Let $f:\mathbb{R^2} \to \mathbb{R^2}$ defined by $f(x,y)=(e^x \cos y,e^x \sin y)$
I have showed that $f$ is a local diffeomorphism by using inverse function theorem, that is $\det(Df)=e^x \gt 0$ for all $x$, so $Df$ is invertible, hence local diffeomorphism. But how do I see this is not a global diffeomorphism?
$f$ is not one to one.
If $f(x,y)=f(x,y+2k\pi)$, for all $k\in\mathbb Z$.
A diffeomorphism is a bijection and $f$ is not!
Computing the jacobian, we get that for all $(x,y) \in \mathbb{R}^2$ $$ df_{(x,y)} = \left( \begin{array}{cc} e^x\cos(y) & -e^x\sin(y) \\ e^x\sin(y) & e^x\cos(y) \end{array} \right). $$ hence $det(df_{(x,y)} )=e^{2x} \neq 0$, that is, by the inverse function theorem, for all $(x,y) \in \mathbb{R}^2$, $f$ is a local diffeomorphism, however, it can not be a global one since $f$ is not even one to one, because clearly $f(x,0)=f(x,2\pi)$.
