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:

  1. The set $\{(0,0),(1,0),(0,1),(1,1)\}$ of four corner points
  2. sets of pairs of points $(x,0),(x,1)$ where $0<x<1$ ;
  3. sets of pairs of point $(0,y), (1,y )$ where $(0,y), (1,y)$ where $0<y<1$;
  4. 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:

  1. the set $S^{n-1}$;
  2. 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.

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    $\begingroup$ If you think of your space as lying flat on the plane (not always possible, alas!) then you are "folding" your space so that identified points coincide. In your second question, the entire surface is just "one big point" which is outside the ball. E.g., if n=2, you get the usual unit disc, but all the points on the circle are "the same". Imagine sewing a thread along the unit circle and then pulling it taut until it is just a single point. That's what you do. $\endgroup$ – Arturo Magidin Oct 9 '10 at 2:30

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:

  1. 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 $2$.
  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:

$$ B^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2 \leq 1 \ , \ x_{n+1} = 0 \right\} $$


$$ S^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2 + (x_{n+1}-1)^2 = 1 \right\} \ . $$

And we try some map of the same kind:

$$ \varphi : B^n \longrightarrow S^n \ , \qquad \varphi (x_1, \dots , x_n ) = (f(r)x_1, \dots , f(r)x_n, 2r) \ . $$


  1. Complete the details.
  2. This is not the same homeomorphism we previously found for $n=1$. Write this general formula for $n=1$.
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There are two things you can do with Topology: (1) try to imagine what's going on intuitively, and (2) write it down rigorously. Both are useful, necessary steps.

For instance, if I understand your first example, you have the unit square $[0,1]^2 \subset \mathbb{R}^2$ with the subspace topology. Let me write the coordinates of its points $(\theta, z) \in [0,1]^2$, for psychological reasons that will become apparent in a minute.

On the intuitive side, what you are doing is bending your square $[0,1]^2$ in such a way that sides $\theta = 0$ and $\theta = 1$ become closer and closer... You put some glue on them, stick them... Ok: and what do you "see" now? -A cylinder, right?

(Except for the fact that in real life you cannot stick just two segments -or it's really difficult-, you can try to do this experiment with a piece of paper and some glue.)

On the rigorous side, you may do the following: you recover that parametrization of the cylinder you have probably seen in some geometry or integral calculus course

$$ \varphi: [0,1]^2 \longrightarrow \mathbb{R}^3 \ , \qquad \varphi (\theta , z) = (\cos (2\pi\theta ), \sin (2\pi\theta ), z) \ . $$

This $\varphi$ is obviously continuous ("obviously" means: "someone taught you this in some calculus course a year ago") and surjective, if we restrict the codomain to the cylinder

$$ C = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 = 1 \ , \ z \in [0,1] \right\} \ . $$

That is, we are looking to $\varphi$ as a map $\varphi : [0,1]^2 \longrightarrow C$. Notice that also this way it is continuous with the subspace topologies on both sides.

Now $\varphi $ is almost injective too, except for those points $(0,z)$ and $(1,z)$ that have the same image

$$ \varphi (0,z) = (1,0,z) = \varphi (1,z) \ , \qquad \text{for all}\qquad z \in [0,1] \ . $$

In order to fix this lack of injectivity, you say: "Ok, let's identify ("identify" is the picky, correct, way to say "glue" in Topology) each $(0,z)$ with its corresponding $(1,z)$ for all $z$." So, you are defining an equivalence relation $\sim$ among the points of the square $[0,1]^2$, the one generated by:

$$ (0,z) \sim (1,z) \qquad \text{for all} \qquad z \in [0,1] \ . $$

You call $X = [0,1]^2/\sim$ the space obtained from the square $[0,1]^2$ after gluing (sorry, "identifying") each $(0,z)$ with its corresponding $(1,z)$.

Let's write $\widetilde{(\theta ,z)} \in X$ the equivalence classes produced by this equivalence relation. Notice that, if $\theta \neq 0, 1$, you just have $\widetilde{(\theta , z)} = \left\{ (\theta , z) \right\}$, but $\widetilde{(0,z)} = \widetilde{(1,z)} = \left\{ (0,z), (1,z) \right\}$. That is, points of $X$ are "the same" as the points of $[0,1]^2$, except for those $(0,z)$ and $(1,z)$ that are now glued togehter ("identified", yes).

