Can one prove that the fundamental group of the circle is $\mathbb Z$ without using covering spaces?

Question

I am curious if there is a decent "bare hands" proof that the fundamental group of $S^1$ is $\mathbb Z$ that does not invoke covering space theory.

One must show two claims. First, that $f(t)=e^{2\pi i t}$ generates the fundamental group. Second, that no power of $f(t)$ is nullhomotopic.

The first claim is not hard to do by hand. It follows from an exercise in chapter 1.1 of Hatcher's textbook, and is also shown in the last chapter of Ahlfors's complex analysis textbook.

The second claim is the hard part. In fact, Ahlfors proves this too. We observe that $$\int_{f^m} \frac{dz}{z} = 2\pi i m,$$ which is zero if and only if $m=0$. Here we use the fact that path integrals are homotopy invariant.

I was wondering if a more direct approach to this second claim exists that does not involve complex analysis. I could not find anything in the texts I have available. May's notes give the covering space proof, and tom Dieck's book gives what looks like a more sophisticated version of the covering space proof using groupoids. Rotman also uses a complex analysis approach.

Answer 1

A version of the Seifert-van Kampen theorem for not necessarily connected spaces was published by me in 1967, see here, and it uses the fundamental groupoid $\pi_1(X,C)$ on a set $C$ of base points. The usual connectivity condition is replaced for a pushout determined by $U \cup V$ by the condition that $C$ meets each path component of $U,V, U \cap V$. You can ask if this method is "bare hands"!

In a sense, the thesis proposed in 1968, maybe not explicitly, is that all of $1$-dimensional homotopy theory is better modelled by groupoids rather than groups.

The book Topology and Groupoids gives a full account of this theory; it is the 2006 3rd edition of a book published in 1968, 1988. See also this mathoverflow discussion, and this reprinted 1971 book Categories and Groupoids by Philip Higgins. A feature of these books is the groupoid construction $U_f(G)$ with object set $Y$ from a groupoid $G$ and a set function $f: Ob(G) \to Y$. This construction includes free groups and free groupoids, and free products of groups. I'll take this opportunity to advertise a small correction to the proof of the Jordan Curve Theorem in T&G: this proof applies groupoid algebra to unions of non-connected spaces.

Alexander Grothendieck wrote in a letter to me in 1983:

" . .... both the choice of a base point, and the $0$-connectedness assumption, however innocuous they may seem at first sight, seem to me of a very essential nature. To make an analogy, it would be just impossible to work at ease with algebraic varieties, say, if sticking from the outset (as had been customary for a long time) to varieties which are supposed to be connected. Fixing one point, in this respect (which wouldn't have occurred in the context of algebraic geometry) looks still worse, as far as limiting elbow-freedom goes!"

[Here is a link to further post 1970 correspondence of Alexander Grothendieck.]

June 10: I should have said from the start that my work on generalisations of the Seifert-van Kampen Theorem for the fundamental group was motivated by the desire to obtain a result which would also yield the fundamental group of the circle. I first extended a method of nonabelian cohomology due to Olum, and this was published in 1965, see here: it gave the fundamental group of a wedge of circles. To my surprise, I then found that the direct method using the fundamental groupoid on a set of base points was both simpler to prove and was more powerful. The cited 1967 paper on that also refers to a paper of Weinzweig, which uses nerves of covers.

The following diagram is another way of looking at the fundamental group of the circle.

Here the groupoid $\mathsf I$ has two objects $0,1$ and one arrow $\iota: 0 \to 1$, and is easily shown to be isomorphic to $\pi_1([0,1],\{0,1\})$.

Nov 17, 2015: I leave the reader to work out how to modify the second diagram if you replace $\{0,1\}$ by $\{0,1,2\}$, $[0,1]$ by $[0,2]$, $S^1$ by $S^1 \vee S^1$, ..... and ....

February 16, 2017 A recent account of the background to and developments of this work is in this article.

