Topology Explained – Tying the Knot: Making Circles from Lines

Imagine, if you will, that you have a piece of string. Here’s a visualization

Stringy

Now imagine that you can “stick” the two ends of the string together, say using some chewed gum or something or other; you’ll end up with something like the following:

Trident Original Flavor; surprisingly good and underrated.

Now with a little bit of straightening out and cleaning up, we get a circle:

Yep, that’s a circle alright. Trust me, I’m a mathematician.

And there you have it; we’ve just completed one of the myriad ways that mathematicians build circles. So what’s the big hullabaloo about then? Well, there’s always a catch. What we just did is what is often referred to as intuitively obvious, and this is a very dangerous phrase indeed. Of course if you take a piece of string and tie the ends together you end up with a circle; this is something that you can very easily test yourself and something that I personally end up doing accidentally every morning while flossing. But we never proved this. Moreover, while it is universally accepted that this works out mathematically, the construction of the circle from a line is fairly obfuscating, and it is not immediately apparent that the space we build is actually the same as a circle. And indeed, the proof that the space one builds is actually the same as the circle is deceptively intricate and lengthy, even though most topologists dismiss it as being intuitively obvious. So let’s take a while and walk through the proof itself, and show that this intuitively obvious fact is anything but.


Quotients

If you’re already familiar with quotient spaces and maps, feel free to skip this section.

Definition: Let X be a topological space, Y be a set, and X\xrightarrow{f}Y a surjective set map. The quotient topology on Y induced by f (or simply quotient topology if context makes this clear) is the finest topology on Y such that f is continuous.

In essence, the quotient topology is the “smallest” topology on a set Y which makes a given map continuous. This is a good intuitive description, but let’s give something a little bit more concrete.

Proposition: Let X be a space, Y a set, and X\xrightarrow{f}Y a surjective set map. The quotient topology on Y induced by f can be characterized in the following way: a subset V\subseteq Y is open in Y if and only if f^{-1}(V) is open in X.

Proof: Take any open V\subseteq Y; by virtue of the fact that the quotient topology makes f continuous, this implies that f^{-1}(V) is open in X. Conversely, suppose that the quotient topology on Y is given by \mathcal{T}. Moreover, suppose for some V\subseteq Y we have that f^{-1}(V) is open in X, but V is not open in Y. Then let \mathcal{T}' denote the collection of subsets of Y that you can obtain via unioning any collection of sets in \mathcal{T} with V or amongst themselves, and by intersecting any finite collection of sets of \mathcal{T} with V or amongst themselves. Then the claim is that \mathcal{T}' is also a topology on Y. I’ll leave the excruciating verification of this to you. However, note that certainly \mathcal{T}\subseteq\mathcal{T}' as any element of \mathcal{T} can be written as the union with itself. Thus \mathcal{T}' is a finer topology on Y than the quotient topology; if we can show that f is still continuous when equipping Y with \mathcal{T}' then we will have reached a contradiction as we are assuming that the quotient topology is the finest topology on Y with this property. Then take any U\in\mathcal{T}'; if U\in\mathcal{T} then there’s nothing to show about f^{-1}(U) as we already know it’s open by definition of the quotient topology. Thus we really need only consider two cases: when U=\left(\bigcup_{\alpha\in I}U_\alpha\right)\cup V where U_\alpha\in\mathcal{T} for each \alpha\in I, and U=\left(\bigcap_{i=1}^nU_i\right)\cap V where U_i\in\mathcal{T} for each i, as these are the only other possibilities for elements of \mathcal{T}'

Then suppose that U=\left(\bigcup_{\alpha\in I}U_\alpha\right)\cup V where U_\alpha\in\mathcal{T} for each \alpha\in I. By definition, we then get that

    \[f^{-1}\left(U\right)=f^{-1}\left(\bigcup_{\alpha\in I}U_\alpha\right)\cup V\right)=\left(\bigcup_{\alpha\in I}f^{-1}(U_\alpha)\right)\cup f^{-1}(V)\]

By our assumptions, however, each f^{-1}(U_\alpha) is open in X by definition of the quotient topology, and f^{-1}(V) is open in X by our assumption on V. Thus f^{-1}(U) is open in X as the union of open sets.

