The Bass-Serre tree of a graph of groups

Let $\mathcal{G}$ be a graph of groups on $X$, let $T$ be a maximal subtree, and write $\mathrm{\Gamma }={\pi }_1(\mathcal{G},T)$. By Vertex groups embed in the fundamental group of a graph of groups, regard each vertex group $G_v$ as a subgroup of $\mathrm{\Gamma }$. For an oriented edge $e$, the injective boundary map identifies $G_e$ with a subgroup of $G_{o(e)}$ and hence of $\mathrm{\Gamma }$. The cosets below use these canonical embeddings.

The Bass-Serre tree $\tilde{X}$ has:

• vertices the left cosets $\mathrm{\Gamma }\mathrm{/}{G}_v$ for vertices $v$ of $X$,
• edges the left cosets $\mathrm{\Gamma }\mathrm{/}{G}_e$ for oriented edges $e$ of $X$.

For an oriented edge $e$, define incidence and edge reversal by

$o(\gamma G_e)=\gamma G_{o(e)},t(\gamma G_e)=\gamma e{G}_{t(e)},\overline{\gamma G_e}=\gamma e{G}_{\bar{e}}\mathrm{.}$

These are independent of the chosen representative $\gamma$: replacing $\gamma$ by $\gamma h$ with $h\in G_e$ leaves the origin coset unchanged because $G_e\le G_{o(e)}$, and leaves the terminus coset unchanged because the edge relation identifies $eh$ with $e{\alpha }_{\bar{e}}(h)$ inside $\mathrm{\Gamma }$. The same relation makes the reversal formula independent of the representative, and applying reversal twice gives $\gamma e\bar{e}{G}_e=\gamma G_e$.