A problem regarding the intersection of a cylinder and a line

Question

Let $N,S$ be the poles of $S^2$. If $C$ is the cylinder with base $x^2+y^2=1$ we define $f:S^2\{N,C\}\to C$ by $f(p)=\vec{op}\cap C$ being $\vec{op}$ the ray from $(0,0,0)$ directed at $p$. Prove that $f$ is a diffeomorphism and show that $f$ transforms parallels in circumferences and meridians in straight lines.

My first idea was to find a better way to write $f$ because $\vec{op}\cap C$ doesn't look too handy. Working with the cartesian equation for the cylinder and the parametric equations for the line seems that I have to find the intersection between $x^2+y^1=1$ and the line given by $\{x=tp_x,y=tp_y,z=tp_z\}$, follows that $t^2p_x^2+t^2p_y^2=1$ which implies $t=\frac{1}{\sqrt{p_x^2+p_y^2}}$ (I'm only considering the $+$ value of $t$ because I'm using the ray $\vec{op}$ so I only care about the upper intersection). Follows that $f$ could be defined by $f(p)=\left(\frac{p_x}{\sqrt{p_x^2+p_y^2}},\frac{p_y}{\sqrt{p_x^2+p_y^2}},\frac{p_z}{\sqrt{p_x^2+p_y^2}}\right)=\frac{1}{\sqrt{p_x^2+p_y^2}}p$

Then I calculated each partial derivative and I got

$$\nabla f(p)=\begin{pmatrix} \frac{p_y^2}{(p_x^2+p_y^2)^{3/2}} & \frac{p_xp_y}{(p_x^2+p_y^2)^{3/2}} & 0 \\ \frac{p_x^2}{(p_x^2+p_y^2)^2} & \frac{p_xp_y}{(p_x^2+p_y^2)^{3/2}} & 0\\ 0 & 0 & \frac{1}{(p_x^2+p_y^2)^{1/2}}\end{pmatrix}$$

And each ones of them looks continuous, but what further argument could I use?, would be enough to say thatgiven the domain of the function the denominator is never zero so each function is continuous in its domain?. And since each function can be differentiated again the $f$ is continuosly differentiable.

Now I have to prove the differentiable continuity of $f^{-1}$. Doing some calculations with the cylidner and the parametric equations above I got $$x^2=t^2p_x^2=1-y^2\; (1)\\y^2=t^2p_y^2=1-x^2 \;(2)\\z=tp_z\; (3)$$

But I couldn't find the expression for $p_x,p_y,p_z$ in terms of $x,y,z$ in order to find $f^{-1}$, and I couldn't think of a way to determine is the statement "$f$ transforms parallels in circumferences and meridians in straight lines" is true.

Answer 1

My natural inclination is to appeal to symmetry about the $z$-axis (which is the content of the wording in the question that parallels map to parallels and meridians map to straight lines). If we use spherical coordinates $(\phi,\theta)$ on the sphere and cylindrical coordinates $(z,\theta)$ on the cylinder, then the mapping $f$ is given by $$f(\phi,\theta)=(\cot\phi,\theta), \quad 0<\phi<\pi, 0\le\theta<2\pi,$$ as we see in this figure:

Now, once you know that these are smooth coordinate systems on the respective surfaces, then it follows easily that $f$ and $f^{-1}$ are smooth, since $$f^{-1}(z,\theta) = (\operatorname{arccot} z,\theta)$$ and both $\cot$ and $\operatorname{arccot}$ are smooth functions.

As is often the case here, it's difficult to know what course you're in and what your acceptable toolbox might be, so feel free to ask questions.