My book is An Introduction to Manifolds by Loring W. Tu. In an exercise, it is claimed that vertical projection is a coordinate map.
How is it a coordinate map when it is not smooth?
Under vertical projection, my answer is the same as Tu's answer at the back of the book (Richard G. Ligo gives an explicit solution here.)
The solution is obtained by computing the Jacobian (or its transpose)
$$\begin{bmatrix} \frac{\partial (x \circ i)}{\partial u} & \frac{\partial (y \circ i)}{\partial u} & \frac{\partial (z \circ i)}{\partial u}\\ \frac{\partial (x \circ i)}{\partial v} & \frac{\partial (y \circ i)}{\partial v} & \frac{\partial (z \circ i)}{\partial v} \end{bmatrix}^T$$
Under vertical projection, we have
$$u(a,b,c)=a,v(a,b,c)=b$$
$$(x \circ i)(a,b,c)=x(a,b,c)=a, (y \circ i)(a,b,c)=y(a,b,c)=b, (z \circ i)(a,b,c)=z(a,b,c)=c$$
Then we get $$u=x \circ i,v=y \circ i, \sqrt{1-u^2-v^2} = z \circ i$$
$$\begin{bmatrix} \frac{\partial u}{\partial u} & \frac{\partial v}{\partial u} & \frac{\partial \sqrt{1-u^2-v^2}}{\partial u}\\ \frac{\partial u}{\partial v} & \frac{\partial v}{\partial v} & \frac{\partial \sqrt{1-u^2-v^2}}{\partial v} \end{bmatrix}^T = \begin{bmatrix} 1 & 0 & \frac{-u}{\sqrt{1-u^2-v^2}}\\ 0 & 1 & \frac{-v}{\sqrt{1-u^2-v^2}} \end{bmatrix}^T$$
Evaluated at $p=(a,b,c)$, we get
$$\begin{bmatrix} 1 & 0 & \frac{-a}{c}\\ 0 & 1 & \frac{-b}{c} \end{bmatrix}^T$$
But according to this, vertical projection, while continuous, is not smooth. Instead I think we should have
$$u(a,b,c)=\frac{a}{1-c},v(a,b,c)=\frac{b}{1-c}$$
Under stereographic projection, we still have
$$(x \circ i)(a,b,c)=x(a,b,c)=a, (y \circ i)(a,b,c)=y(a,b,c)=b, (z \circ i)(a,b,c)=z(a,b,c)=c$$
But now we have
$$\frac{2u}{u^2+v^2+1}=x \circ i,\frac{2v}{u^2+v^2+1}=y \circ i, \frac{u^2+v^2-1}{u^2+v^2+1} = z \circ i$$
Then we get
$$\begin{bmatrix} \frac{\partial (x \circ i)}{\partial u} & \frac{\partial (y \circ i)}{\partial u} & \frac{\partial (z \circ i)}{\partial u}\\ \frac{\partial (x \circ i)}{\partial v} & \frac{\partial (y \circ i)}{\partial v} & \frac{\partial (z \circ i)}{\partial v} \end{bmatrix}^T = \begin{bmatrix} \frac{\partial (\frac{2u}{u^2+v^2+1})}{\partial u} & \frac{\partial (\frac{2v}{u^2+v^2+1})}{\partial u} & \frac{\partial \frac{u^2+v^2-1}{u^2+v^2+1}}{\partial u}\\ \frac{\partial (\frac{2u}{u^2+v^2+1})}{\partial v} & \frac{\partial (\frac{2v}{u^2+v^2+1})}{\partial v} & \frac{\partial \frac{u^2+v^2-1}{u^2+v^2+1}}{\partial v} \end{bmatrix}^T$$
Questions
Should we use stereographic projection instead of vertical projection because vertical projection cannot be used because vertical projection is not smooth while stereographic projection is smooth?
Where can I find the Jacobian to stereographic projection, so I can check my answer? I'm not sure of the terminology and so do not know the keywords to search. Besides checking if my partial derivative computations are correct, I would like to know if these partial derivatives are the correct ones to compute.
I think the answer is
$$\begin{bmatrix} 1-c-a^2 & -ab & a(1-c)\\ -ab & 1-c-b^2 & b(1-c) \end{bmatrix}^T$$
But I think wolfram gives me
$$\begin{bmatrix} 2-a^2 & 0 & 2a\\ 0 & 2-b^2 & \pm 2 i a \end{bmatrix}^T$$
The last entry tells you I'm still trying to figure out how to use wolfram.