TITLE: Why does the Jacobian show a transformation is one to one?
QUESTION [8 upvotes]: Consider $f: \mathbb{R}^2 \to \mathbb{R}^2$, where $f$ is given by
$$f =\begin{bmatrix}
f_1(x,y)\\ 
f_2(x,y)
\end{bmatrix}$$
Where we may assume $f$ is of class $C^1$ (continuously differentiable), if $J$ is the associated Jacobian of $f$ and if $\det J \neq 0$ for every $x,y \in\mathbb{R}^2$, why does that imply $f$ is one to one?
I understand this idea for single variable functions, but not so convinced in multivariables.
EDIT,
Thank you for the counterexample, but let me add one more restriction. Let $f$ be one to one  defined on an open set $A$, so $f: A \to \mathbb{R}^2$

REPLY [7 votes]: The best you can say is that there is a neighborhood on which your function is one-to-one; this is the Inverse Function Theorem, which applies when the Jacobian is non-degenerate.