TITLE: Quotient topologies and equivalence classes
QUESTION [14 upvotes]: I'm currently studying the notion of a quotient topology. The one thing I'm having trouble with understanding is what we're actually doing to the points as we're identifying them. Say we have a $[0,1] \times [0,1]$ in $E^2$ (euclidean $2$-space) with the subspace topology and we partition $X$ into:

The set $\{(0,0),(1,0),(0,1),(1,1)\}$ of four corner points
sets of pairs of points $(x,0),(x,1)$ where $0<x<1$ ; 
sets of pairs of point $(0,y), (1,y )$ where $(0,y), (1,y)$ where $0<y<1$;
sets consisting of a single point $(x,y)$ where $0<x<1$ and $0<y<1.$ 

then our quotient space will be the torus, but this will set-speaking be:
$\{0,0),(1,0),(0,1),(1,1)\}$, sets of pairs of points as described above for $(x,0),(x,1),$ sets of pairs of points as described above for $(y,0),(y,1),$ and $(x,y)$ where $0<x<1, 0<y<1$.
I wonder whether there's a any natural way to see how to "place" the points? This example is of course kinda easy, but just given this set, is there any way to figure out how it looks geometrically? If we'd have say the set consisting of all points from a fixed distance from a fixed point, we'd identify it as a circle, is there anything similar here?
And now, for the last question here:
Let $B^n$ denote the unit ball in $n$-dimensional euclidean space, and let $S^{n-1}$ denote its boundary. Consider the partition of $B^n$ which has as members:

the set $S^{n-1}$;
the individual points of $B^n-S^{n-1}$. 

OK, when we now identify $S^{n-1}$ it collapses into one point, right? Exactly what does it collapse into and where do we "put it"?
Hope you can shed some light on this.

REPLY [17 votes]: As for your second example, Ross Milikan's intuition is right. Let's see how we can arrive at the conclusion that $B^n/S^{n-1} \cong S^n$ examining what happens in low dimensions.
For $n=1$, the (closed) unit ball is the interval $B^1 = [-1,1]$ and its boundary the "sphere" $S^0 = \left\{ -1, 1\right\}$. Now, if we try to imagine what happens to $B^1$ when we glue ("identify") the points $-1$ and $1$, we see the segment $[-1,1]$ bending, points $-1$ and $1$ approaching one to each other... Our glue provider has already put some of it on them, so when they meet, they stick together and we get... a circumference! So $B^1 / S^0 \cong S^1$.
Ok, before going to higher dimensions: how can we prove this? Well, it turns out that, with some adaptations, we already know how to do it. First, notice that $B^1 \cong [0,1]$ and here $S^0 = \left\{ 0,1 \right\}$. So we resort again to a parametrization; this time, of the circumference
$$
\varphi : [0,1] \longrightarrow \mathbb{R}^2 \ ,\qquad \varphi (\theta ) = (\cos (2\pi\theta) , \sin (2\pi \theta) )
$$
and proceed along the lines of the former cylinder example: $\varphi$ identifies the points of the "sphere" $S^0$, $\varphi (0) = (1,0) = \varphi (1)$, so $\varphi$ induces a well-defined map
$$
\widetilde{\varphi} : B^1/S^0 \longrightarrow  S^1 \ , \qquad \widetilde{\varphi}(\widetilde{\theta}) = \varphi (\theta) \ ,
$$
which is continuous and bijective. Now use the GTET (Greatest Trick of Elementary Topology) you already know to conclude that $\widetilde{\varphi} $ is a homeomorphism  $B^1/S^0 \cong S^1$.
So far, so good. What's next? Let's take a look at $n=2$. The unit ball $B^2$ is now a disk and its boundary the circumference $S^1$. We put some glue all over $S^1$ and we bend the disk and at the same time shrink the circumference making it become smaller and smaller till it's just one point. What do we get? A sphere!
So we may suspect that $B^2 / S^1 \cong S^2$. In order to prove it, we may try to reproduce our "visual" homeomorphism with formulae. Let's put our disk $B^2$ in $\mathbb{R}^3$ in the $xy$-plane, centered at $(0,0,0)$:
$$
B^2 = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 \leq 1 \ , \ z= 0  \right\} \ .
$$
And we put our sphere centered at $(0,0, 1)$:
$$
S^2 = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 + (z-1)^2 = 1 \right\}  \ .
$$
So the South Pole is $(0,0,0)$ -the center of the disk $B^2$-, the North Pole is $(0,0,2)$ and the equator lies in the plane $z= 1$.
Our "visual" homeomorphism bends the disk over the sphere and in doing this shrinks its boundary till it is identified with the North Pole of the sphere. So, let's try to do the same with a map like
$$
\varphi : B^2 \longrightarrow S^2 \ , \qquad \varphi (x,y) = (f(r)x , f(r)y, 2r) \ .
$$
Here $r= + \sqrt{x^2+y^2}$. Let's analyse what this kind of map is doing:

