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Let $h:]0,+\infty[ \times ]-1,1[ \to \mathbb{R}\times ]0,+\infty[$ be defined by :

$h((x,y)) = (u,v) = (xy, x\sqrt{1-y^2})$

I need to show that h is bijective.

What I did:

I showed that the jacobian of $h$ in a point $(a,b) \in ]0,+\infty[ \times ]-1,1[$ is :

$ DH_{(x,y)}(a,b)=-a(\frac{2b^2}{\sqrt{1-y^2}}+\sqrt{1-b^2}) <0 \neq 0 $

Is that enough to say that h is bijective or it is only injective?

Is there a need to show that $h^{-1}(u,v) \in ]0,+\infty[ \times ]-1,1[$ ?

Many thanks!

EDIT:

I calculated the inverse:

$h((x,y)) = (u,v) \Leftrightarrow \left\{ \begin{array}{@{}l} y=\frac{u}{\sqrt{u^2+v^2}}\\ x=\sqrt{u^2+v^2} \end{array} \right.$

Now I guess that means h is bijective ?

Thank you!

Conjecture
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    By merely computing the Jacobian, you cannot prove bijectivity. There are functions which are not surjective, even though the determinant of the Jacobian is everywhere nonzero (think about an function which is 'asymptotic', like an arctangent or something). It is however true that a function with nowhere vanishing determinant of its Jacobian is injective, as long as the domain is path-connected. This follows from the mean value theorem. For your case, I would probably just compute an inverse explicitly. – Thomas Bakx Apr 13 '18 at 16:16
  • Thank you! will compute the inverse and edit my question – Conjecture Apr 13 '18 at 16:20
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    This looks correct. Observe that the inverse map you wrote down is indeed a two-sided inverse, so that h is bijective. – Thomas Bakx Apr 15 '18 at 13:24

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