In doing this, we also get a natural map ("projection", "identification")

$$ \pi : [0,1]^2 \longrightarrow X \ , \qquad \pi (\theta , z) = \widetilde{(\theta , z)} \ , $$

which is continuous by definition of the quotient topology on $X$.

Back to our $\varphi$: since it identifies the same points as $\sim$ does, $\varphi$ induces a well-defined map

$$ \widetilde{\varphi} : X \longrightarrow C \ , \qquad \widetilde{\varphi}\widetilde{(\theta , z)} = \varphi (\theta ,z) \ . $$

Now, since we have glued the points with the same image by $\varphi$, this new $\widetilde{\varphi}$ is bijective, but it's also continuous, because of the universal property of the quotient topology and the fact that $\widetilde{\varphi} \circ \pi = \varphi$. Right?

Last step. A continuous bijective map doesn't need to be a homeomorphism, as you probably know: the inverse $\widetilde{\varphi}^{-1}$ is not necessarily continuous.

What could we do? Well, you can try to write down an explicit formula for $\widetilde{\varphi}^{-1}$ and check its continuity directly, but this is hard and painful, so nobody does it.

Instead, everybody resorts to the following marvelous, great, second to none trick (the best one you can buy in elementary Topology; btw, "trick" means "proposition": you can prove it, of course):

"$X$ is compact, because it is a quotient of a compact space (namely, $[0,1]^2$). $C$ is Hausdorff, because it is a subspace of a Hausdorff space (namely, $\mathbb{R}^3$).

The GTET (Greatest Trick in Elementary Topology)

Now, I have a continuous, bijective map $\widetilde{\varphi} : X \longrightarrow C$ between a compact space and a Hausdorff one. Hence $\widetilde{\varphi}$ is a homeomorphism."

Isn't that fantastic? :-) (Don't forget it the next time you need to prove that some map is a homeomorphism: it will save your life.)

So, you can also prove rigorously your first intuition: your space is a cylinder, $X \cong C$.

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  • 2
    $\begingroup$ A great answer! $\endgroup$ – Jyotirmoy Bhattacharya Oct 9 '10 at 3:14
  • $\begingroup$ @Jyotirmoy. Glad you like it. :-) $\endgroup$ – Agustí Roig Oct 9 '10 at 3:32
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    $\begingroup$ Definitely one of the best answers I have ever seen here. $\endgroup$ – Ivo Terek Aug 22 '15 at 17:50
  • $\begingroup$ Thank you, Ivo Terek. $\endgroup$ – Agustí Roig Aug 23 '15 at 18:19
  • $\begingroup$ I guess it will be better to write $[0,1]=I$ $\endgroup$ – user210387 Sep 14 '15 at 12:38

Trying to place points in topology is risky. The geometric intuition it gives can be very useful, but can also be very misleading. Formally, a topology is defined by the open sets regardless of any capability to imagine them. See "Counterexamples in Topology" for examples of the problems you face here.

Collapsing points can change the shape of a space considerably. I think your specific example of collapsing the boundary of $B^n$ gives you $S^n$ as you have a compact n-dimensional manifold without boundary in which every loop is contractible to a point, but I am willing to be corrected.

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  • $\begingroup$ $B^n/S^{n-1}=S^n$ by definition -- this is certainly not the only compact boundaryless simply-connected n-manifold! $\endgroup$ – Aaron Mazel-Gee Oct 9 '10 at 23:50
  • $\begingroup$ I thought that was the heart of the Poincare conjecture, just solved. $\endgroup$ – Ross Millikan Oct 10 '10 at 1:58
  • $\begingroup$ The Poincare conjecture also assumes (in addition to what you've got) that the space is a homology sphere, i.e. $H_0(M)=H_n(M)=\mathbb{Z}$ is its only nontrivial homology. In fact, Poincare originally conjectured that this condition was sufficient, but then he found a counterexample which is now called the Poincare dodecahedral space (cf. wikipedia) -- it's a 3-dimensional homology sphere, but it has nontrivial fundamental group so of course it isn't truly a sphere. $\endgroup$ – Aaron Mazel-Gee Oct 10 '10 at 2:17
  • $\begingroup$ ($H_1$ is the abelianization of $\pi_1$, so in this case $\pi_1\not= 0$ but $H_1=\pi_1^{ab.} = \pi_1/[\pi_1,\pi_1]=0$.) He reformulated his conjecture accordingly, of course. If you take the Cartesian product of any two compact boundaryless simply-connected manifolds, you'll get a counterexample to your claim. $\endgroup$ – Aaron Mazel-Gee Oct 10 '10 at 4:16

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