"One-Line" Proof: Fundamental Group of the Circle


Once upon a time I wrote a six-part blog series on why the fundamental group of the circle is isomorphic to the integers. (You can read it here, though you may want to grab a cup of coffee first.) Last week, I shared a proof* of the same result. In one line. On Twitter. I also included a fewer-than-140-characters explanation. But the ideas are so cool that I'd like to elaborate a little more.

As you might guess, the tools are more sophisticated than those in the original proof, but they make frequent appearances in both topology and category theory, so I think it's worth a blog post. (Or six. Heh.) To keep the discussion at a reasonable length, I'll have to assume the reader has some familiarity with algebraic topology and basic category theory. But even if some of the words sound foreign, I encourage you to read as much as you can! My hope is that this post will whet your appetite to study further.

So without further ado, I present

Theorem: The fundamental group of the circle is isomorphic to ℤ.



Let's take a closer look at each of the three isomorphisms.

The Loop-Suspension Adjunction

There are two important functors in topology called based loop $\Omega$ and reduced suspension $\Sigma$:
The loop functor $\Omega$ assigns to each pointed space $X$ (that is, a space with a designated basepoint) the space $\Omega X$ of based loops in $X$, i.e. loops that start and end at the basepoint of $X$. On the other hand, $\Sigma$ assigns to each $X$ the (reduced) suspension $\Sigma X$ of $X$. This space is the smash product of $X$ with $S^1$. In general it might not be easy to draw a picture of $\Sigma X$, but when an $n$-dimensional sphere, it turns out that $\Sigma S^n$ is homeomorphic to $S^{n+1}$ for $n\geq 0$. So for $n=1$ the picture is
The loop-suspension adjunction is a handy, categorical result which says that $\Omega$ and $\Sigma$ interact very nicely with each other: up to homotopy, maps out of suspension spaces are the same as maps in to loop spaces. More precisely, for all pointed topological spaces $X$ and $Y$ there is a natural isomorphism
Here I'm using the notation $[A,B]$ to indicate the set of homotopy classes of basepoint-preserving maps from $A\to B$. (Two based maps are homotopic if there is a basepoint-preserving homotopy between them. This is an equivalence relation, and the equivalence classes are given the name homotopy classes.) This, together with the observation that the $n$th homotopy group $\pi_n(X)$ is by definition $[S^n,X]$, yields the following:
And that's the first isomorphism above!**

Remark: The $\Omega-\Sigma$ adjunction is just one example of a general categorial construction. Two functors are said to form an adjunction if they are - very loosely speaking - dual to each other. I had planned to blog about adjunctions after our series on natural transformations but ran out of time! In the mean time, I recommend looking at chapter 4 of Emily Riehl's Category Theory in Context for a nice discussion.

The Homotopy Equivalence

Next, let's say a word about why $\Omega S^1$ and $\mathbb{Z}$ are homotopy equivalent. This equivalence will immediately imply the second isomorphism above since $\pi_0$ (and in fact each $\pi_n$) is a functor, and functors preserve isomorphisms. (That is, $\pi_0$ sends homotopy equivalent spaces to isomorphic sets.***) Now, why are $\Omega S^1$ and $\mathbb{Z}$ homotopy equivalent? It's a consequence of the

Claim: A homotopy equivalence between fibrations induces a homotopy equivalence between fibers.

Eh, that was a mouthful, I know. Let's unwind it. Roughly speaking, a map $p:E\to B$ of topological spaces is called a fibration over $B$ if you can always lift a homotopy in $B$ to a homotopy in $E$, provided the initial "slice" of the homotopy in $B$ has a lift. And the preimage $p^{-1}(b)\subset E$ of a point in $b$ is called the fiber of $b$. So the claim is that if $p:E\to B$ and $p':E'\to B$ are two fibrations over $B$, and if there is a map between them that is a homotopy equivalence (We'd need to properly define what this entails, but it can be done.) then there is a homotopy equivalence between fibers $p^{-1}(b)$ and $p'^{-1}(b)$.

Example #1

The familiar map $\mathbb{R}\to S^1$ that winds $\mathbb{R}$ around the circle by $x\mapsto e^{2\pi ix}$ is a fibration, and the fiber above the basepoint $1\in S^1$ is $\mathbb{Z}$. Incidentally, we proved this in the original six-part series that I mentioned earlier!

Example #2

The based path space $\mathscr{P}S^1$ of the circle gives another example. This is the space of all paths in $S^1$ that start at the basepoint $1\in S^1$. The map $\mathscr{P}S^1\to S^1$ which sends a path to its end point is a fibration. What's the fiber above $1$? By definition, a path is in the fiber if and only if it starts and ends at 1. But that's precisely a loop in $S^1$! So the fiber above 1 is $\Omega S^1$.
These examples give us two fibrations over the circle: $\mathbb{R}\to S^1$ and $\mathscr{P} S^1\to S^1$. And it gets even better. Both $\mathbb{R}$ and $\mathscr{P}S^1$ are contractible and therefore homotopy equivalent! By the claim above, $\Omega S^1$ and $\mathbb{Z}$ must be homotopy equivalent, too. This gives us the second isomorphism above.

Pretty neat, right? If you're interested in the details of the claim and ideas used here, take a look at J. P. May's A Concise Course in Algebraic Topology, chapter 7.5. By the way, there is a dual notion to fibrations called cofibrations. (Roughly: a map is a cofibration if you can extend - rather than lift - homotopies.) And both of these topological maps have abstract, categorical counterparts -- also called (co)fibrations -- which play a central role in model categories.

