Constructing the Riemann Sphere Problem:
In the construction of the Riemann sphere, we begin with the sphere $\mathbb{S}^2$ with two charts: 


*

*the stereographic projection $\sigma_N : \mathbb{S}^2 \setminus \{N\} \to \mathbb{R}^2  \cong \mathbb{C}$ from the North pole, $N$, given by
$$
\sigma_N (x_1, x_2, x_3) := \frac{(x_1, x_2)}{1-x_3},
$$

*the stereographic projection $\sigma_S : \mathbb{S}^2 \setminus \{S\}  \to \mathbb{R}^2 \cong \mathbb{C}$ from the South pole, $S$, given by
$$
\sigma_S (x_1, x_2, x_3) := \frac{(x_1, x_2)}{1+x_3}.
$$
Question:
How does one show that the transition function of the two charts is:
$$\sigma_1 \circ \sigma_0^{-1} (z) = z^{-1}$$
Remark:
By elementary calculations we see that:
$$\sigma_S(x_1,x_2,x_3)=\sigma_N(x_1,x_2,-x_3)$$
$$\sigma_N(x_1,x_2,x_3)=\sigma_S(x_1,x_2,-x_3)$$
 A: Let the stereographic projection $\sigma_0 : \mathbb{S}^2 \setminus \{N\} \to \mathbb{C}$ from the North pole, $N$, be given by
$$
\sigma_0 (x_1, x_2, x_3) := \frac{x_1 + ix_2}{1-x_3},
$$
and the stereographic projection $\sigma_1 : \mathbb{S}^2 \setminus \{S\}  \to \mathbb{R}^2 \cong \mathbb{C}$ from the North pole, $S$, be given by
$$
\sigma_1 (x_1, x_2, x_3) := \frac{x_1 - ix_2}{1+x_3}
$$
(notice how the orientation has been reversed via complex conjugation). The inverse of $\sigma_0$ is given by
$$
\sigma_0^{-1} (z) = \sigma_0^{-1} (u+iv) = \frac{(2u,2v,|z|^2-1)}{|z|^2+1}.
$$
The computation now shows that indeed the transition function is
$$
\sigma_1 \circ \sigma_0^{-1} (z) = \frac{1}{z}.
$$
Notice moreover that
$$
\sigma_1^{-1}(z) = \frac{(2u,2v,1-|z|^2)}{|z|^2+1},
$$
so a computation shows that
$$
\sigma_0 \circ \sigma_1^{-1} (z) = \frac{1}{z^\ast}.
$$
A: As commented by Ben Blum-Smith, it is impossible that one of two expressions $\sigma_1 \circ \sigma_0^{-1}(z)$ and $\sigma_0 \circ \sigma_1^{-1}(z)$ produces $\varphi(z) = \frac{1}{z}$ and the other $\psi(z) = \frac{1}{z^*}$ because $(\sigma_1 \circ \sigma_0^{-1})^{-1} = \sigma_0 \circ \sigma_1^{-1}$. In fact, we have $\varphi^{-1} = \varphi$ and $\psi^{-1} = \psi$.
Where is the mistake in Rick's answer?

*

*Writing $\mathbb R^2 =  \mathbb C$ and $\mathbb R^3 =  \mathbb C \times \mathbb R$ Ricks definitions are
$$\sigma_0 (z, x_3) = \frac{z}{1-x_3}, \tag{1}$$
$$\sigma_1 (z, x_3) = \frac{z^*}{1+x_3} \tag{2*}.$$
Whereas $\sigma_0$ is the stereographic projection $\sigma_N$ from the north pole, $\sigma_1$ is not the stereographic projection $\sigma_S$ from the south pole. In fact, $\sigma_1 = \sigma_S^*$. The correct formula for stereographic projection from the south pole is
$$\sigma_1 (z, x_3) = \frac{z}{1+x_3} \tag{2}.$$


*We get
$$\sigma_0^{-1} (z) = \frac{(2z,|z|^2-1)}{|z|^2+1}. \tag{3}$$
Based on $(2^*)$ we get
$$\sigma_1^{-1}(z) = \frac{(2z^*,1-|z|^2)}{|z|^2+1} . \tag{4*}$$
This differs from Ricks's formula
$$\sigma_1^{-1}(z) = \frac{(2z,1-|z|^2)}{|z|^2+1} \tag{4}$$
which is based on $(2)$.


*In principle it is irrelvant if we work with $(2)$ or $(2^*)$. But working with  $(2^*)$ and $(4)$ is not correct; this produces the inconsistent result
$$\sigma_1 \circ \sigma_0^{-1} (z) = \frac{1}{z} , \sigma_0 \circ \sigma_1^{-1} (z) = \frac{1}{z^*} .$$


*Working with $(1)$ and $(2^*)$ gives
$$\sigma_1 \circ \sigma_0^{-1} (z) = \sigma_0 \circ \sigma_1^{-1} (z) = \frac{1}{z} .$$
Thus we get holomorphic transition functions.


*Working with $(1)$ and $(2)$ gives
$$\sigma_1 \circ \sigma_0^{-1} (z) = \sigma_0 \circ \sigma_1^{-1} (z) = \frac{1}{z^*} .$$
Thus we do not get holomorphic transition functions.


*This shows that the charts based on the two genuine stereographic projections do not have holomorphic transition functions. If we want to have holomorphic transition functions, we have to take the complex conjugate of one of these charts.
