I know that $\mathbb{P^1(C)} \cong \mathbb{P^1(C)} \cup \{N\} $, where $N$ is the north-pole of the sphere, is homeomorphic to the sphere $S^2$ thanks to the stereographic projection, but I am not sure if the explicit projection could be the following: $$f: S^2 \longrightarrow \mathbb{P^1(C)} \\(x,y,z) \mapsto \left( \frac{x+iy}{1-z},\frac{x-iy}{1+z} \right) .$$

Is it the right approach?

  • $\begingroup$ This is not defined at $(0,0,1)$. Both $S^2$ and $\mathbb CP^1$ can be obtained by gluing $\mathbb C$ along two copies of $\mathbb C^\times$ identified by inversion. This should help you defined a homeomorphism using these covers. $\endgroup$ – Pedro Tamaroff Aug 23 '18 at 11:20
  • $\begingroup$ @PedroTamaroff What if I keep $f$ but I map (0,0,1) to the infinite point of $\mathbb{C}P^1$? $\endgroup$ – Phi_24 Aug 23 '18 at 13:16
  • $\begingroup$ What you want is to modify one denominator to be $1+z$ instead of $1-z$, I think. $\endgroup$ – Pedro Tamaroff Aug 23 '18 at 13:18
  • $\begingroup$ @PedroTamaroff you're right, I edited because I wrote wrongly the first time, but still in this case if I put (0,0,1) the first coordinate is not defined, isn't it? $\endgroup$ – Phi_24 Aug 23 '18 at 13:24
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    $\begingroup$ Why don't you use homogeneous coordinates in $\mathbb{C}P^1$, e.g. $[\frac{x}{y}: z ] = [x :yz]$ ? $\endgroup$ – Maxime Ramzi Aug 23 '18 at 15:54

The map you provide will not work. Even restricted to the sphere without the north pole, it is not even injective. Indeed, $(1,0,0)$ and $(-1,0,0)$ go to the same point, since $$f(1,0,0)=[1:1]=[-1:-1]=f(-1,0,0).$$

To answer the question, let's try going step by step:

The stereographic projection $\mathrm{Steo}:S^2\backslash\{N\} \to \mathbb{R}^2$ is given by $$(x,y,z) \mapsto \left(\frac{x}{1-z},\frac{y}{1-z} \right).$$ When you identify $\mathbb{R}^2 \simeq \mathbb{C}$, you have a formula which appears in your attempt (I'll still call the stereographic projection by the same name): \begin{align*} \mathrm{Steo}:S^2\backslash\{N\} &\to \mathbb{C} \\ (x,y,z) &\mapsto \frac{x+iy}{1-z}. \end{align*} Now, $\mathbb{C}$ embeds naturally in $\mathbb{C}P^1$ via \begin{align*} g:\mathbb{C} &\to \mathbb{C}P^1\\ z &\mapsto [z:1]. \end{align*} This strategy is alluded to in the comments by Max. Now, we have that this $g$ misses a single point: $[1:0]$. This is due to the fact that if $b \neq 0$, then $[a:b]=[ab^{-1}:1]$ (and if $b=0$, $[a:0]=[aa^{-1}:0]=[1:0]$, where we recall that $a$ can't be zero if $b$ is zero).

So we have the explicit map \begin{align*} g \circ \mathrm{Steo}:S^2 \backslash\{N\} &\to \mathbb{C}P^1 \backslash \{[1:0]\}\\ (x,y,z) &\mapsto \left[\frac{x+iy}{1-z}:1\right], \end{align*} which is an homeomorphism. There is the problematic missing north pole, and missing $[1:0]$. However, they are not a problem at all. Indeed, we can extend $g$ to have domain $S^2$ and codomain $\mathbb{C}P^1$ by sending $(0,0,1) \mapsto [1:0]$, which is essentially the uniqueness of the one-point compactification.

So, the final mapping becomes: \begin{align*} g \circ \mathrm{Steo}:S^2 &\to \mathbb{C}P^1\\ (x,y,z) &\mapsto \left[\frac{x+iy}{1-z}:1\right], \quad z \neq 1 \\ (0,0,1) &\mapsto [1:0], \quad z=1. \end{align*} Since you only want an homeomorphism, this escaping via general topology is enough. If you want to check for differentiability etc, you will need to follow through Pedro's suggestion in the comments, which is essentially considering the other map analogous to $g$ which helps covering the $[1:0]$ problematic case via a chart.

  • $\begingroup$ Any particular reason you don't avoid the case distinction using $$(x,y,z)\mapsto\left[x+iy:1-z\right]\;?$$ $\endgroup$ – MvG Aug 23 '18 at 21:03
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    $\begingroup$ @MvG I was thinking the same, but if $z=1$ then $x=y=0$ and so this would map to $[0:0]$. Which suggests to me that something is wrong, but I haven't read thoroughly yet. $\endgroup$ – Servaes Aug 23 '18 at 21:05
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    $\begingroup$ @Servaes: Ah yes, you are right. Since we don't have enough direction information to pick a representative for the pole, I see that this case distinction makes sense here. Thanks for pointing this out. $\endgroup$ – MvG Aug 23 '18 at 21:11
  • $\begingroup$ I found out that using $\alpha[x+iy:1-z]+\beta[1+z:x-iy]$ one can put the problematic point (the one which maps to $[0:0]$) anywhere on the sphere, but can't get rid of it altogether. Interesting. $\endgroup$ – MvG Aug 23 '18 at 21:54

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