Manifold #15. The Cylinder as a Quotient Space매니폴드 #15. 몫공간으로 만든 원기둥

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers Lesson 15: The cylinder as a quotient space. In the playlist this lecture sits at the very end, but in content it follows the previous post directly. This time we run the quotient machinery properly for the first time and prove that gluing two sides of a square gives a cylinder.

What do we want to show?

Roll a sheet of paper and join two opposite edges, and you get a cylinder. The goal is to translate this picture into the language of topology. The paper is the square $[0,1] \times [0,1]$ with the subspace topology from the plane. We write the gluing as an equivalence relation.

So the point $(0, b)$ on the left side is tied to the point $(1, b)$ on the right side at the same height, and every other point is tied only to itself. Call the quotient space $X = [0,1]^2 / \!\sim$ and the canonical projection $p : [0,1]^2 \to X$. This $X$ is a set of sets, so it is not yet a cylinder. The goal is that $X$ is homeomorphic to the cylinder.

We define the cylinder inside three-dimensional space.

It carries the subspace topology of $\R^3$. Then we can define a map $\varphi : [0,1]^2 \to C$ by $\varphi(a, b) = (\cos 2\pi a, \sin 2\pi a, b)$. Roll it yourself. The toy below curls the paper into the shape of this map.

Push the Glue slider all the way to the right and the paper becomes a cylinder, and the distance between the marked points $(0, b)$ and $(1, b)$ drops to 0. That is $\varphi(0, b) = \varphi(1, b)$. This picture is still not a proof. The picture is the plan, and the three steps below are the proof.

Q. A rolled-up sheet is already a cylinder, so why is a separate proof needed?
A. Rolling only brings the two edges close together, and the topology of the quotient space is fixed separately by the definition. We have to check that the open sets of the quotient space match the open sets of the cylinder exactly before we can say they have “the same shape”.

Step 1: build a one-to-one correspondence

$\varphi$ is a continuous surjection. Let us see where it takes equal values. If $\varphi(a, b) = \varphi(a', b')$, the heights agree, so $b = b'$, and $\cos 2\pi a = \cos 2\pi a'$ and $\sin 2\pi a = \sin 2\pi a'$ force $a - a'$ to be an integer. Since $a, a' \in [0, 1]$, either $a = a'$ or $\{a, a'\} = \{0, 1\}$. So the groups of points that $\varphi$ sends to the same value are exactly the equivalence classes above.

Therefore $\varphi$ gives one value per class and different values on different classes. This gives a one-to-one correspondence

Note that the value does not depend on which representative of the class we pick. In the toy above, the two points merging into one is what it means for this map to be well defined.

Step 2: show it is continuous

To check that a map out of a quotient space is continuous, we use the following property.

Let $p : A \to Y$ be a quotient map and $f : A \to Z$ a continuous map that takes the same value on each equivalence class of $p$. Then there is exactly one map $\tilde{f} : Y \to Z$ with $f = \tilde{f} \circ p$, and this $\tilde{f}$ is continuous.

The proof is short. Take an open set $W$ of $Z$. Then $p^{-1}(\tilde{f}^{-1}(W)) = f^{-1}(W)$, and the right side is open because $f$ is continuous. By the definition of the quotient topology, $\tilde{f}^{-1}(W)$ is open. This is sometimes called the universal property. Apply it with $A = [0,1]^2$ and $f = \varphi$, and we get at once that $\tilde{\varphi}$ is continuous.

Step 3: show the inverse is continuous

A continuous one-to-one correspondence is not automatically a homeomorphism. We have to check separately that the inverse is continuous. Here the following theorem helps.

A continuous one-to-one correspondence from a compact space to a Hausdorff space is a homeomorphism.

To use this theorem we check two things.

• $X$ is compact. $X = p([0,1]^2)$, the square is compact, and $p$ is continuous, so this follows from the fact that the continuous image of a compact set is compact.
• $C$ is Hausdorff. It is a subspace of $\R^3$.

Let us also recall why the theorem holds. A closed set $F$ of $X$ is a closed subset of a compact space, hence compact. Its image $\tilde{\varphi}(F)$ under the continuous map is compact, and a compact set in the Hausdorff space $C$ is closed. So $\tilde{\varphi}$ is a closed map, and since it is a one-to-one correspondence, the preimage under the inverse, which is the image under $\tilde{\varphi}$, of a closed set is closed, so the inverse is continuous.

Putting the three steps together, $\tilde{\varphi} : X \to C$ is a homeomorphism. The quotient space from gluing two sides of a square is homeomorphic to the cylinder.

Q. Why do we need the compact condition and the Hausdorff condition, one each?
A. Compactness keeps closed sets of the domain compact, and the Hausdorff property makes compact sets closed in the target. Only together do closed sets go to closed sets, and that makes the inverse continuous. Without compactness there is a counterexample. The map $t \mapsto (\cos 2\pi t, \sin 2\pi t)$ from the interval $[0, 1)$ to the circle is a continuous one-to-one correspondence that is not a homeomorphism, because the domain is not compact. (This counterexample is added for understanding and was not covered in the lecture.)

What this proof shows

Back to the first question. The sentence “roll the paper and join two edges to get a cylinder” is now exact in one line. The quotient space $[0,1]^2 / \!\sim$ is homeomorphic to the cylinder $C$ in $\R^3$. Homeomorphic spaces share every topological property, such as connectedness and compactness. The skeleton of the proof is this.

1. Take the groups of points that the gluing map $\varphi$ sends to the same value as the equivalence classes.
2. Get continuity of $\tilde{\varphi}$ from the universal property.
3. Get continuity of the inverse from compactness and the Hausdorff property.

These three steps repeat almost unchanged when we build other shapes.

As an extra remark, this cylinder is the closed cylinder with round rims at both ends. Points near a rim have neighborhoods that look like a half plane, so it belongs to the manifolds with boundary. We return to this when the definition of a manifold arrives. (This paragraph is also added and was not covered in the lecture.)

In the next lecture we flip the direction of the gluing to make the Möbius strip and see how far this proof carries over unchanged.