Show that $X/$~ is homeomorphic to $S^2$.

Question

Let $S^2$ be the 2-sphere, and let $X$ be $\mathbb{C}^2 \setminus \left\{(0,0)\right\}$.

And define an equivalence relation ~ on X by

$(x_1, y_1)$ ~ $(x_2, y_2)$ if and only if there is t $\in \mathbb{C} \setminus \left\{0\right\}$ such that $(x_1, y_1)=(tx_2, ty_2)$.

Then show that $X/$~ is homeomorphic to $S^2$.

Here is what I thought:

For this relation, I thought I need to find a surjective continuous map $\ f : X \rightarrow S^2$ which is constant if $(x_1, y_1)$ ~ $(x_2, y_2)$.

Then this induces continuous map $\ \bar{f} : X/$~ $\rightarrow S^2$.

I guess this will be the homeomorphism what I’m trying to find.

Such $f$ probably satisfies the following property,

$f(x_1, y_1) = f(tx_2, ty_2)$ for any $t \in \mathbb{C} \setminus \left\{0\right\}$.

But I couldn’t come up with such a continuous map $\ f$.

Could you give a hint? Or any suggestion?

Thank you.

Answer 1

If one remembers that the 2-sphere is a one point compactification of the plane, the result can be proved as follows. Let $(z,w)$ be the two coordinates on $\bf C^2$. Note that the image of the line $w=0$ is just a point in our quotient space. Remove this point. Then we have to study the quotient $Q$ of the set of pairs $(z,w)$ such that $w\not =0$ by homothety. But if $w\not = 0$, $(z,w)\sim ({z\over w},1)$, so that the map $\bf C \to \bf C^2$, $x\to (x,1)$ induces an homeomorphism between $\bf C$ and $Q$. Being the one point compactification of $Q$ (a plane), it is a sphere.

Answer 2

Consider the map $X \to \mathbb{C}\cup \{\infty\}$ given by $(x,y)\to \frac{x}{y}$ for $y\neq 0$ and $(0,y)\to \infty$. So long as you are happy $\mathbb{C}\cup\{\infty\}$ is homeomorphic to $S^2$.

Answer 3

Since $\mathbb{C} \cup \left\{\infty\right\}$ is homeomorphic to $S^2$, it suffices to show that $X/$~ is homeomorphic to $\mathbb{C} \cup \left\{\infty\right\}$.

Let $\phi : X \rightarrow \mathbb{C} \cup \left\{\infty\right\}$ be defined by

$(x,y) \mapsto \frac{x}{y}$ if $y \ne 0$, $(x,y) \mapsto \infty$ if $y=0$

Then $\phi$ is surjective, since for any $t \in \mathbb{C}$, it maps $(t,1)$ to $t$, and also it maps $(x,0)$ to $\infty$.

And To show that $\phi$ is continous, it suffices to show that for open $V$ containing $\infty$, $\phi^{-1}(V)$ is open in $X$.

Let $x \in \phi^{-1}(V)$, then there exists $\mathbb{C} \setminus C$ containing $\phi(x)$, which is open in $\mathbb{C} \cup \left\{\infty\right\}$

So $x \in \phi^{-1} (\mathbb{C} \setminus C) \subseteq \phi^{-1}(V)$, and $\phi^{-1}(\mathbb{C} \setminus C)$ is open in $X$.

This actually shows that $\phi$ is a quotient map. ($V$ is open in $\mathbb{C} \cup \left\{\infty\right\}$ if and only if $\phi^{-1}(V)$ is open in $X$)

Then $\phi$ induces a homeomorphism $\bar{\phi} : X/$~ $\rightarrow \mathbb{C} \cup \left\{\infty\right\}$.

So it’s done.