Otherwise, suppose that U=\left(\bigcap_{i=1}^nU_i\right)\cap V where U_i\in\mathcal{T} for each i. Then

    \[f^{-1}(U)=f^{-1}\left(U=\left(\bigcap_{i=1}^nU_i\right)\cap V\right)=\left(\bigcap_{i=1}^nf^{-1}(U_i)\right)\cap f^{-1}(V)\]

But again, by similar reasoning each f^{-1}(U_i) and f^{-1}(V) are open, so f^{-1}(U) is then open as the intersection of finitely many open sets.

Thus, to recap our argument, we’ve shown that for any U\in\mathcal{T}', f^{-1}(U) is open in X, hence f is continuous when Y is equipped with the topology {\mathcal{T}'. However, \mathcal{T}' is finer than that quotient topology, \mathcal{T}, which is the finest topology on Y such that f is continuous, a contradiction. Thus no such V\subseteq Y where f^{-1}(V) is open in X but V is not open in Y can exist. As such, if f^{-1}(V) is open in X, we must have that V is open in Y, completing the proof.

\varnothing

Let’s talk about the most common way quotient topologies come up. Suppose that we have a space X, and an equivalence relation \sim on X. Let X/\sim denote the set of the equivalence classes of \sim, and let X\xrightarrow{\pi}X/\sim be the “canonical” surjection sending an element x to its equivalence class [x], i.e. \pi(x)=[x]. Then we can equip X/\sim with the quotient topology induced by \pi; it turns out that this is most often the “natural” topology to put on the space X/\sim. This comes up most often in two contexts: “gluing” points together, and “collapsing” regions. Creating the circle from an interval is an example of the former; you “glue” the ends of the interval together to form the circle. For the latter, you often take a connected region in a space X, say A, and collapse it to a single point to form the quotient X/A, where you can think about this notation as saying we are “modding out” the region A and “reducing” it to a single point. I may eventually do a post or two about this construction, likely in the context of the ambiguity behind the notation \mathbb{R}/\mathbb{Z}.

In any event, there are a few more results we should talk about with regards to quotients. 

Definition: Let X,Y be spaces. Given a surjective continuous map X\xrightarrow{f}Y, we say that f is a quotient map if for any V\subseteq Y, f^{-1}(V) being open in X implies that V is open in Y. In light of the preceding proposition, we see that f is then a quotient map if and only if Y has the quotient topology induced by f.

Proposition: Let X\xrightarrow{f}Y be a quotient map. Let Z be a space, and let Y\xrightarrow{g}Z be a set map, so that we are in the following situation:

Mmm, Commutative Diagrams

Then the map g is continuous if and only if the composition g\circ f is continuous.

Proof: Suppose first that g is continuous. Then g\circ f is also continuous, as the composition of continuous maps. Conversely, suppose that g\circ f is continuous. Take any W\subseteq Z open; as g\circ f is continuous, this implies that \big(g\circ f\big)^{-1}(W) is open in X. However,

    \[\big(g\circ f\big)^{-1}(W)=f^{-1}\big(g^{-1}(W)\big)\]

As such, by our work on the quotient topology, if f^{-1}\big(g^{-1}(W)\big) is open in X, we know that g^{-1}(W) is open in Y. As W\subseteq Z was arbitrary, this implies that g is continuous.

\varnothing


Setup and Construction

Let’s take a moment to establish some notation and facts. First, we’ll set the unit interval to be I=[0,1]. Second, the unit circle will be \mathbb{S}^1=\left\{(x,y)\in\mathbb{R}^2:\sqrt{x^2+y^2}=1\right\}.

