Show that every graph product of trivial groups is free.
Solution
Let \(\Gamma\) be the underlying graph of the graph of groups. Assume every vertex group and every edge group is trivial.
1. Every classifying space used for a vertex or edge group is a point.
A \(K(1,1)\) is a point. Thus each vertex space in the graph-of-spaces construction is a point, and each edge mapping cylinder is just an interval joining the two endpoint vertex spaces, or a loop interval when the edge is a loop.
2. Consequently the graph-of-spaces realization \(K\Gamma\) is naturally homeomorphic to the underlying graph \(\Gamma\).
Replacing every vertex group by a point and every edge group by a point leaves exactly one \(0\)-cell for each vertex of \(\Gamma\) and one \(1\)-cell for each edge, attached according to the incidence data of \(\Gamma\). This is precisely the geometric realization of the graph.
3. The fundamental group of each connected component of a graph is free.
Choose a maximal tree \(T\) in a connected component. Collapsing \(T\) to a point is a homotopy equivalence and leaves a wedge of one circle for each edge not in \(T\). Hence the fundamental group is a free group on these remaining edges.
4. Therefore every graph product of trivial groups is free.
By 2 its fundamental group is the fundamental group of the underlying graph, and 3 shows this group is free.