Tree of groups $(\mathcal{G},T)$

Question

Let $G$ be act on $\Gamma$ with a fundamental domain $T$ where $T$ is tree. We construct tree of groups $(\mathcal{G},T)$ with the following structure: $$\text{for every } v\in V(T),\,\,G_v=\operatorname{Stab}_G(v) $$

$$\text{for every } e\in E(T),\,\,G_e=\operatorname{Stab}_G(e) $$

Assume that $G_T$ is the direct limit of the system $(\mathcal{G},T)$. With using the universal property of the definition of the direct limit, we get the map $\phi\colon G_T\mapsto G$.

Now my question is:

If $\Gamma$ is connected, then why can we conclude that $\phi$ is a surjective map?

Answer 1

We cannot conclude that $\phi$ is surjective, because it is not true in general. Here is a counterexample.

• $\Gamma = \mathbb{R}$ with vertex set $\mathbb{Z}$.
• $G = \mathbb{Z}$ acting on $\mathbb{R}$ by translation.
• $T = [0,1]$ is a fundamental domain.

The stabilizer of each vertex $0$ and $1$ of $T$ is trivial, and the stablizer of the unique edge $[0,1]$ in $T$ is trivial. So $G_T$ is the trivial group. There is no map from the trivial group onto $\mathbb{Z}$.

Answer 2

This is precisely the statement of Lemma 4 in Chapter 4.1 of Serre's Trees book. In terms of your specific problem in the lemma is a little too general and you can set $Y=T$. However, the previous Lemma 2 which is being generalized is too specific. This book is a beautiful read and strongly recommended!

Note that this is an answer to your question in terms of Bass-Serre-Theory. That is also why Lee Mosher's answer is incorrect, since in his example there exists no fundamental domain.