Polar coordinates in plane. Consider the following open set of $\mathbb{R}^2$:
$M=\mathbb{R}^2-\{0,0\}$.
Let $\psi:\mathbb{R}_+\times\mathbb{R}\to M\\(\rho\cos(\theta),\rho\sin(\theta))$
a)
i)$\psi$ is of $C^\infty$ class
ii)$\psi$ is not injective
iii)$\det\nabla(\psi,(\rho\theta))=\rho^2>0$, so $\psi$ is an immersion
b)
Let:
$I_1=]-\pi,\pi[,\:\:\:A_1=\mathbb{R}_+\times I_1,\:\:\:\psi:A_1\to M,(\rho,\theta)\to\psi(\rho,\theta)\\I_2=]0,2\pi[,\:\:\:A_2=\mathbb{R}_+\times I_2,\:\:\:\psi:A_2\to M,(\rho,\theta)\to\psi(\rho,\theta)$.
Therefore:
i)$\psi(A_1)=U_1=\mathbb{R}^2-\{(x,0):x\leqslant 0\}$ is an open set of $M$.
ii)$\psi(A_2)=U_2=\mathbb{R}^2-\{(x,0):x\geqslant 0\}$ is an open set of $M$.
Questions:
1) Why is (0,0) excluded from $\mathbb{R}^2$? If I take $\psi((0,0))=0$. What is the problem? Why $M=\mathbb{R}^2-\{0,0\}$?
2) I do not understand $\psi(A_1)=U_1=\mathbb{R}^2-\{(x,0):x\leqslant 0\}$. Why is the negative x axis excluded? In the interval $]-\pi,\pi[$, $cos(\theta)$ can be zero. Why exclude the negative part of the x-axis?
Thanks in advance!
1) The point $(0,0)$ is excluded because $\psi((0,\theta))=(0,0)$ for each $\theta\in\mathbb{R}$, while we would like that $\psi$ is injective as a map defined on $\mathbb{R}^{+} \times ]0,2\pi[$ (or similar intervals with length $2\pi$, so you can consider for example $]-\pi,\pi[$, if you prefer).
2) The negative axis $\{x\leq 0\}$ is the image of $\mathbb{R}^+ \times \{\pi+2k\pi\}_{k\in\mathbb{Z}}$, but $\pi+2k\pi\notin I_1$ for each $k\in\mathbb{Z}$. A similar argument can be used for $A_2$.