \cos y \\ e^x \sin y \end{pmatrix} $$

Let $G = \mathbb{R^2}$.

How can one prove that $ \det \nabla f > 0$ for all $x \in G$, but f is not injective?

So, $\nabla f$ would be $\begin{pmatrix} \cos (y) e^x & -e^x \sin(y) \\ \sin (y)e^x & e^x \cos(y)\end{pmatrix}$.

The determinant would then be:

$$ (\cos (y) e^x \cdot e^x \cos(y)) - (-e^x \sin(y) \cdot \sin (y)e^x)$$

but then?

Answer 1

To see is not injective just take $(0, \pi /4) \text{ and } (0, 2\pi + \pi/4)$.

The determinant will be $e^{2x }> 0$ for all $x$.