Van Kampen counterexamples

Question

The van Kampen theorem states that if I have a space $X = \cup_\alpha A_\alpha$ where $A_\alpha$ are all path connected and if for any given $\alpha, \beta, \gamma $ we get $A_\alpha \cap A_\beta \cap A_\gamma$ is also path connected then the kernel of $\Phi:*\Pi_1(A_\alpha) \to \Pi_1(X)$ is generated by loops in the intersection of two to $A$s and not cancelling with its inverse. That is to say objects of the following form: suppose we have some loop $\gamma$ in $A_\alpha \cap A_\beta$ then it shows up in $\Pi_1(A_\alpha)$ as $\gamma_\alpha$. It also shows up in $\Pi_1(A_\beta)$ as $\gamma_\beta$ and the inverse of this loop is in $\Pi_1(A_\beta)$ as $\gamma^{-1}_\beta$.

The statement of the theorem is then that $\ker\Phi$ is generated by objects of the form $\gamma_\alpha \gamma^{-1}_\beta$

My question is if I relax $A_\alpha \cap A_\beta \cap A_\gamma$ being path connected and just demand that any pair $A_\alpha \cap A_\beta$ be path connected, what else can I get in the kernel?

Answer 1

A counterexample to the conjecture is given on p.44 (in my edition) of Hatcher's book on algebraic topology. You really do need the 3-fold intersection condition.

However there is a method of dealing with this which I published in 1967, namely to use many base points and so use the fundamental groupoid $\pi_1(B,S)$ on a set $S$ chosen according to the geometry of the situation: there is discussion of this idea in this mathoverflow entry. The idea was thought of in order to get a theorem which would compute at the same time also the fundamental group of the circle, a rather important example in topology, as well as a myriad of other cases. Why not?

The most general theorem of this type seems to be the following, from this paper, of 1984. Notice that for convenience we write also $\pi_1(U,S)$ for $\pi_1(B, U \cap S)$.

Theorem Let $(B_\lambda)_{\lambda\in \Lambda}$ be a family of subspaces of $B$ such that the interiors of the sets $B_\lambda$ $(\lambda\in \Lambda)$ cover $B$, and let $S$ be a subset of $B$. Suppose $S$ meets each path-component of each one-fold, two-fold, and each three-fold intersection of distinct members of the family $(B_\lambda)_{\lambda\in \Lambda}$. Then there is a coequalizer diagram in the category of groupoids: $$\bigsqcup_{\lambda,\mu\in\Lambda}\pi_1(U_\lambda\cap U_\mu, S)\rightrightarrows^\alpha_\beta \bigsqcup _{\lambda\in \Lambda}\pi_1(U_ \lambda,S)\to^\gamma\pi_1(B,S) $$ in which $\bigsqcup$ stands for the coproduct (disjoint union) in the category of groupoids, and $\alpha$, $\beta$, and $\gamma$ are determined by the inclusion maps $U_\lambda\cap U_\mu\to U_\lambda$, $U_\lambda\cap U_\mu\to U_\mu$, and $U_\lambda\to B$, respectively.

The proof involves verifying the universal property of a coequaliser, and so does not involve knowing how to construct coequalisers of groupoids. However this is discussed in the 1971 book by Philip Higgins "Categories and Groupoids" referred to in the mathoverflow discussion. I do not attempt here to find the "groupoid error term" in the case the 3-fold condition is not satisfied.

Grothendieck has supported the idea of "many base points" in his 1984 "Esquisse d'un programme" Section 2. Part of his comment is, in translation, "... people still obstinately persist, when calculating with fundamental groups, in fixing a single base point,,, ".