TITLE: "Jacobian Conjecture" for $k[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$?
QUESTION [5 upvotes]: Is there exist a similar conjecture to the famous Jacobian Conjecture with $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$ instead of 
$\mathbb{C}[x_1,\ldots,x_n]$?
Namely, let $f$ be $\mathbb{C}$-algebra endomorphism of $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$, denote 
$f_i:= f(x_i)$, and further assume that the Jacobian of $\{f_1,\ldots,f_n\}$ is in $\mathbb{C}-\{0\}$. Is such $f$ an automorphism?
(I guess that first one should be familiar with the group of automorphisms of 
$\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$, see this question).
Edit: I also wonder if there exists any nice connection between the Jacobian Conjecture and my above conjecture (which is not exactly phrased yet); for example, are the two conjectures equivalent?

REPLY [15 votes]: Counterexample: the endomorphism of the product of two punctured lines (complement of the curve of equation $xy=0$ in the plane) given by $$(x,y)\mapsto f(x,y)=\left(\frac{x}{y},y^2\right)$$
We have
$$\begin{pmatrix}\partial_1f_1 & \partial_2f_1\\ \partial_1f_2 & \partial_2f_2\end{pmatrix}=\begin{pmatrix}\frac{1}{y} & -\frac{x}{y^2}\\ 0 & 2y\end{pmatrix},$$
so the Jacobian is constant equal to 1.
It's not an automorphism, since both $(1,1)$ and $(-1,-1)$ map to $(1,1)$.