# Understanding $\varphi: S^2\setminus \{(0,1)\}\to \mathbb C\mathbb P^1$ given by $(z,\alpha) \mapsto \left[\frac{2z}{1-\alpha}: 1\right]$

I am reading this MSE answer. Writing $$S^2=\big\{(z,\alpha) \in \mathbb{C} \times \mathbb{R}: |z|^2+\alpha^2=1\big\},$$ the author claims that \begin{align*} \varphi: S^2\setminus \{(0,1)\} & \to \mathbb{CP}^1 \\ (z,\alpha) & \mapsto \left[\frac{2z}{1-\alpha}: 1\right] \end{align*} extends smoothly to a map $$\varphi: S^2\to \Bbb C\Bbb P^1$$ sending $$(0,1)$$ to $$[1:0]$$. Moreover, $$\varphi$$ is a diffeomorphism. I'd like some help with understanding this statement.

1. Why is the map defined by $$(z,\alpha) \mapsto \left[\frac{2z}{1-\alpha}: 1\right]$$ on $$S^2\setminus\{(0,1)\}$$? The choice seems quite unmotivated.

2. Why does $$\varphi$$ extend smoothly to a map on $$S^2$$?

Lastly, I believe showing that $$\varphi$$ is a diffeomorphism would involve explicitly writing $$\varphi^{-1}$$ and composing with charts to obtain that $$\varphi$$ and $$\varphi^{-1}$$ are smooth maps. Thanks!

Consider the Stereographic projection: \begin{align} \phi:S^2\smallsetminus (0,0,1)&\longrightarrow \mathbb R^2\\ (x,y,\alpha)&\longmapsto\left(\frac{x}{1-\alpha},\frac{y}{1-\alpha}\right) \end{align} This can be identified with a map to $$\mathbb C$$, by composing the $$\mathbb R$$ vector space isomorphism $$(x,y)\mapsto x+iy$$. Hence, if we identify $$S^2$$ as a subset of $$\mathbb C^2$$, then the stereographic projection becomes something of the form: \begin{align} \phi:S^2\smallsetminus (0,1)&\longrightarrow \mathbb C\smallsetminus 0\\ (z,\alpha)&\longmapsto\frac{z}{1-\alpha} \end{align} so you can just compose this with the chart for most of $$\mathbb {CP}^1$$, and then argue that this extends to a smooth map on the whole of $$S^2$$.

To show it is a diffeomorphism, it suffices to show that this is bijective, and that the differential is a linear isomorphism everywhere, which it is. This fact follows from the inverse function theorem (as smoothness and continuity is local)

This is essentially contained in youler's answer to Diffeomorphism of $\mathbb{C}P^1$ and $S^2$.

$$S^2$$ has a smooth atlas consisting of the stereographic projections $$\varphi_1 : S^2 \setminus \{N\} \to \mathbb C = \mathbb R^2, \varphi_1(z,\alpha) = \frac{z}{1-\alpha},$$ $$\varphi_2 : S^2 \setminus \{S\} \to \mathbb C, \varphi_2(z,\alpha) = \frac{z}{1+\alpha}.$$

Concerning the stereorgraphic projection see for example Two equal smooth structures on $S^n$.

$$\mathbb{CP}^1$$ has a smooth atlas consisting of the maps $$\psi_1 : \mathbb{CP}^1 \setminus \{[0:1]\} \to \mathbb C,\psi_2[z:w]=w/z ,$$ $$\psi_2 : \mathbb{CP}^1 \setminus \{[1:0]\} \to \mathbb C, \psi_2[z:w]=\overline{z/w}.$$ The inverses of the $$\psi_i$$ are $$\psi_1^{-1}(\omega) = [1:\omega],$$ $$\psi_2^{-1}(\zeta) = [\overline \zeta:1].$$ This gives diffeomorphisms $$\delta_1 = \psi_1^{-1} \circ \varphi_1 :S^2 \setminus \{N\} \to \mathbb{CP}^1 \setminus \{[0:1]\} ,$$ $$\delta_2 = \psi_2^{-1} \circ \varphi_2 :S^2 \setminus \{S\} \to \mathbb{CP}^1 \setminus \{[1:0]\} .$$ youler showed that they fit together to a diffeomorphism $$\Delta : S^2 \to \mathbb{CP}^1$$ such that $$\Delta(N) = [0:1]$$ and $$\Delta(S) = [1:0]$$.

Thus $$\delta_1$$ is the restriction of the diffeomorphism $$\Delta$$. In other words, $$\delta_1$$ extends to a diffeomorphism via $$N \mapsto [0:1]$$. We have

$$\delta_1(z,\alpha) = \left[1 : \frac{z}{1-\alpha}\right] = [1-\alpha : z] .$$

This is not exactly the map $$\varphi$$ in your question. However, the map $$\tau : \mathbb{CP}^1 \to \mathbb{CP}^1, \tau[z:w] = [2w:z]$$, is easily seen to be a diffeomorphism such that $$\tau[1:0] = [0:1]$$ and $$\tau[0:1] = [1:0]$$ (use the above charts $$\psi_i$$ to see that the maps $$\tau$$ and $$\tau^{1}[w:z] = [z:w/2]$$ are smooth).

This proves the claim in your question.