Inverse Function Theorem and global inverses We learnt the Inverse Function Theorem for multi-variable functions, and it only dealt with "local" inverses, not "global" inverses. Is my interpretation of a global inverse just that there exists an inverse around ALL points in the domain?  
Here is the question related.  

Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ by
  $$
f(x,y) = \begin{pmatrix} e^x\cos y \\ e^x\sin y \end{pmatrix}.
$$
  Show that $f$ has a local $C^1$ inverse at every point in $\mathbb{R}^2$, but it has not any global inverse on $\mathbb{R}^2$.

What I did was basically found the Jacobian matrix of $f$, and showed that its determinant is zero iff $x$ approaches infinity. So for every $(x,y)$ with $x$ finite, then Jacobian of $f$ is invertible and hence there is a local $C^1$ inverse. But with large $x$, there is no inverse, so no global inverse?
 A: The idea is that the inverse function theorem guarantees, on a neighbourhood of each $(x,y)$, the existence of a local inverse (because the Jacobian is never zero, and the function is $C^1$). However there cannot be a global inverse; note, for example, that $f$ is not injective. 
A: Before getting started, let us go back on some definitions:

Definition. Let $U$ be an open subset of $\mathbb{R}^n$ and let $f\colon U\rightarrow V\subseteq\mathbb{R}^n$ be a $C^1$-mapping, then $f$ is a $C^1$-diffeomorphism or globally invertible if and only if there exists $g\colon V\rightarrow U$ a $C^1$-mapping such that:
  $$f\circ g=\textrm{id}_{V}\textrm{ and }g\circ f=\textrm{id}_U.$$ 

In other words, $f$ is a bijection whose inverse is smooth. This is not to be confused with:

Definition. The mapping $f$ is said to be a local diffeomorphism or locally invertible if and only if when retricted to an open subset of $U$, $f$ is a diffeomorphism onto its image.

Please notice that being globally invertible does not mean being locally invertible around any point.
As a reminder, let us state the inverse function theorem once again:

Theorem. Let $U$ be an open subset of $\mathbb{R}^n$, let $x$ be a point of $U$ and let $f\colon U\rightarrow\mathbb{R}^n$ be a $C^1$-mapping. Assume that $\mathrm{d}_xf\colon\mathbb{R}^n\rightarrow\mathbb{R}^n$ is an invertible linear map, then there exists $V$ an open neighbourhood of $x$ such that $f\colon V\rightarrow f(V)$ is a $C^1$-diffeomorphism.

In our case, for all $(x,y)\in\mathbb{R}^2$, the matrix of $\mathrm{d}_{(x,y)}f$ in the canonical basis of $\mathbb{R}^2$ is:
$$\begin{pmatrix}e^x\cos(y)&-e^x\sin(y)\\e^x\sin(y)&e^x\cos(y)\end{pmatrix}.$$
Its determinant is $e^{2x}$ which is nonzero for all $(x,y)$. Theorefore, according to the theorem, for all $(x,y)\in\mathbb{R}^2$, $f$ is a $C^1$-diffeomorphism in a neighborhood of $(x,y)$.
Assume by contradiction that there exists $g\colon\mathbb{R}^2\rightarrow\mathbb{R}^2$ a $C^1$-mapping such that:
$$g\circ f=\textrm{id}_{\mathbb{R}^2}.$$
Then, let $(X,Y)\in(\mathbb{R}^2)^2$ such that $f(X)=f(Y)$, then $g(f(X))=g(f(Y))$ i.e. $X=Y$ and $f$ is injective. However, $f$ is non injective since $f(0,0)=(1,0)=f(0,2\pi)$ and $(0,0)\neq (0,2\pi)$, a contradiction. Whence, $f$ is not a $C^1$-diffeomorphism.
Maybe it will give more insight to notice that if one identifies $\mathbb{R}^2$ with $\mathbb{C}$ through $(x,y)\mapsto x+iy$, then the considered mapping is the complex exponential, $z\mapsto e^z$ which is invertible on any horizontal strip of length at most $2\pi$.
