Show that $\mathbb{R}^2/{\sim}$ is homeomorphic to the sphere $S^2$.

Question

$\mathbb{R}^2/{\sim}$ is the smallest equivalence relation such that $P\sim Q$ for all $P,Q$ with $\|P\|_2,\|Q\|_2 \geq 1$. Show that $\mathbb{R}^2/{\sim}$ is homeomorphic to the sphere $S^2$.

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Attempt:

$\mathbb{R}^2/{\sim}=A\cup B$ where $A:=\{\{P\}:\|P\|_2<1\}$ and $B:=\{\{P:\|P\|_2\geq1\}\}$. I think $A$ corresponds to a unit open disk and we can deform it to the sphere substracted a point $x$ and also $x$ corresponds to the element of $B$. But I cannot explicitly write the homeomorphism between $\mathbb{R}^2/{\sim}$ and $S^2$.

Edit: Formally we have $\mathbb{R}^2/{\sim}\supseteq A\simeq \mathbb{B}^2\simeq\mathbb{R}^2\simeq S^2-\{x\}\subseteq S^2$ where $\mathbb{B}^2$ is the unit open disk in $\mathbb{R}^2$. Clearly we can extend the homeomorphism between $A$ and $S^2-\{x\}$ to the bijection from $\mathbb{R}^2/{\sim}$ to $S^2$. How can we show this bijection is actually a homeomorphism?

Answer 1

If you aren't completely comfortable working with quotient spaces, the following is very useful.

Universal property of the quotient space

Let $X,Y$ be topological spaces, and let $\sim$ be an equivalence relation on the points of $X$. Then we have a natural continuous map $p\colon X\to X/{\sim}$ given by projecting a point of $X$ on to its equivalence class.

Now, let $g\colon X/{\sim}\to Y$ be a continuous map. By composing with $p$, we get a continuous map from $X$ to $Y$:

$$ g\circ p\colon X\xrightarrow{p}X/{\sim}\xrightarrow{g}Y $$

Write $f=g\circ p$. Now $f$ isn't just any old continuous map from $X$ to $Y$; it has a special property; namely, that if $x\sim y$ then $f(x)=f(y)$. [This is because $p(x)=p(y)$.]

Why is this interesting? Well, it gives us a characterization of the continuous maps from $X/{\sim}\to Y$:

if $f\colon X\to Y$ is a continuous map such that

(*) $f(x)=f(y)$ whenever $x\sim y$

then there is a unique continuous map $g\colon X/{\sim}\to Y$ such that $f=g\circ p$.

In other words, the following map of sets is a bijection:

\begin{align} \{\textrm{Continuous maps }g\textrm{ from }X/{\sim}\textrm{ to }Y\}&\to\{\textrm{Continuous maps }f\textrm{ from }X\textrm{ to }Y\textrm{ satisfying (*)}\} \\ g&\mapsto g\circ p \end{align}

So, coming up with a continuous map from $X/{\sim}$ to $Y$ is the same thing as coming up with a continuous map from $X$ to $Y$ that is constant on equivalence classes.

This particular example

Here, you want to define a homeomorphism from $\mathbb R^2/{\sim}$ to $S^2$, where $\sim$ identifies all points whose norm is at least $1$. By the above discussion, we need to specify a continuous map from $\mathbb R^2$ to $S^2$ that is constant outside the open unit disc. Clearly, in order to do this, it is enough to define a continuous map from the closed disc $D^1=\{P\in\mathbb R^2\;\colon\;\|P\|\le 1\}$ to $S^2$ and then extend it by the constant value everywhere else.

Can you think of a suitable map? It should be surjective, and it should be injective except on the boundary of the disc, where it should take some constant value.

Hint: Remember that the closed disc $D^1$ can be parameterized by polar coordinates $(\rho,\vartheta)$, and that the sphere $S^2$ can be parameterized by spherical polar coordinates $(\phi, \theta)$.

Answer 2

Instead of $S^2$, one can work with the Riemann sphere $\widehat{\mathbb C}=\mathbb R^2\cup\{\infty\}$ and show that $\mathbb R^2/ {\sim}$ is homeomorphic to $\widehat{\mathbb C}$.

Consider the following map : $$\begin{array}{rrcl}\varphi: & \mathbb R^2 & \longrightarrow & \widehat{\mathbb C} \\ & x & \longmapsto & \left\{ \begin{array}{cc} \dfrac{x}{1-\|x\|} & \text{ if } \|x\|<1 \\ \infty & \text{ if } \|x\|\geq 1\end{array}\right. \end{array} $$

$\varphi$ is continuous and constant on each equivalence classes thus it definies a well defined continuous map $\overline{\varphi}$ from $\mathbb R^2/{\sim}$ to $\widehat{\mathbb C}$.

Moreover, one can easily check that :

1. $\overline{\varphi}$ is now bijective,
2. $\mathbb R^2/{\sim}$ is Hausdorff.
3. Let $\pi:\mathbb R^2\rightarrow \mathbb R^2/{\sim}$ be the canonical projection and $B$ be the closed unit ball in $\mathbb R^2$. Then $\pi(B)=\mathbb R^2/{\sim}$ and thus $\mathbb R^2/{\sim}$ is a compact space.

Hence, $\overline{\varphi}$ is a continuous bijective map from a compact space into a Hausdorff space so, it is a homeomorphism.

Answer 3

The sphere $S^2$ is the set of all points $$\left\{\pmatrix{\cos \phi\\ \sin\phi\cos\theta\\ \sin\phi\sin\theta}\middle| \array{0\le\phi\le\pi,\\0\le\theta\le2\pi} \right\}$$ The plane $R^2$ can be written as the set $$\left\{\pmatrix{r\cos\sigma\\ r\sin\sigma}\middle| \array{0\le r\\ 0\le\sigma\le 2\pi}\right\}$$ There is a function $R^2\to S^2$ sending $$\pmatrix{r\cos\sigma\\ r\sin\sigma}\mapsto\pmatrix{\cos(\pi\min(r,1))\\ \sin(\pi\min(r,1))\cos\sigma\\ \sin(\pi\min(r,1))\sin\sigma}$$ Can you show this map is continuous and induces a homeomorphism $R^2/{\sim} \to S^2$ ?