# Identifying the Quotient Space in $\mathbb{R}^3$

If I have:

$X = \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1 \}$ and $A = \{ (x,y,0) \in \mathbb{R}^3 : x^2 + y^2 = 1 \}$

identify the quotient space $X/A$ as a more familiar topological space and show its homeomorphic to $X/A$.

-
Here $X$ is a sphere and $A$ is its equator. Then the quotient space is two spheres glued together at one point. –  Matt N. Feb 27 '13 at 7:32
What would the function be of that? –  user64013 Feb 27 '13 at 7:56
Topological balloon tricks. –  copper.hat Feb 27 '13 at 8:23
@user64013 I'm not entirely competent and I'm on the run right now but if there is still no answer when I'm back I'll try to write something useful. –  Matt N. Feb 27 '13 at 10:34
add comment

## 2 Answers

You could map the unit sphere $X$ to the set $Y=\{(v,w,z)\in\mathbb R^3\mid v^2+w^2+(z\pm\frac12)^2=\frac14\}$ which is the union of two spheres with radius $\frac12$ and centers $(0,0,\pm\frac12)$. The easiest way is to map a point $(x,y,z)$ to the point $(v,w,z)$ with same third coordinate. When you imagine those sets, you'll see that $v$ and $w$ are just dilations of $x$ and $y$ by the same factor $\lambda(|z|)$ which only depends on $|z|$. Try to derive a formula for this $\lambda(z)$ by looking at the equations $x^2 + y^2 + z^2 = 1$ and $v^2+w^2+(z\pm\frac12)^2=\frac14$.

-
add comment

Show that open disk $D^2$ is homeomorphic to sphere $S^2$ with one point removed. Let $D^2$ be the open disk of radius $1$ centered at the origin in $\mathbf R^3$ and let $S^2$ be the sphere of radius $1/2$ with the south pole at the origin. Let $N$ denote its north pole. Define map $\varphi : D^2 \to S^2 \setminus \{N\}$ as $$\begin{cases} (x,y,0) \mapsto (x,y, 1/2 - \sqrt{1/4 - (x^2 + y^2)}) & x^2 + y^2 \le 1/2 \\ (x,y,0) \mapsto (1-x, 1-y, 1/2 + \sqrt{1/4 - (x^2 + y^2)}) & x^2 + y^2 > 1/2 \end{cases}$$

Then $\varphi$ is bijective, continuous and open:

Injective: If $(x,y,0) \neq (x', y', 0)$ then either $x\neq x'$ or $y\neq y'$. Hence in both cases, $\|(x,y,0)\| , \|(x', y', 0)\| \le 1/2$ and $\|(x,y,0)\| , \|(x', y', 0)\| > 1/2$ it is immediate that $\varphi ((x,y,0)) \neq \varphi ((x',y',0))$. If $\|(x,y,0)\| \le 1/2$ and $\|(x', y', 0)\| > 1/2$ then $x^2 + y^2 \neq x'^2 + y'^2$ and hence in this case also $\varphi ((x,y,0)) \neq \varphi ((x',y',0))$.

Surjective: Let $(x,y,z) \in S^2 \setminus\{N\}$. If $z \le 1/2$ then $\varphi ((x,y,0)) = (x,y, 1/2 - \sqrt{1/4 - (x^2 + y^2)}) = (x,y,z)$ since $z = 1/2 - \sqrt{1/4 - (x^2 + y^2)}$. Similarly if $z > 1/2$.

Continuous: It is clear that $\varphi$ is continuous if $x^2 + y^2 \le 1/2$ and if $x^2 + y^2 > 1/2$. Hence it is enough to verify that $$\lim_{x^2 + y^2 \to 1/2} (x,y, 1/2 - \sqrt{1/4 - (x^2 + y^2)}) = (1-x_0, 1-y_0, 1/2 + \sqrt{1/4 - (x_0^2 + y_0^2)})$$ for $x_0^2 + y_0^2 = 1/2$. This holds since by symmetry we may choose to take the limit along $x=y$. Then $$\lim_{2x^2\to 1/2} (x,x, 1/2 - \sqrt{1/4 - 2x^2 )}) = (1-1/2, 1-1/2, 1/2 + \sqrt{1/4 - (1/4)}) = (1/2, 1/2, 1/2)$$

Open: Similarly we show that $\varphi^{-1}$ is continuous.

To prove $S^2 \sqcup S^2 / \sim$ is homeomorphic to $X/A$ note that $X/A$ consists of two open disks (all the points whose equivalence class only contains one point) and a point (all points of $A$ are one equivalence class). Similarly, the space $S^2 \sqcup S^2 / \sim$ consist of two sphere (with point removed) and a point. Then there are homeomorphism $\varphi_1$ from one disk in $X/A$ to one sphere minus a point in $S^2 \sqcup S^2 / \sim$ and similarly for the other disk a $\varphi_2$. Let $\varphi$ be the map $[a] \mapsto [N,S]$ where $[a]$ is the equivalence claas of $A$ in $X/A$ and $[N,S]$ is the equivalence class of the north and south pole in $S^2 \sqcup S^2 / \sim$. Then $\varphi_1 \sqcup \varphi \sqcup \varphi_2$ is a homeomorphism.

-
add comment