Van Kampen lemma

Statement

A key basic result in the theory is the so-called Van Kampen lemma which states the following:

1. Let $\mathcal D\,$ be a Van Kampen diagram over the presentation (†) with boundary label w which is a word (not necessarily freely reduced) in the alphabet A ∪ A⁻¹. Then w=1 in G.
2. Let w be a freely reduced word in the alphabet A ∪ A⁻¹ such that w=1 in G. Then there exists a reduced Van Kampen diagram $\mathcal D\,$ over the presentation (†) whose boundary label is freely reduced and is equal to w.

First observe that for an element w ∈ F(A) we have w = 1 in G if and only if w belongs to the normal closure of R in F(A) that is, if and only if w can be represented as

$w=u_1s_1u_1^{-1}\cdots u_n s_nu_{n}^{-1} \text{ in } F(A),$ (♠)

where n ≥ 0 and where s_i ∈ R_∗ for i = 1, ..., n.

Part 1 of Van Kampen's lemma is proved by induction on the area of $\mathcal D\,$. The inductive step consists in "peeling" off one of the boundary regions of $\mathcal D\,$ to get a Van Kampen diagram $\mathcal D'\,$ with boundary cycle w and observing that in F(A) we have

$$w=usu^{-1} w',\,$$

where s∈R_∗ is the boundary cycle of the region that was removed to get $\mathcal D'\,$ from $\mathcal D\,$.

The proof of part two of Van Kampen's lemma is more involved. First, it is easy to see that if w is freely reduced and w = 1 in G there exists some Van Kampen diagram $\mathcal D_0\,$ with boundary label w₀ such that w = w₀ in F(A) (after possibly freely reducing w₀). Namely consider a representation of w of the form (♠) above. Then make $\mathcal D_0\,$ to be a wedge of n "lollipops" with "stems" labeled by u_i and with the "candys" (2-cells) labelled by s_i. Then the boundary label of $\mathcal D_0\,$ is a word w₀ such that w = w₀ in F(A). However, it is possible that the word w₀ is not freely reduced. One then starts performing "folding" moves to get a sequence of Van Kampen diagrams $\mathcal D_0, \mathcal D_1, \mathcal D_2,\dots\,$ by making their boundary labels more and more freely reduced and making sure that at each step the boundary label of each diagram in the sequence is equal to w in F(A). The sequence terminates in a finite number of steps with a Van Kampen diagram $\mathcal D_k\,$ whose boundary label is freely reduced and thus equal to w as a word. The diagram $\mathcal D_k\,$ may not be reduced. If that happens, we can remove the reduction pairs from this diagram by a simple surgery operation without affecting the boundary label. Eventually this produces a reduced Van Kampen diagram $\mathcal D\,$ whose boundary cycle is freely reduced and equal to w.