When going from $r = 0$ to $r=1$, we want the points $\varphi (x,y)$ go up the sphere, from height $z=0$ (South Pole) to $z=2$ (North Pole). So $z$ has to climb up at double speed as $r$ goes from the center of the disk to its boundary. Hence we put that $2$.
To make things easier, we want the projections of the points $\varphi (x,y)$ onto the $xy$-plane to lie in the same direction as the point of the disk $(x,y)$ they come from. So this projection should be something like $(\lambda x, \lambda y)$. At the same time this lambda will vary with the distance $r$ to the center of the disk in order that $\varphi (x,y)$ remains on the sphere. So we try this $(f(r)x , f(r)y)$.

To find which $f(r)$ we need, we impose the condition that $\varphi (x,y) $ lies on the sphere:
$$
1 = x^2 + y^2 + (z-1)^2 = f(r)^2 r^2 + (2r-1)^2
$$
And we get $f(r) = 2 \sqrt{\frac{1-r}{r}}$. So our map is
$$
\varphi (x,y) = 
\begin{cases}
\left( 2 \sqrt{\frac{1-r}{r}}x , 2 \sqrt{\frac{1-r}{r}}y, 2r  \right) & \text{if}\ (x,y) \neq (0,0) \\\
(0,0,0) & \text{if}\ (x,y) = (0,0) \ .
\end{cases}
$$
Exercise. Prove that $\varphi$ is continuous.
Now, we proceed much in the same way as in the cylinder example. By construction, $\varphi$ is clearly surjective. But not injective: all the points in the boundary $r=1$ of the disk  go to the North Pole. So we quotient out this boundary and, as before, $\varphi$ induces a bijective continuous map
$$
\widetilde{\varphi} : B^2/S^1 \longrightarrow S^2 \ , \qquad \widetilde{\varphi}\widetilde{(x,y)} = \varphi (x,y) \ .
$$
We apply again our nuclear weapon GTET and we are done.
All right, what about higher dimensions? -We've got no geometric or visual intuition there (at least, I've got none; if you think you can "see" what happens in, say, $\mathbb{R}^4$, congratulations, but just in case do a visit to your psychiatrist and tell him/her). Anyway, formulae allow us to go beyond our visual intuition. We put our ball and sphere inside $\mathbb{R}^{n+1}$ as before:
$$
B^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2 \leq 1 \ , \ x_{n+1} = 0 \right\}
$$
and
$$
S^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2   + (x_{n+1}-1)^2 = 1 \right\} \ .
$$
And we try some map of the same kind:
$$
\varphi : B^n \longrightarrow S^n \ , \qquad \varphi (x_1, \dots , x_n ) = (f(r)x_1, \dots , f(r)x_n, 2r) \ .
$$
Exercises.

Complete the details.
This is not the same homeomorphism we previously found for $n=1$. Write this general formula for $n=1$.