May's A Concise Course in Algebraic Topology gives the following statement of the Seifert-van Kampen theorem for fundamental groupoids $\Pi(X)$ (section 2.7):

Theorem (van Kampen). Let $\mathcal{O} = \{ U \}$ be a cover of a space $X$ by path connected open subsets such that the intersection of finitely many subsets in $\mathcal{O}$ is again in $\mathcal{O}$. Regard $\mathcal{O}$ as a category whose morphisms are the inclusions of subsets and observe that the functor $\Pi$, restricted to the spaces and maps in $\mathcal{O}$, gives a diagram $\Pi | \mathcal{O} : \mathcal{O} \to \text{Gpd}$ of groupoids. The groupoid $\Pi(X)$ is the colimit of this diagram.

As far as I can tell, however, his proof makes no use of the hypothesis that $\mathcal{O}$ consists of path connected subsets. Am I correct in thinking this? The motivation here is to be able to compute e.g. the fundamental group of the circle $S^1$ by writing it as the union of two intervals with disconnected intersection.

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    $\begingroup$ Do we need this rather so we can revert back to the fundamental group from the colimit, i.e. to guarantee that each $\Pi(U)$ has skeleton $\pi_1(U,x)$? $\endgroup$
    – Andrew
    Sep 18, 2012 at 5:12
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    $\begingroup$ This seems likely – May immediately goes on to talk about how to recover the fundamental group version of van Kampen. As far as I can tell the proof of the cited theorem does not make use of path-connectedness. On the other hand, computing the colimit of a diagram of possibly-disconnected groupoids is not fun! $\endgroup$
    – Zhen Lin
    Sep 18, 2012 at 5:16
  • $\begingroup$ I agree with Andrew. Have you checked out Ronnie Brown - "Groupoids and Van Kampen's theorem" (1967)? (I'm sure he'll eventually be along to answer anyhow!) $\endgroup$
    – Juan S
    Sep 18, 2012 at 6:33
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    $\begingroup$ But it's precisely because they're not compatible with equivalences that $\pi_1(S^1)$ doesn't just collapse to the trivial group! (There's some homotopy-theoretic nonsense about fibrant/cofibrant replacements that can be invoked here to make things work, but I won't pretend to even understand what is going on.) $\endgroup$
    – Zhen Lin
    Sep 18, 2012 at 7:15
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    $\begingroup$ @Qiaochu Yuan: For the cover with 2 open subsets, that result was in the 1967 paper referred to by Juan S. See also my answer below. $\endgroup$ Sep 18, 2012 at 16:33

1 Answer 1


Higgins' downloadable book Categories and groupoids has quite a lot on computing colimits of groupoids. The point is that the groupoid van Kampen theorem has the probably optimal theorem of this type in

  • R. Brown and A. Razak, A van Kampen theorem for unions of non-connected spaces, Archiv. Math. 42 (1984) 85-88. pdf

This involves the fundamental groupoid $\pi_1(X,A)$ on a set of base points, and for this and a general open cover, one needs that $A$ meets each path-component of each $1$-, $2$-, $3$-fold intersection of sets of the cover. This answers a particular point in the question. The case $A=X$ is the theorem as stated in the question, and the special case when $A$ is a singleton is in most texts. The reduction to $3$-fold intersections essentially relies on the idea of Lebesgue covering dimension.

This result translates the problem from topology into algebra; a particular fundamental group, if one wants it, is kind of hidden in the middle of this colimit of groupoids. One then has to do various combinatorial things such as choosing trees in components of graphs, to find the fundamental group. These methods are directly related to methods of use in combinatorial group theory, so one should think of them, including the notion of covering morphism of groupoids used in Higgins' book, as a form of combinatorial groupoid theory. HNN extensions of groups can also be seen as pushouts of groupoids.

One of the useful tools in Higgins' book is the following: given a groupoid $G$ with object set $X$ and a function $f:X \to Y$ construct a groupoid $U_f(G)$ with object set $Y$. This construction yields free groups, and free products of groups, as special cases. In Chapter 9 of Topology and Groupoids this construction is related to making identifications on a discrete subset of a topological space. For example, one might want to form the circle $S^1$ by identifying $0,1$ in the unit interval $[0,1]$.

These ideas usefully generalise to higher dimensions, via Higher Homotopy Seifert-van Kampen Theorems: see for example Part I of Nonabelian Algebraic Topology for results on second relative homotopy groups. There is more to be said ...

Later: I realise I did not answer the question as to the purpose of the generalisation. As suggested, the immediate purpose was to have a theorem which yielded the fundamental group of the circle, which is, after all, THE basic example in algebraic topology. It also gave easily additional examples: for example, let $X$ be the space formed by identifying all corresponding points of two copies of the interval $[-1,1]$ except for the point $0$. Thus $X$ is a non Hausdorff space. From the groupoid version with $A$ consisting of the two points $\pm 1/2$, we obtain that the fundamental group of $X$ is the integers.

Grothendieck in Section 2 of his 1984 "Esquisse d'un Programme" emphasises that choosing a single base point will often destroy any symmetry in the situation. Consider the following union of five open sets:


One is in a "Goldilock's" situation. Choosing all points as base points is too large for comfort. Choosing one base point is too small. But choosing eight base points is just about right!

Situations like this arise in combinatorial group theory. In general, one chooses the set of base points according to the geometry of the given situation.

The proof given in the paper referred to is by verification of the universal property, and does not require knowledge that the category of groupoids admits colimits, nor any specific method of constructing them.

For me, this work led to the impression that ALL of $1$-dimensional homotopy theory was better expressed in terms of groupoids rather than groups.

The proof also has the advantage of generalising to higher dimensions, once one has the appropriate higher dimensional homotopical gadgets. (It did take me 9 years to find, in conversation with Philip Higgins, the right $2$-dimensional gadget.)

July 28, 2015: For further discussion on this area, see this mathoverflow discussion.

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    $\begingroup$ There is further comment to be made on my answer. The result proved by May does not require any further assumption, since it deals with $\pi_1(X)=\pi_1(X,A)$ with $A=X$. However to get a specific result, one usually has to reduce $A$ to a subset appropriate to the geometry, and then the proof depends on making choices of paths to base points lying in certain intersections. The easiest proof does I think go back to the version by Crowell, and that proof has the advantage of generalising to higher dimensions. $\endgroup$ Sep 19, 2012 at 9:23

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