Next we want to go through the process of “building” the circle via identifying the endpoints of I. To do this, we make use of the quotient topology. Define a relation \sim on I by

    \[x\sim y\Longleftrightarrow \begin{array}{c} x=y \\ \text{or} \\ x=0 \text{ and }y=1 \\ \text{or} \\ x=1\text{ and }y=0\end{array}\]

I will leave it to you to verify that this is an equivalence relation on I with the following properties: for any x\in I with 0<x<1, x is only related to itself, and the only other relation is that 0\sim 1. Thus for any x\in I with 0<x<1 we have that [x]=\{x\}, and [0]=\{0,1\}=[1].

Now let C=I/\sim, and equip C with the quotient topology induced by the “canonical” surjection I\xrightarrow{\pi}C. The claim is that we are now done; the space C is simply the circle, \mathbb{S}^1. Now hopefully you see my issue with topologists; we have constructed this vague space C and are simply claiming that, yes, indeed C is indistinguishable from the unit circle \mathbb{S}^1 itself. As an undergraduate, this bothered me quite a bit as the elements of C are not even ordered pairs, so how could this be \mathbb{S}^1? Too often professors would reply that, “ah of course C is not \mathbb{S}^1 on the nose, it is simply homeomorphic to \mathbb{S}^1.” When pressed to show that these two spaces are homeomorphic, most just fell back on the argument that, hey, it’s obvious that they’re the same space, so clearly they’re homeomorphic. Hopefully now you’re starting to see some of my frustration. In any case, let’s show that C is homeomorphic to \mathbb{S}^1.


C\cong\mathbb{S}^1

This one is a real doozy, and part of the reason I hate that people generally state that this result is “obvious”. The proof of this requires multiple parts, which we prove at first, and then bring it all together in one final proof. The fundamental result we need is the following theorem:

Theorem: Let X,Y be spaces and X\xrightarrow{f}Y a continuous bijection. If X is compact, and Y is Hausdorff, then f is a homeomorphism.

Proof: It suffices to prove that f is an open map; an open bijection has a continuous inverse, hence as f is already continuous this will imply that f is a homeomorphism. Then take any U\subseteq X open. As U is open, U^C is closed; however every closed subspace of a compact space is also compact, so as X is compact, U^C is compact. The image of a compact space under a continuous map is compact, hence f\big(U^C\big) is a compact subspace of Y, as we assume that f is continuous. But Y is Hausdorff, and every compact subspace of a Hausdorff space is closed, thus f\big(U^C\big) is a closed subspace of Y. Note that f\big(U^C\big)=\big(f(U)\big)^C, so \big(f(U)\big)^C is closed in Y. By definition, this implies that f(U) is open in Y. As U\subseteq X was arbitrary, this implies that f is an open map and hence that f is a homeomorphism.

\varnothing

With this in hand, our overall structure/goal is obvious: show that C is compact, \mathbb{S}^1 is Hausdorff, and then build a continuous bijection C\xrightarrow{f}\mathbb{S}^1.

Lemma: \mathbb{S}^1 is Hausdorff.

Proof: This follows immediately as any subspace of a Hausdorff space is also Hausdorff (prove this yourself if you find yourself asking why), \mathbb{S}^1\subseteq\mathbb{R}^2, and \mathbb{R}^2 is Hausdorff (as you can always take sufficiently small open balls about two distinct points).

\varnothing

Lemma: C is compact.

Proof: Recall our quotient map: I\xrightarrow{\pi}C; this is a continuous surjection, so that \pi(I)=C. However, \pi is continuous, and I=[0,1] is compact, so \pi(I)=C is also compact as the image of a compact space under a continuous map.

