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Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the function defined by $f(r,\theta)=(r\cos\theta,r\sin\theta).$ Then for which of the open subset $U$ of $\mathbb{R}^2$ given below, $f$ restricted to $U$ admit 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. $U=\{x,y \in\mathbb{R}^2:x<-1,y<-1\}$

It is clear 1 is not true as $\sin ,\cos$ are periodical function. Similarly 3 is not true. What about 2nd and 4th ? please help.Thanks in advance.

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To admit an inverse $f$ restricted to $U$ must be bijective, in particular one-one. $1$ and $2$ are not True since $f(1,2) = f(1,2+2\pi)$. $4$ is not true since $f(-2,-2\pi) = f (-2,-4\pi)$. $3$ is not True since $f(0,1/2) = f (0,1/4)$.

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  • $\begingroup$ its means its incorrect question? $\endgroup$ – neelkanth Dec 23 '15 at 4:40
  • $\begingroup$ f(1, 2) = f(1, 1+2$\pi$), how come? $\endgroup$ – user268307 Dec 23 '15 at 4:42
  • $\begingroup$ there should be $f(1,2+2\pi)$ $\endgroup$ – neelkanth Dec 23 '15 at 4:43
  • $\begingroup$ That's what I am thinking. $\endgroup$ – user268307 Dec 23 '15 at 4:44
  • $\begingroup$ oh. yes ..i am sorry for that mistake, $\endgroup$ – Samiron Dec 23 '15 at 4:44
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In options 2 and 4, The determinant of jacobian= $r$ is nonzero everywhere. Thus inverse function theorem guarantees that, for every point $p$ in $U$, there exists a neighborhood about $p$ over which function is invertible. This does not mean function is invertible over whole domain $U$: in this case $f$ is not even injective since it is periodic : $f(x,y)=f(x,y+2 \pi)$. (And injection is necessary for inverse.)

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  • $\begingroup$ I think all options are incorrect??? $\endgroup$ – neelkanth Nov 27 '18 at 17:39
  • $\begingroup$ @neelkanth of course all options are incorrect. Actually many got confused that using inverse function theorem, options 2 and 4 should be correct. That's why I wrote this. $\endgroup$ – ramanujan Nov 27 '18 at 17:53

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