The Integers are Discrete

The third isomorphism is relatively simple: we just have to think about what $\pi_0(X)$ really is. Recall that $\pi_0(X)=[S^0,X]$ consists of homotopy classes of basepoint-preserving maps $S^0\to X$. But $S^0$ is just two points, say $-1$ and $+1$, and one of them, say $-1$, must map to the basepoint of $X$. So a basepoint-preserving map $S^0\to X$ is really just a choice of a point in $X$. And any two such maps are homotopic when there's a path between the corresponding points! So $\pi_0(X)$ is the set of path components of $X$.
It follows that $\pi_0(\mathbb{Z})\cong\mathbb{Z}$ since there are $\mathbb{Z}$-many path-components in $\mathbb{Z}$. And that's precisely the third isomorphism above!

And with that, we conclude





Okay, okay, I suppose with all the background and justification, this isn't an honest-to-goodness one-line proof. But I still think it's pretty cool! Especially since it calls on some nice constructions in topology and category theory.

Well, as promised in my previous post I'm (supposed to be) taking a small break from blogging to prepare for my oral exam. But I had to come out of hiding to share this with you - I thought it was too good not to!

Until next time!



* I first learned of this proof from my advisor while taking a course in K-theory last semester. I've been meaning to blog about it ever since!

** You might worry that $\pi_0(\Omega S^1)$ is just a set with no extra structure. But it's actually a group! To see this, note that there is a multiplication on $\Omega S^1$ given by loop concatenation. It's not associative, but it is up to homotopy. (So loops spaces are not groups. They are, however, $A_\infty$ spaces.) So in general, sets of the form $[X,\Omega Y]$ are groups. For more, see May's book section 8.2.

*** Yes, sets. Not groups. In general, $\pi_0(X)$ is merely a set (unlike $\pi_n(X)$ for $n\geq 1$ which is always a group). But we're guaranteed that $\pi_0(\mathbb{Z})$ is a group since it's isomorphic to $\pi_0(\Omega S^1)$ (and see the second footnote).



One of my students recently said to me, "I'm not good at math because I'm really slow." Right then and there, she had voiced what is one of many misconceptions that folks have about math.

But friends, speed has nothing to do with one's ability to do mathematics. In particular, being "slow" does not mean you do not have the ability to think about, understand, or enjoy the ideas of math.

Let me tell you....

Last weekend, I spent seven (SEVEN!) hours trying to understand two-and-a-half (TWO-AND-A-HALF!) sentences of a page-long proof. (My brain just couldn't make it to the end of the third sentence.) And this past weekend, it took me two (TWO!) full days (DAYS!) to understand three (THRE--okay, I'll stop shouting) little equations.

But now I finally understand the proof, and now I know what those little equations really mean. Even better, I understand them both thoroughly because of (not in spite of) all those hours that I spent with them.

So, as I wanted to tell my student:

Who cares how long it takes you?

We can never focus on and enjoy the mathematical scenery around us
if we're too busy concerned with racing our neighbors. 

Don't forget, dear friends: 

Speed is not equivalent to mathematical ability.



On a very related note, I am currently preparing for my oral exam*, so free time is a rarity these days. For that reason, I'm going to take a little break from blogging over the next few months. But I do hope to have lots of cool things to write about when my exam is over! In the mean time, I plan to continue sharing math-y things on Facebook, Twitter, and Instagram. See you in a few months!

*This means I'm learning about a specialized topic now, and will soon give an oral presentation (a seminar talk) in front of some faculty members. Afterwards, I'll be asked a bunch of questions about the topic to ensure I'm ready to move on to the dissertation stage of grad school. No pressure, really ;)



Physicist Freeman Dyson once observed that there are two types of mathematicians: birds -- those who fly high, enjoy the big picture, and look for unifying concepts -- and frogs -- those who dwell on the ground, find beauty in the scenery close by, and enjoy the details.

Of course, both vantage points are essential to mathematical progress, and
I often tend to think of myself as more of a bird.
(I'm, uh, bird-brained?)

Well one day I complained to my advisor, John Terilla -- you can see him doing cool math in that video above! -- about one of my classes which at the time felt a bit froggy:


Me: I'm having a hard time enjoying this class. I don't want to do a bunch of detailed computations on these objects. I'd rather spend my time learning category theory. I want to soar high and study the full landscape!

John: Well, yes, the sky is nice. But the ground -- and the ocean -- is nice, too. In fact, the best place to be is where the sky and the ocean meet, where the tips of the waves turn white.

That's the only place where you can go surfing.

That's where you want to be.


Snap, y'all.

I think I've a new respect for the little details.

Let's go surfing!


Group Elements, Categorically

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The answer to that question is the topic for today's post, written by guest-author Arthur Parzygnat.

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A category 𝖢 consists of some data that satisfy certain properties...

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Now rather than give you a list of definitions--which are easy enough to find and may feel a bit unmotivated at first--I thought it would be nice to tell you what category theory is in the grand scheme of (mathematical) things. You see, it's very different than other branches of math....

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Whether they raised their hand during a lecture and gave a "wrong" answer, received a less-than-perfect score on an exam or quiz, or felt completely confused during a lesson, I tried to emphasize that things aren't always as bad as they seem...

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