\varnothing

With this all done, we simply need to build a continuous bijection C\xrightarrow{f}\mathbb{S}^1. We construct the map here, then prove it has the desired properties. First, define a map C\xrightarrow{\phi}[0,2\pi) by

    \[\phi\big([x]\big)=\left\{\begin{array}{cl} 2\pi x & : 0<x<1 \\ 0 & :x=0,1\end{array}\right.\]

Note that this is well-defined, for if 0<x<1 then [x]=\{x\} so there’s only one choice of representative, and likewise \phi\big([0]\big)=0=\phi\big([1]\big). Notice that this is also a bijection; it is clearly surjective by construction, for if we take any y\in[0,2\pi, then \left[\frac{y}{2\pi}\right] maps to y under \phi. Moreover, it is injective. For suppose that [x],[y]\in C are such that \phi\big([x]\big)=\phi\big([y]\big). There are two cases to consider: \phi\big([x]\big)=0 and \phi\big([x]\big)\ne0. Suppose the latter first; this then implies that 0<x,y<1 and 2\pi x=2\pi y, so that x=y and hence [x]=[y]. In the former, this implies that x=0,1 and y=0,1; but regardless of the possible combinations we see that this implies x\sim y so that [x]=[y]. Thus in either case we see that if \phi\big([x]\big)=\phi\big([y]\big), then [x]=[y] so that \phi is injective, and hence bijective.

Next, define a map [0,2\pi)\xrightarrow{\psi}\mathbb{S}^1 by \psi(\theta)=\big(\cos(\theta),\sin(\theta)\big). Then this is again a bijection, the verification of which I leave to you (it’s really just algebra involving \sin‘s and \cos‘s, but it’s annoying algebra so you do it).

Composing \psi and \phi, we thus get a bijection C\xrightarrow{\psi\circ\phi}\mathbb{S}^1, call it \gamma=\psi\circ\phi. To finish the proof, it will suffice to prove that \gamma is continuous. To do this, recall from our work on quotients that it will suffice to prove that the composition I\xrightarrow{\gamma\circ\pi}\mathbb{S}^1 is continuous. However, the claim is that the composition \gamma\circ\pi is given by

    \[\big(\gamma\circ\pi\big)(x)=\big(\cos(2\pi x),\sin(2\pi x)\big)\]

For x\in I with 0\le x<1, this is immediate if you track down all of our definitions. However, \sin,\cos are 2\pi periodic so even though we formally have that

    \[\big(\gamma\circ\pi\big)(1)=\gamma\big([1]\big)=\psi\Big(\phi\big([1]\big)\Big)=\psi(0)=\big(\cos(0),\sin(0)\big)\]

we in turn get that

    \[\big(\gamma\circ\pi\big)(1)=\big(\cos(0),\sin(0)\big)=\big(\cos(2\pi\cdot 1),\sin(2\pi\cdot 1)\big)\]

as needed. However, \gamma\circ\pi is then continuous, as is is simply the restriction of the map I\to\mathbb{R}^2 given by x\longmapsto\big(\cos(2\pi x),\sin(2\pi x)\big) to \mathbb{S}^1, which is continuous as its component functions, \cos(2\pi x),\sin(2\pi x) are both continuous.

Thus as \gamma\circ\pi is continuous, we then know that \gamma is continuous; as such we have found a continuous bijection C\xrightarrow{\gamma}\mathbb{S}^1. As C is compact and \mathbb{S}^1 is Hausdorff, this implies by the first theorem that \gamma is in fact a homeomorphism, and hence that C\cong\mathbb{S}^1.


Conclusion

And there you have it folks, a circle. This process underlines my fundamental displeasure with topology; so much of what is deemed an “intuitive” proof is omitted, and further is anything but “intuitive”. At the same time, arguing like this is necessary in topology; should every topologist be asked to argue every result as I did above, we would probably be decades, if not centuries, behind where we are now. Just think about it; it took us probably four or five pages worth of work, presupposing a good basic knowledge of point-set topology, to simply argue that tying the ends of a string together gives you a circle. Think about how much work it would take then to prove something like the Hairy Ball Theorem. And this is why hyper-specialization in math is not only good, but necessary; there’s no way in Hell I would ever research topology, so let’s be glad it’s some other poor bastards’ jobs to do it.

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