# "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

## Proof:

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

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!**

### 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

## QED

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).

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