# Constructing the Riemann Sphere

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

1. the stereographic projection $\sigma_0 : \mathbb{S}^2 \setminus \{N\} \to \mathbb{R}^2 \cong \mathbb{C}$ from the North pole, $N$, given by $$\sigma_0 (x_1, x_2, x_3) := \frac{(x_1, x_2)}{1-x_3},$$ and
2. the stereographic projection $\sigma_1 : \mathbb{S}^2 \setminus \{S\} \to \mathbb{R}^2 \cong \mathbb{C}$ from the North pole, $S$, given by $$\sigma_1 (x_1, x_2, x_3) := - \sigma_0 (-x_1, -x_2, -x_3) = \frac{(-x_1, -x_2)}{1+x_3}.$$

Question: How does one choose the correct orientations of the two charts to ensure that the transition function is $$\sigma_1 \circ \sigma_0^{-1} (z) = z^{-1}?$$

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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 $$\sigma_0 \circ \sigma_1^{-1} (z) = \frac{1}{z^\ast}.$$