# On which open set $f$ admits an inverse? [duplicate]

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?

• Hint For (2), consider $f(1, 2 \pi)$ and $f(1, 4 \pi)$. – Travis Willse Dec 23 '15 at 2:14
• oh, then $f$ is not one to one – S. Pitchai Murugan Dec 23 '15 at 2:16
• 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". – Travis Willse Dec 23 '15 at 2:20
• Geometrically, one can grasp the intuition that $f$ is a diffeomorphism from $]0,+\infty[\times]-\pi,\pi[$ to $\mathbb{R}^2\times(]-\infty,0]\times\{0\})$. – C. Falcon Dec 23 '15 at 2:22
• Are you sure that the regions $U$ defines the domain for $f(r,\theta)$, or does (3) mean, for example, $r<1$? – 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 $\pi<\theta<\frac{3\pi}{2}$