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

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:

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

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 be a topological space, be a set, and a surjective set map. The **quotient topology on induced by ** (or simply **quotient topology** if context makes this clear) is the finest topology on such that is continuous.

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

**Proposition**: Let be a space, a set, and a surjective set map. The quotient topology on induced by can be characterized in the following way: a subset is open in if and only if is open in .

*Proof*: Take any open ; by virtue of the fact that the quotient topology makes continuous, this implies that is open in . Conversely, suppose that the quotient topology on is given by . Moreover, suppose for some we have that is open in , but is not open in . Then let denote the collection of subsets of that you can obtain via unioning any collection of sets in with or amongst themselves, and by intersecting any finite collection of sets of with or amongst themselves. Then the claim is that is *also* a topology on . I’ll leave the excruciating verification of this to you. However, note that certainly as any element of can be written as the union with itself. Thus is a *finer* topology on than the quotient topology; if we can show that is still continuous when equipping with then we will have reached a contradiction as we are assuming that the quotient topology is the finest topology on with this property. Then take any ; if then there’s nothing to show about as we already know it’s open by definition of the quotient topology. Thus we really need only consider two cases: when where for each , and where for each , as these are the only other possibilities for elements of

Then suppose that where for each . By definition, we then get that

By our assumptions, however, each is open in by definition of the quotient topology, and is open in by our assumption on . Thus is open in as the union of open sets.

Otherwise, suppose that where for each . Then

But again, by similar reasoning each and are open, so is then open as the intersection of finitely many open sets.

Thus, to recap our argument, we’ve shown that for any , is open in , hence is continuous when is equipped with the topology . However, is finer than that quotient topology, , which is the finest topology on such that is continuous, a contradiction. Thus no such where is open in but is not open in can exist. As such, if is open in , we must have that is open in , completing the proof.

Let’s talk about the most common way quotient topologies come up. Suppose that we have a space , and an equivalence relation on . Let denote the set of the equivalence classes of , and let be the “canonical” surjection sending an element to its equivalence class , i.e. . Then we can equip with the quotient topology induced by ; it turns out that this is most often the “natural” topology to put on the space . 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 , say , and collapse it to a single point to form the quotient , where you can think about this notation as saying we are “modding out” the region 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 .

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

**Definition**: Let be spaces. Given a surjective continuous map , we say that is a **quotient map** if for any , being open in implies that is open in . In light of the preceding proposition, we see that is then a quotient map if and only if has the quotient topology induced by .

**Proposition:** Let be a quotient map. Let be a space, and let be a set map, so that we are in the following situation:

Then the map is continuous if and only if the composition is continuous.

*Proof*: Suppose first that is continuous. Then is also continuous, as the composition of continuous maps. Conversely, suppose that is continuous. Take any open; as is continuous, this implies that is open in . However,

As such, by our work on the quotient topology, if is open in , we know that is open in . As was arbitrary, this implies that is continuous.

## Setup and Construction

Let’s take a moment to establish some notation and facts. First, we’ll set the **unit interval** to be . Second, the **unit circle** will be .

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

I will leave it to you to verify that this is an equivalence relation on with the following properties: for any with , is only related to itself, and the only other relation is that . Thus for any with we have that , and .

Now let , and equip with the quotient topology induced by the “canonical” surjection . The claim is that we are now done; the space is simply the circle, . Now hopefully you see my issue with topologists; we have constructed this vague space and are simply claiming that, yes, indeed is indistinguishable from the unit circle itself. As an undergraduate, this bothered me quite a bit as the elements of are not even * ordered pairs*, so how could this be ? Too often professors would reply that, “ah of course is not *on the nose*, it is simply homeomorphic to .” 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 is homeomorphic to .

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 be spaces and a continuous bijection. If is compact, and is Hausdorff, then is a homeomorphism.

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

With this in hand, our overall structure/goal is obvious: show that is compact, is Hausdorff, and then build a continuous bijection .

**Lemma**: 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), , and is Hausdorff (as you can always take sufficiently small open balls about two distinct points).

**Lemma**: is compact.

*Proof*: Recall our quotient map: ; this is a continuous surjection, so that . However, is continuous, and is compact, so is also compact as the image of a compact space under a continuous map.

With this all done, we simply need to build a continuous bijection . We construct the map here, then prove it has the desired properties. First, define a map by

Note that this is well-defined, for if then so there’s only one choice of representative, and likewise . Notice that this is also a bijection; it is clearly surjective by construction, for if we take any , then maps to under . Moreover, it is injective. For suppose that are such that . There are two cases to consider: and . Suppose the latter first; this then implies that and , so that and hence . In the former, this implies that and ; but regardless of the possible combinations we see that this implies so that . Thus in either case we see that if , then so that is injective, and hence bijective.

Next, define a map by . Then this is again a bijection, the verification of which I leave to you (it’s really just algebra involving ‘s and ‘s, but it’s annoying algebra so you do it).

Composing and , we thus get a bijection , call it . To finish the proof, it will suffice to prove that is continuous. To do this, recall from our work on quotients that it will suffice to prove that the composition is continuous. However, the claim is that the composition is given by

For with , this is immediate if you track down all of our definitions. However, are periodic so even though we *formally* have that

we in turn get that

as needed. However, is then continuous, as is is simply the restriction of the map given by to , which is continuous as its component functions, are both continuous.

Thus as is continuous, we then know that is continuous; as such we have found a continuous bijection . As is compact and is Hausdorff, this implies by the first theorem that is in fact a homeomorphism, and hence that .

## 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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