Graph of groups

Remarks

Up to isomorphism, $\pi_1(\mathbb{G}, T)$ does not depend on the choice of $T$, and each $G_v$ embeds in it.

Example: $Y$ has two vertices $u, v$ and one pair of edges $e, \bar e$ from $u$ to $v$, with $G_u = A$, $G_v = B$ and $G_e = C$. Then $T = Y$, all $t_e$ are trivial, and the relations identify $\alpha_{\bar e}(c) \in A$ with $\alpha_e(c) \in B$; so $\pi_1(\mathbb{G}, T)$ is the amalgamated free product $A *_C B$.

Example: $Y$ has one vertex $v$ and a loop $e$, with $G_v = A$ and $G_e = C$. Then $T = \set{v}$, and $\pi_1(\mathbb{G}, T)$ is generated by $A$ and $t = t_e$ with $t \, \alpha_e(c) \, t^{-1} = \alpha_{\bar e}(c)$; this is the HNN extension of $A$ along the isomorphism $\alpha_e(C) \to \alpha_{\bar e}(C)$.

Non-example: if some map $\alpha_e$ is not injective, clause 4 fails, and the data is not a graph of groups; the vertex groups need not embed in the resulting quotient.

When all vertex and edge groups are trivial, $\pi_1(\mathbb{G}, T)$ is the free group on the edges of $Y$ outside $T$, one generator for each pair $e, \bar e$. This is the fundamental group of $Y$ as a topological graph.