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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!

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Kindly asking if these are you homework? :-) – Babak S. Jun 11 '13 at 8:31
No, its nottttt – Karen Jun 11 '13 at 8:35
@Karen Please review my edit. – BU982T Jun 11 '13 at 8:41
What specifically is giving you trouble? – Ataraxia Jun 11 '13 at 8:45
I dont understand this topic, that is why I can't answer them – Karen Jun 11 '13 at 8:48

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 $f(a,b)=f(-a,-b)$.

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I'm sorry, I still dont understand for the last question... – Karen Jun 11 '13 at 9:26
@Karen: It would be helpful if you could clarify which parts you don't understand. – Fredrik Meyer Jun 11 '13 at 9:50
Why f cannot be globally invertible? What is injective? – Karen Jun 11 '13 at 10:05
i dont know how to tell that f is not a globally invertible.. – Karen Jun 11 '13 at 10:06
and for part iv, is the inverse is (1/8 1/2) (-1/8 1/2)? i dont know how to show it here in 2x2 matrix – Karen Jun 11 '13 at 10:08

@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.

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oh i see! now i get what's injective! :) – Karen Jun 11 '13 at 13:29
Thank you!!!!!!! – Karen Jun 11 '13 at 13:30
you are welcome – math Jun 11 '13 at 14:34
Does this apply the same to part (ii)? – Karen Jun 11 '13 at 14:43
How to tell that f is locally invertible at any point (x,y)=(0,0)? – Karen Jun 11 '13 at 14:44

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