# Brouwer's Fixed Point Theorem (Proof)

Today I'd like to talk about Brouwer's Fixed Point Theorem. *Literally!* It's the subject of this week's episode on PBS Infinite Series. Brouwer's Fixed Point Theorem is a result from topology that says no matter how you stretch, twist, morph, or deform a disc (so long as you don't tear it), there's always one point that ends up in its original location.

**Brouwer's Fixed Point Theorem**

Every continuous function from a disk to itself has a fixed point.

As you can see in the video, I chose to focus on a *proof *of the theorem, rather than elaborating on its meaning or its applications*.* The mathematics behind the proof are just as fascinating as the theorem itself! This proof uses a "portal" between topology and algebra -- a fancier version of the interplay between geometry and algebra that one learns in high school mathematics. In today's blog post, I'd like to present the proof again and fill in some details that I left out of the video. In particular, I'd like to be more explicit about the "portal," which as I mention, is really a **functor!**

*What IS a functor, exactly? *It's very much like a bridge between different realms (or categories) of mathematics.

*I've written an introduction to functors here. Incidentally, Example #2 in that link is*

*precisely*the functor featured in the video! It's called

**the fundamental group**and is denoted $\pi_1\colon \mathsf{Top}\to\mathsf{Group}$. As I explain there, $\pi_1$ assigns a group $\pi_1(X)$ to each topological space $X$. The elements of $\pi_1(X)$ are really (homotopy classes of) maps of the circle into $X$. For now, it's enough to think of $\pi_1(X)$ as a "hole-detector," which keeps track of loops in $X$. You can find much more detail in a book on algebraic topology (such as Allen Hatcher's or Peter May's) or this friendly primer on Jeremy Kun's blog. We'll only need to consider when $X$ is a disk $D^2$ or a circle $S^1$. In that case

$\pi_1(D^2)\cong 0$ and $\pi_1(S^1)\cong \mathbb{Z}.$

Intuitively, $\pi_1(D^2)\cong 0$ because every loop in a disc can be wiggled down to a point. *Topologically, there are no fundamental loops in a disc.* On the other hand, $\pi_1(S^1)\cong \mathbb{Z}$ because, well, a circle IS just a loop, which we can further loop around *itself* zero, one, two, three,... times clockwise (the positive integers) or counterclockwise (the negative integers). For a rigorous proof of this isormorphism, check out the books referred to above or take a look at this mini-series on Math3ma. You can also find the proof in this tweet, which was later expanded into this blog post.

In addition to assigning groups to topological spaces, $\pi_1$ also preserves the relationships between them. In other words, it assigns a group homomorphism $\pi_1(f)\colon \pi_1(X)\to \pi_1(Y)$ to every continuous function $f\colon X\to Y$ with the properties that
$$\pi_1(f\circ g)=\pi_1(f)\circ\pi_1(g) \qquad\text{and}\qquad \pi_1(\text{id}_{X})=\text{id}_{\pi_1(X)}.$$
In other words, *functors preserve compositions* and *sends the identity map of $X$ to the identity map of its fundamental group.* This is known as *functoriality* and is the “key feature” alluded to in the video when I said

“Any scenario that we can construct between a circle and a disc, must correspond to an identical scenario between the integers and the number 0. In short, whatever happens in the land of topology should be mirrored in the land of algebra.”

This interplay between topology and algebra is heart of this proof of Brouwer's Fixed Point Theorem, which we are now ready to state.

## The Proof

If Brouwer's Fixed Point Theorem is not true, then there is a continuous function $g\colon D^2\to D^2$ so that $x\neq g(x)$ for all $x\in D^2$. This allows us to construct a function $h$ from $D^2$ to its boundary $S^1$ by drawing a ray from $g(x)$ to $x$. This ray intersects with a point, which we can label $h(x)$, on the boundary circle of the disc. Th assignment $x\mapsto h(x)$ is continuous: a small change in $x$ corresponds to a small change in $h(x)$. Moreover, $h(x)=x$ for every $x\in S^1$.

Let $i\colon S^1\to D^2$ denote the *inclusion* of the circle into the disc. That is, $i(x)=x$ for all $x\in S^1$. Then $h(i(x))=x$ for all $x\in S^1$ and so
$$h\circ i = \text{id}_{S^1}.$$

Applying $\pi_1$, we obtain a composition of group homomorphisms $$\pi_1(h)\circ\pi_1(i)=\text{id}_{\mathbb{Z}}.$$

(Here is where we're using functoriality: $\pi_1(h\circ i) = \pi_1(h)\circ\pi_1(i)$ and $\pi_1(\text{id}_{S^1}) = \text{id}_{\pi_1(S^1)}=\text{id}_\mathbb{Z}$.) Notice that $\pi_1(i)\colon \mathbb{Z}\to 0$ is the constant map at 0; it assigns every integer to 0. On the other hand, $\pi_1(h)$ is injective: it assigns 0 to exactly one integer (0, in fact). YET their composition must assign each integer to itself! That is, the composition $\pi_1(h)\circ\pi_1(i)$ must be *equal* to $\text{id}_\mathbb{Z}$, the identity on $\mathbb{Z}$.
But this is impossible!

Conclusion? The function $h$ cannot exist, and so there must be at least *one* point $x\in D^2$ so that $g(x)=x$. In other words, Brouwer's Fixed Point Theorem is true.

## The Takeaway

The takeaway is that

**we cannot map the set of all integers to zero and then reverse the process.**

We don't even need group theory to see this. It follows from the definition of a function! And thanks to the correspondence between topology and algebra, namely:

*#1) a disc is the topologist's version of the number zero*, and

*#2) the boundary of a disk (i.e. a circle) is the topologist's version of the set of integers,*

we can obtain an analogous, topological statement:

**we cannot map the boundary of a disc to the disc and then reverse the process,**

at least, *not without tearing the disc* (i.e. without violating continuity). If however Brouwer's Fixed Point Theorem is *not true*, then we *CAN *perform such a reversal using the map, *h*.

*That *is the contradiction.

And *this (*as my advisor likes to say) is algebraic topology *par excellence.*