# "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.

**Proof:**

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

## The Loop-Suspension Adjunction

*based loop*$\Omega$ and

*reduced suspension*$\Sigma$:

*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

*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

*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:

*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

__Claim__: *A homotopy equivalence between fibrations induces a homotopy equivalence between fibers. *

**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

### Example #2

*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 $\Omega S^1$.

*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

*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 $X$.

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!

*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).