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Given a connected graph of groups $\mathcal G$ (where edge maps are embeddings), by a subgraph we mean a graph of groups obtain by omitting some vertices, some edges, and replacing the remaining vertex groups by some subgroups containing the remaining embedded edge groups (which includes the possibility of leaving some as-is).

Suppose $y_0$ is a vertex of $\mathcal H$ and $x_0$ is the corresponding vertex of $\mathcal G$. Then we have a natural homomorphism $\pi_1(\mathcal H,y_0)\to \pi_1(\mathcal G,x_0)$.

What are some natural sufficient conditions on $\mathcal H$ for this to be an embedding? Is there a more general fact that tells us when it happens? I'm also interested in references.

A non-example following a suggestion by Andy Putman in another (now deleted) question of mine is the following: let $G$ be a non-trivial group and consider a graph of groups with three vertices of order $2$ with vertex group $G$, two edge groups $G$ (with isomorphisms as edge map) and one trivial edge group.

Then deleting a vertex corresponding to the $G$ with two isomorphisms yields $G*G$ as the fundamental group, which maps to $G\leq G*\mathbf Z$ in the fundamental group of the whole graph.

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    $\begingroup$ I believe the natural homomorphism is always an embedding. In the proposed counterexample, it is true that $G \ast G$ embeds in $G \ast \bf Z$ (since $G \ast t G t^{-1}$ is a subgroup of $G \ast \langle t \rangle$). I think you can also allow replacing edge groups with subgroups, provided that each edge group in $\mathcal H$ is the intersection of the corresponding edge group in $\mathcal G$ and a vertex group in $\mathcal H$ (separately for each endpoint). $\endgroup$
    – David Wärn
    Commented Mar 27 at 22:05
  • $\begingroup$ @DavidWärn: Yes, I knew that $G*G$ embeds into $G*\mathbf Z$, but is the induced homomorphism the embedding? For some reason, I thought it is not, but maybe you're right. If what you say is true, do you have a reference for this? This seems like something that ought to be written somewhere, if true. $\endgroup$
    – tomasz
    Commented Mar 27 at 22:10
  • $\begingroup$ If you allow to remove an edge without removing its vertices, there are easy counterexamples to injectivity: if the edge is not a self-loop and is label by a nontrivial group $C$, at the level of these two vertices we get the canonical non-injective map $A\ast B\to A\ast_C B$. The general setting where one should get injectivity should be when one considers a subset of vertices and one keeps at least all edges between them that are not self-loops. The most significant case is that of an induced graph, and is covered by HJRW's answer. $\endgroup$
    – YCor
    Commented Mar 28 at 16:13
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    $\begingroup$ @YCor: I’m puzzled by your comment, which seems misleading. The subgraph doesn’t have to be induced, and deleting edges is fine. However, it is true that, if a vertex group $G_v$ is replaced by a subgroup $H_v$, then any remaining incident edge $e$ must be labelled by the full intersection $G_e\cap H_v$. Since the question specifies that $H_v$ should contain $G_e$, this condition js satisfied here. $\endgroup$
    – HJRW
    Commented Mar 29 at 3:55
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    $\begingroup$ @YCor there is a difference between the graphs "two disconnected vertices, labelled by $A$ and $B$" and "two vertices labelled by $A$ and $B$ connected by an edge labelled by the trivial group". The former graph is non-connected and has two fundamental groups: $A$ and $B$, while the latter has one fundamental group $A \ast B$. Only the former is a subgraph of "two vertices connected by an edge labelled by $C$" (because you are not allowed to change edge groups arbitrarily). Indeed we do have embeddings from $A$ and $B$ to $A \ast_C B$. $\endgroup$
    – David Wärn
    Commented Mar 29 at 10:35

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As mentioned in comments, if $\mathcal{H}$ is a subgraph of a graph of groups $\mathcal{G}$, with the natural induced structure, then the map

$H=\pi_1(\mathcal{H})\to G=\pi_1(\mathcal{G})$

induced by inclusion is always injective.

Probably the easiest way to see this is to note that any element of $H$ that is in normal form with respect to $\mathcal{H}$ is also in normal form with respect to $\mathcal{G}$, and so also non-trivial in $G$ - see Serre's book Trees for more on normal forms, which you absolutely need if you want to work with graphs of groups.

The statement that $H$ embeds into $G$ can also be thought of as an instance of Britton's lemma, which says that vertex groups embed - just contract $\mathcal{H}$ to a vertex labelled by $H$.

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