Computing `$\pi_1 S^1$` using groupoids

Question

I believe it is possible to compute $\pi_1 S^1$ by applying the groupoid version of the Seifert-Van Kampen Theorem (in the version presented in May's Concise Course) to a covering of the circle by three arcs. Is there an account like this somewhere in the literature? Ideally I'd like a discussion that a student familiar with May's book would be able to read. (May doesn't take a 2-categorical approach to groupoids, and so he does not discuss the fact that a diagram of groupoids that is a point-wise equivalence induces an equivalence of colimits. This is rather important for computations.)

Edit: this last statement is false in general! I was thinking of homotopy colimits. The relevant (correct) fact appears in Ronnie Brown's book: retracts of pushouts are pushouts. This is the means by which one compares the Van Kampen theorem for the full fundamental groupoid - as in May's book - with the Van Kampen theorem for the fundamental groupoid on a set of basepoints.)

Answer 1

I can only point to the place where this was originally done (or rather, the latest edition thereof):

Topology and Groupoids by Ronnie Brown

It's a fantastic textbook and easy to read (and cheap, if you buy the electronic copy - the best £5 I've spent). Ideally what you'd do is calculate the equivalent subgroupoid $\Pi_1(S^1,\{a,b,c\})$ where $a,b,c$ are three points in $S^1$, one in each intersection of opens.

Answer 2

A rather belated comment on these! I like the comparison between the circle $S^1$ as obtained from the unit interval $[0,1]$ by identifying $0$ and $1$ in the category of spaces, and the group of integers $\mathbb Z$ as obtained from the groupoid $\mathcal I$, which has objects $0$ and $1$ and exactly one arrow $\iota:0 \to 1$, by identifying $0$ and $1$ in the category of groupoids.

I got hold of the idea in the 1960s from writing the first edition of this book that all of 1-dimensional homotopy theory was better expressed in terms of groupoids rather than groups This led to the question: are groupoids useful in higher homotopy theory? Is the 1-dimensional case a ``one-off''? or not?

I liked the more exciting prospect, but it took 9 years to get with Philip Higgins in 1974 a good definition in dimension 2, namely the homotopy double groupoid of a pair of spaces, and a 2-dimensional van Kampen theorem.

June 2, 2018 More background to this area is in the paper published this year:

Modelling and Computing Homotopy Types:I

One of the points made is that the usual base point approach seems to assume all interesting spaces are path connected; yet the circle is a counterexample since it is nicely represented as a union of non connected spaces. The algebra of groupoids nicely copes with the more general case; one main point of my 1967 paper on the groupoid version of the van Kampen theorem was to obtain by such a method the fundamental group of the circle!

Note also that if the theorem is proved by verification of the universal property, then the proof for $\pi_1(X,C)$ is no more difficult than the proof for $\pi_1(X,c)$; the case of many sets is given in this paper.

Answer 3

The reference to Brown is probably the best one for the moment. Unfortunately, May's book doesn't seem to be so useful because, even in the theorem about groupoids, there is a connectedness assumption on the interesections of the open sets covering the space. Moreover, once one has a general pushout theorem for groupoids, one needs to know how to compute the isotropy groups of the given groupoid, which May doesn't explain.

André Gramain has written a short account of that, Le théorème de van Kampen.

For a good space, the theory of coverings gives an equivalence of category between coverings and sets with an action of the fundamental group. (This determines the group.) This gives various way of computing the fundamental group(oid) of a space via descent theory. In the setting of schemes, Grothendieck gives the relevant theorems and formulae in a few lines in SGA 1.

I can phrase Denis-Charles's answer above in a slightly more elementary way, using the formulation via coverings.

The circle $S^1$ is the interval $[0,1]$ with endpoints attached; therefore, a covering of $S^1$ can be described as a covering of $[0,1]$ together with an identification of the fibers at $0$ and $1$. We thus have a covering $A\times [0,1]$, with a bijection of $A\times\{0\}$ with $A\times\{1\}$. That is, a set $A$, with a bijection of~$A$. That is, a set $A$ with an action of the group $\mathbf Z$. So $\pi_1(S^1)=\mathbf Z$.