Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be a function $f(r,\theta)=(r\cos\theta,r\sin\theta).$Then for which of the open subsets $U$ of $\mathbb{R}^2$ given below, $f$ restricted to $U$ admits an inverse?

  1. $U=\mathbb{R}^2$

  2. $U=\{(x,y)\in\mathbb{R}^2:x>0,y>0\}$

  3. $U=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$

  4. $U=\{(x,y)\in\mathbb{R}^2:x<-1,y<-1\}$

Here the Jacobian of $f$ is $r=\sqrt{x^2+y^2}$ and it is not vanishing on option 2 and 4. Can I say option 2 and 4 are correct?

  • $\begingroup$ Hint For (2), consider $f(1, 2 \pi)$ and $f(1, 4 \pi)$. $\endgroup$ – Travis Willse Dec 23 '15 at 2:14
  • $\begingroup$ oh, then $f$ is not one to one $\endgroup$ – S. Pitchai Murugan Dec 23 '15 at 2:16
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    $\begingroup$ Yes, exactly. Are you sure this problem is correct as stated? The options for $U$ and the labeling of the variables suggest that this problem might be somehow "backward". $\endgroup$ – Travis Willse Dec 23 '15 at 2:20
  • $\begingroup$ Geometrically, one can grasp the intuition that $f$ is a diffeomorphism from $]0,+\infty[\times]-\pi,\pi[$ to $\mathbb{R}^2\times(]-\infty,0]\times\{0\})$. $\endgroup$ – C. Falcon Dec 23 '15 at 2:22
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    $\begingroup$ Are you sure that the regions $U$ defines the domain for $f(r,\theta)$, or does (3) mean, for example, $r<1$? $\endgroup$ – David Dec 23 '15 at 4:48

For 1st part counterexample is $f(1,\theta)=f(1,\theta +2\pi)$

For second part take $f(1,\theta)=f(1,\theta +2\pi)$ where $0<\theta<\frac{\pi}{2}$

For third part $f(\frac{1}{2},\theta)=f(\frac{1}{2},\theta+2\pi)$

For 4th part as $f(2,\theta)=f(2,\theta +2\pi)$ where $\pi<\theta<\frac{3\pi}{2}$


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