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: $$\sigma_S(x_1,x_2,x_3)=\sigma_N(x_1,x_2,-x_3)$$ $$\sigma_N(x_1,x_2,x_3)=\sigma_S(x_1,x_2,-x_3)$$