Jacobian Matrix? Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be given by $f(x,y)=(x^4-y^4,xy)$. 
(i) Evaluate the Jacobian of $f$, and its Jacobian determinant. 
(ii) Show that $f$ is locally invertible at any point $(x,y) = (0,0)$. 
(iii) What is the Jacobian of $f^{-1}$ at the point $(0,1) = f((1,1))$? 
(iv) Is $f$ globally invertible on $\mathbb R^2 \backslash(0,0)$?
For part i) Jacobian determinant is $4(x^2+y^2)$. 
I'm at loss at part ii, iii and iv. Please help me out...
Lots of thanks!
 A: HINT This is the inverse function theorem. It says that if the Jacobian determinant is non-zero at some point $x$, then the function has a local inverse: there exists a neighbourhood $U \ni x$ such that $f|U$ is injective, and such that $(f|U)^{-1}$ is continous and differentiable (if $f$ is).
There are lots more details at the Wikipedia page, and a good example too.
For the last question: $f$ cannot be globally invertible, because it is not injective. For any pair $(a,b) \in \mathbb{R}^2$, we have $f(a,b)=f(-a,-b)$.
A: Put $u=x^4-y^4$ and $v=xy$. Then the Jacobian is the determinant of the matrix
$$
\begin{bmatrix}
\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}
\end{bmatrix}
=\begin{bmatrix}
4x^3&-4y^3\\
y&x&\\
\end{bmatrix}.$$
So the Jacobian equals $4(x^4+y^4)$. Hence you have (i). And (ii) follows immediately, since it is clear that the Jacobian is $0$ if and only is $x=0=y$. Also (iv) follows.
According to the inverse function theorem, the matrix inverse of the Jacobian matrix of an invertible function is the Jacobian matrix of the inverse function. So $J(1,1)=8$ and hence of the inverse function $f^{-1}$ it is $\frac{1}{8}$.
A: @Karen, try to understand what injective means at first as this is a key point for inverse function theorem. 
$f(x)$ is injective if $$f(a)=f(b) \Rightarrow a=b.$$ For an example, $f(x)=x^2$ is not injective since $f(-1)=f(1)$ does not imply 1=-1. Thus $f(x)=x^2$ is not globally invertible.
