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Find diffeomorphism from $\mathbb{R}^2$ onto

$U:= \{(u,v): \ u>v>0, \ uv<1\}$

$V:= \{u,v): \ u>v^2, \ u+v<6\}$ (or from $U,V$ onto $\mathbb{R}^2$).

2 Answers 2

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We have in our utility belt: $\begin{matrix}f\colon&\mathbb R&\to &\mathbb R_{>0}\\ &x&\mapsto& \exp(x)\end{matrix}$ and $\begin{matrix}g\colon&\mathbb R_{>0}&\to& (0,1)\\&x&\mapsto &\frac2\pi\arctan x.\end{matrix}$ This allows us to produce $\begin{matrix}\mathbb R^2&\to& U\\(x,y)&\mapsto& \frac{g(f(x)f(y))}{f(x)f(y)}\cdot(f(x),f(y))\\&&=\left(\frac2\pi\arctan(\exp(x+y))\exp(-y),\frac2\pi\arctan(\exp(x+y))\exp(-x)\right)\end{matrix}$

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For $U \longrightarrow \mathbb{R}^2$ try $(u,v)\mapsto \left(\ln u, \ln \left(\dfrac{1}{u\cdot v}-1\right)\right)$.

For $V \longrightarrow \mathbb{R}^2$ try
$(u,v)\mapsto \left(\ln\left(\dfrac{9}{u}-1\right), \ln \left(\dfrac{2\sqrt{u}}{v+\sqrt{u}}-1\right)\right)$, for $u \in (0,4]$ and
$(u,v)\mapsto \left(\ln\left(\dfrac{9}{u}-1\right), \ln \left(\dfrac{6-u+\sqrt{u}}{v+\sqrt{u}}-1\right)\right)$, for $u \in [4,9)$.

These can be obtained as compositions of:

  • $(0,1) \longrightarrow \mathbb{R} , \ \ t \mapsto \ln\left(\dfrac{1}{t}-1\right)$,
  • $(0,a) \longrightarrow (0,1) , \ \ t \mapsto \dfrac{t}{a} ,$
  • $(a,b) \longrightarrow (0,b-a) , \ \ t \mapsto t-a $.