4
$\begingroup$

There's the following example from Introduction to Topological Manifolds by John Lee, the sphere $\mathbb{S}^2$ is homeomorphic to the closed disk $\bar{\mathbb{B}}^2 \subseteq \mathbb{R}^2$ modulo the equivalence relation generated by $(x, y) \sim (-x, y) \in \partial \bar{\mathbb{B}}^2$. (The claim is thus $\mathbb{S}^2 \cong \mathbb{\bar{B}}^2/ {\sim}$)

Now my intuition failed me here, because the Torus, $\mathbb{T}^2$, which is used as the prototypical example to demonstrate the power and simplicity of quotient spaces, where "gluing" ends of a square to form first a cylinder, and then finally a torus make sense. That "gluing" intuition didn't work for me in this case, if anything by that argument the quotient space we'd end up with above would be a semi-circle sitting in $\mathbb{R}^2$.

But to prove this claim the book explicitly constructs a quotient map $f : \mathbb{\bar{B}}^2 \to \mathbb{S}^2$ where $f$ is a rather involved function, and since $f$ makes the same identitifcations, namely being that on the boundary of the disk we have $f(x, y) = f(-x, y)$, and by the uniqueness of quotient space we prove our claim.

It seems most of the quotient spaces I've come across are of this nature. Which is for any quotient space $X /{\sim}$ there is a quotient map say $f$ (which is not the canonical mapping $q : X \to X / {\sim}$ defined by $q(x) = [x]$, sending an element to its equivalence class), which makes the same identifications as the equivalence relation. We then then appeal to the uniqueness of quotient spaces, and because $f$ makes the same identifications as $q$, we have $f[X] = X /{\sim}$. But since $X / {\sim}$ is easier to state and work with we use that instead.

Am I correct in this observation? Are all quotient spaces of this nature?

If so, what can we do in the case of the Klein bottle, $K$, obtained by identifying the edges of the square $I \times I$ according to $(0, t) \sim (1, t)$ and $(t, 0) \sim (1-t, 1)$ for $t \in [0, 1]$. $K$ is certainly a quotient space, but it is seemingly not of the nature of the above examples, mainly because I don't know a quotient map $f$ (as used above), that makes the same identifications as $\sim$.

What stops me from simply defining $K$ (or any quotient space like $\mathbb{S}^2$ above) as the quotient space generated by the relation instead of having to do the laborious calculations to find a quotient map $f$ that makes the same identifications as the relation?

$\endgroup$
  • 5
    $\begingroup$ I don't really understand the question. Are you having trouble visualizing the gluing? Have you ever seen someone roll up a dumpling from a circle of dough? It's like that. If you see a semicircle then you're either only looking at the unit circle as opposed to the closed ball, or gluing the ball as opposed to only its boundary. $\endgroup$ – Qiaochu Yuan Sep 26 '17 at 3:46
  • 1
    $\begingroup$ Notice this typographical difference: $$X/\sim$$ $$\text{versus} $$ $$X/{\sim}$$ The latter is the right way to do it. When you write $5+3$ there's a certain amount of space to the left and right of the plus sign that you don't see in $5{+}3,$ and the same applies to $A\sim B$ as opposed to $A{\sim}B.$ But in an expression like $X/{\sim}$ that extra space becomes inappropriate, so do it that way. (I edited the question accordingly.) $\qquad$ $\endgroup$ – Michael Hardy Sep 26 '17 at 3:48
1
$\begingroup$

For the $S^2 \cong D^2/\partial D^2$, you can start by noting that $D^1/\partial D^1 \cong S^1$, which is easier to see: you take a strand and glue the edges together. The higher dimensional analogue works the same way, but you take the entire boundary and bring it to a point, kind of like a knapsack.

As for your question about these "laborious maps," I think there is a fundamental misunderstanding about the quotient space. One should first think of it as an equivalence relation on a set, $(X, \sim)$, and a canonical map $x \mapsto [x]$, that sends an element to its equivalent class. This map can always be defined set theoretically. The Quotient space is a space $X/{\sim}$ along with a map $\phi:X \to X/{\sim}$ that is continuous. To make it continuous, we endow $X/{\sim}$ with the coarsest topology that accomplishes this goal. Hence, one should think of it as a topological space and a map that makes all of this happen.

The property of "coarsest" can be summarized by saying that whenever you have a map $f:X \to Y$ so that $x \sim y \implies f(x)=f(y)$, there exists a map $$\tilde{f}: X/{\sim} \longrightarrow Y$$ so that $\tilde{f} \circ \phi=f$.

The construction of a "laborious map" usually involves finding some $f:X \to Y$ so that $\tilde{f}$ is a homeomorphism, which would show that the quotient is homeomorphic to $Y$, or that some topological space can be identified with a quotient of $X$.

Here is an example: $D^1/\partial D^1 \cong S^1$

Consider the surjective map $f: [0,2\pi] \to S^1$ given by $x \mapsto e^{ix}$. Clearly, the identification $x \sim y \iff x,y \in \partial S^1$ implies that $f(x)=f(y)$, so $f$ factors through the quotient space, inducing $\tilde{f}$. However, this is the only place where $f$ was noninjective, so the induced map is injective. However, $\tilde{f}$ is now a continuous bijection whose domain is compact and codomain Hausdorff, and so it is a homeomorphism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.