TITLE: $(e^x \cos y,e^x \sin y)$ is local diffeomorphism not global
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: 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 $f(x,0)=f(x,2\pi)$.