April, 2020 See also my answer to this mathoverflow question on using more than one base point for the Van Kampen Theorem. For the circle, it seems like a Goldilocks situation: one base point is too small; the whole fundamentaal groupoid is too big; but two base points are just right!

More generally, what is wrong with a connected space which is the union of $50$ open sets the intersection of any two of which has at least $200$ pathcomponents? The algebraic theory of groupoids is a natural extension of the theory of groups, available for over half a century.

Answer 2

There is a purely topological definition of winding number in the plane that involves neither covering spaces nor complex analysis. Let $H_0$, $H_1$, $H_2$, an $H_3$ be the half-planes $x>0$, $y>0$, $x<0$, and $y<0$, respectively.

Given a map $\gamma\colon [0,1]\to \mathbb{R}^2-\{0\}$ with $\gamma(0) = \gamma(1) = (1,0)$, a mesh for $\gamma$ is a pair of sequences $t_0=0<t_1 < \cdots < t_k=1$ and $n_1,\ldots,n_k \in \{0,1,2,3\}$ such that $\gamma\bigl([t_{i-1},t_i]\bigr) \subset H_{n_i}$ for all $i$. Clearly any closed curve $\gamma$ has at least one mesh.

Given a mesh for $\gamma$, the winding number of $\gamma$ is defined by $$ w(\gamma) \;=\; \sum_{i=1}^k \Delta(n_{i-1},n_i)\qquad\text{where}\qquad \Delta(m,n) = \begin{cases} -1 & \text{if }n-m\equiv -1\;(\text{mod }4), \\ 0 & \text{if }n-m\equiv 0\;(\text{mod }4), \\ 1 & \text{if }n-m\equiv 1\;(\text{mod }4). \end{cases} $$ It is possible to show that $w(\gamma)$ does not depend on the mesh chosen, using the fact that any two meshes have a common refinement.

One can use this to prove combinatorially that any closed curve in $\mathbb{R}-\{0\}$ is homotopic to $e^{2\pi int}$ for some value of $n$. This involves using homotopy to eliminate any stuttering in the sequence $n_1,\ldots,n_k$, i.e. subsequences of the form $m,n,m$ for $n\ne m$, and removing points $t_i$ to eliminate repeated pairs $n,n$. Eventually, one gets a mesh without any stuttering or repetition, which will have the same sequence $\{n_i\}$ as a mesh for some $e^{2\pi int}$. But any two curves having meshes with the same sequence $\{n_i\}$ are homotopic.

Next, one can prove that any two homotopic curves have the same winding number. Specifically, suppose that $\Phi\colon [0,1]\times[0,1]\to \mathbb{R}^2-\{0\}$ is a homotopy of curves, with $\Phi(0,t) = \Phi(1,t) = (1,0)$ for all $t\in[0,1]$. If we let $1/n$ be a Lebesgue number for the covering $\bigl\{\Phi^{-1}(H_0),\Phi^{-1}(H_1),\Phi^{-1}(H_2),\Phi^{-1}(H_3)\bigr\}$, then each of the $\frac{1}{n}\times \frac{1}{n}$ subsquares of the $n\times n$ grid in the unit square maps to a single $H_i$. Then one can give some fairly simple combinatorial arguments that the winding number is the same for each row of squares, and therefore winding number is a homotopy invariant.

Answer 3

Using category theory one can show if a functor preserves finite products and terminal objects then it sends group objects to group objects. Fundamental group is such a functor from category of topological spaces to category of groups.

A group object of category of topological spaces is a topological group. A group object of groups is an abelian group. So basically this says, fundamental group of a topological group is abelian. Since circle is a topological group its fundamental group is abelian.

Also one can show first homology of a topological space is abelianization of its fundamental group. (In 2.A of Hatcher's book you can see proof of this fact). So basically fundamental group of circle is same as its first homology.

Now using cellular homology it is quite easy to show that the first homology of circle is integers, hence its fundamental group is integers.

As far as I know any of the above facts does not use covering spaces.

Note: After I wrote this answer I realized one user posted same answer in the comments. I just wanted to keep it since it has a bit more details.