Consider the graph of groups having one vertex, \(\mathbb{Z}\), and one edge, the map \(\mathbb{Z} \to \mathbb{Z}\) that is multiplication by 2, realized by the 2-sheeted covering space \(S^1 \to S^1\). Show that \(\pi_1(K\Gamma)\) has presentation \(\langle a, b \mid bab^{-1}a^{-2} \rangle\) and describe the universal cover of \(K\) explicitly as a product \(T \times \mathbb{R}\) with \(T\) a tree. [The group \(\pi_1(K\Gamma)\) is the first in a family of groups called Baumslag–Solitar groups, having presentations of the form \(\langle a, b \mid ba^mb^{-1}a^{-n} \rangle\). These are HNN extensions \(\mathbb{Z} *_\mathbb{Z}\).]
Solution
Let the vertex group be \[ A=\langle a\rangle\cong\mathbb Z \] and let the edge group be another copy \[ C=\langle c\rangle\cong\mathbb Z. \] For the loop edge, take the two endpoint injections \[ \iota_0(c)=a, \qquad \iota_1(c)=a^2. \]
1. The graph-of-groups fundamental group has presentation\[ \pi_1(K\Gamma) \cong \langle a,b\mid bab^{-1}=a^2\rangle. \]
Choose the unique vertex as base vertex and let \(b\) be the stable letter corresponding to the loop edge. The HNN-extension presentation identifies the two images of the edge group by \[ b\iota_0(c)b^{-1}=\iota_1(c). \] Substituting the two endpoint maps gives \[ bab^{-1}=a^2, \] which is equivalent to the relator \[ bab^{-1}a^{-2}. \]
2. The universal cover of the graph of spaces is a tree of spaces whose vertex spaces are copies of \(\mathbb R\) and whose edge spaces are strips \(\mathbb R\times I\).
The universal cover of the vertex circle \(K(A,1)=S^1\) is \(\mathbb R\). The universal cover of the edge circle \(K(C,1)=S^1\) is also \(\mathbb R\), so the lifted mapping cylinder for an edge is a strip \[ \mathbb R\times I. \] The incidence graph of these lifted vertex and edge spaces is the Bass–Serre tree \(T\) of the HNN extension. It is a tree because the full graph-of-spaces cover is simply connected and the usual normal-form theorem rules out cycles in the incidence graph.
3. The tree \(T\) is trivalent.
At one end of an edge, the edge subgroup maps onto the full vertex group, so the index is \[ [A:\iota_0(C)]=1. \] At the other end its image is \(2\mathbb Z\), so \[ [A:\iota_1(C)]=2. \] Thus at each vertex there is one incident edge of the first type and two of the second type, for total valence three.
4. Each boundary attachment of a lifted edge strip to a lifted vertex line is a homeomorphism \(\mathbb R\to\mathbb R\).
The identity endpoint map lifts to an affine map \[ t\longmapsto t+c. \] The degree-two map \(S^1\to S^1\) lifts to an affine map \[ t\longmapsto 2t+c. \] Both are homeomorphisms of the universal covering line.
5. The universal cover is homeomorphic to\[ \boxed{T\times\mathbb R}. \]
Choose an affine coordinate on one vertex line. Since \(T\) is a tree, proceed outward from this vertex. Whenever an edge strip joins a vertex line whose coordinate has already been chosen to a new vertex line, choose the affine coordinate on the new line so that both boundary identifications of the strip become the identity in the \(\mathbb R\) coordinate. There is no compatibility obstruction because a tree has no cycles.
After making these choices, every vertex space is \[ \{v\}\times\mathbb R \] and every edge strip is \[ e\times\mathbb R, \] with the obvious product gluings. Their union is exactly \(T\times\mathbb R\). Since \(T\) and \(\mathbb R\) are contractible, this space is simply connected and is therefore the universal cover.
6. Thus\[ \boxed{\pi_1(K\Gamma)=\langle a,b\mid bab^{-1}a^{-2}\rangle} \]and its universal cover is the product of a trivalent tree with a line.
Combine 1 and 5.