Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.)

share|improve this question
1  
The book of Tammo Tom Dieck "Algebraic Topology" takes a related but different approach. –  j.c. Oct 29 '10 at 7:09
2  
The isomorphism $\pi_1(S^1)\simeq \mathbf{Z}$ is just a triviality:if we define $S^1$ as the pushout of the diagram of simplicial sets $\Delta^0 \leftarrow \partial\Delta^1\to \Delta^1$, then we see immediately that the set of maps from $S^1$ to the nerve of a groupoid $G$ is simply the set of arrows $x\to y$ in $G$ such that $x=y$. Therefore, the groupoid $\pi_1(S^1)$ is canonically isomorphic to the group $\mathbf{Z}$ (seen as a groupoid with one object), simply because they both share the same universal property. –  Denis-Charles Cisinski Nov 14 '10 at 0:13
1  
Denis-Charles, what exactly do you mean by "the groupoid $\pi_1 S^1$"? Are you referring to a combinatorial definition of the fundamental groupoid of a simplicial set? I'm familiar with Kan's combinatorial description of the homotopy groups of a Kan complex, and I guess you could mean something similar? Most importantly, how easy is it to identify this object with the fundamental groupoid of the geometric realization? Milnor's original proof that Kan's homotopy groups agree with the homotopy groups of the realization used the Van Kampen theorem. –  Dan Ramras Nov 14 '10 at 6:01
    
Using the version of Van Kampen in Brown's book, you can even compute the fundamental group of the circle by using a cover with two intervals: just pick one basepoint in each component of the intersection. –  Omar Antolín-Camarena Nov 14 '10 at 14:43
    
Omar, yes that's right! May's book has a seemingly unnecessary connectedness assumption on the intersections of sets in the cover, which is why I was thinking of 3 arcs. –  Dan Ramras Nov 14 '10 at 18:10

3 Answers 3

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.

share|improve this answer
    
This is what I had in mind. I did look in Chapter 9 of Brown's book (Chap. 9 is free) and didn't see this. (Brown seems to compute $\pi_1 S^1$ in some other way in Corollary 9.1.5; I haven't digested his notation yet, though.) Maybe this argument appears somewhere else? I'll have to get myself the full book, I guess. –  Dan Ramras Oct 29 '10 at 5:50
    
Actually this isn't how he computes it, but it should be possible to do so. –  David Roberts Oct 29 '10 at 5:52
    
I second David's recommendation. –  jd.r Oct 29 '10 at 11:33
    
Chapter 6 is the place to look, not Chapter 9. –  Dan Ramras Nov 13 '10 at 20:59

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 $\pi_1(S^1)=\mathbf Z$.

share|improve this answer
    
May's path connectedness assumption seems to be unnecessary. I don't see anywhere in his proof that path connectedness is used (am I missing something?). The covering space perspective sounds a lot like the way Quillen computed the fundamental group of the Q-construction. –  Dan Ramras Nov 14 '10 at 18:10
    
It is indeed unnecessary. But the computation of the isotropy groups of pushouts is definitely lacking without the assumption... –  ACL Nov 14 '10 at 23:50

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.

share|improve this answer
    
Further comment: although I was attracted by the retraction argument, it turned out that it did not work easily for any family of open sets, and did not work in higher dimensions. So I returned to the original argument of Crowell and this is given in the paper: R.Brown and A. Razak, ``A van Kampen theorem for unions of non-connected spaces'', Archiv. Math. 42 (1984) 85-88, which also gives the best possible connectivity condition, that the set $A$ of base points meets each path component of each $3$-fold intersections of the cover. This is related to the Lebesgue Covering Dimension. –  Ronnie Brown Jan 13 '13 at 22:19
    
I also prefer, contrary to Gramain and other authors, to give a proof by verification of the required universal property, rather than relying on a specific construction of a pushout, or colimit, of groups, or groupoids. One reason is to have a proof which generalises to higher dimensions, given the appropriate gadget, and so obtain new results and methods in homotopy theory. –  Ronnie Brown Nov 8 '13 at 18:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.