Brouwer's fixed point theorem using category theory and "a computation involving covering spaces, that $\pi(S^1) = \mathbb Z$"

Question

My latest folly has been to open a book about Category Theory (CT in Context). My scribbles in the following picture reveal my consternation in the highlighted blue line. "a computation involving covering spaces," and the line following. On the next page, the proof concludes,

...which carries the generator $1\in \mathbb Z$ to itself $(0\ne 1)$. This contradiction proves that the retraction $r$ cannot exist, and so $f$ must have a fixed point.

Questions:

a. I don't follow how we arrive that the functor $\pi_1(S^1)=\mathbb Z$ , or that $\pi_1(D^2) = 0$.

b. the composite endomorphism of $\mathbb Z$ must be zero because "it factors through the trivial group." What is the meaning of this sentence?

Answer 1

I think you're missing the point of the book, at least for your question a. The intent is to show how functoriality can be used in a concrete situation, not to provide a complete self-contained proof of the Brouwer theorem. The fact that $\pi_1(S^1)=\mathbb{Z}$ and $\pi_1(D^2)=0$ are explicitly stated as "black box" statements. If you want to see proofs, you can take a look at any book covering fundamental groups, those are probably the first two computations you will encounter (for $D^2$, it follows easily from the fact that a disk is contractible, and for $S^1$ one has to work out a notion of degree of a map $S^1\to S^1$, in some form or another).

Question b is more in line with the spirit of the book. The whole point of the proof is to convince you that if you can construct $f$ with no fixed point, then you can use it to define maps $S^1\to D^2\to S^1$ whose composition is the identity, and then showcase how functoriality is useful in such a situation. Indeed, since you "know" that $\pi_1(D^2)=0$ (or at least you are supposed to accept that fact if you don't know it), then the induced maps on the level of fundamental groups have the form $\mathbb{Z}\to 0\to \mathbb{Z}$, so since there is the trivial group in the middle, the composition must be the zero map. But functoriality also guarantees that the identity map (of topological spaces) is sent to the identity map (of groups), which gives a contradiction.

To summarize my point: this is not intended as a proof which teaches you something about topology, but as a proof which teaches you how functoriality works in practice.

Answer 2

That $\pi_1(S^1) = \mathbb{Z}$ is a standard fact that you will find in basically any topology book. The basic idea is that you can "walk around the loop $n$ times", which defines a group homomorphism $\mathbb{Z} \rightarrow \pi_1(S^1)$. Here walking around $-n$ times means walking around the loop $n$ times in the other direction. Now one shows that this is an isomorphism by using the unique path lifting of covering spaces; here the universal cover $p \colon \mathbb{R} \rightarrow S^1$.

The computation $\pi_1(D^2) = 0$ is also very standard. Here you can for example show that the fundamental group of every star-shaped subset of $\mathbb{R}^n$ is trivial (which is a nice exercise; so really try it).

Both of these results should at least be quite clear intuitively if you feel comfortable with the concept of homotopies.

"Factors through the trivial group":

We have that $r \circ i = \text{id}_{S^1}$, which yields $\pi_1(r) \circ \pi_1(i) = \text{id}_{\pi_1(S^1)}$, which is the identity on the integers as $\pi_1(S^1) = \mathbb{Z}$. But now, since $\pi_1(D^2) = 0$, the map $\pi_1(i)$ can only be the zero homomorphism, so that $\pi_1(r) \circ \pi_1(i)$ also is the zero homomorphism, which is a contradiction (the identity on the integers is not the zero homomorphism).

Also, the book is not trying to be self-contained here, but rather to show the power of functors.