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An undirected graph is sometimes defined as a pair of sets $V$ and $E$ (vertices and oriented edges), together with two maps $i,f: E\to V$ (sending a directed edge its initial/final vertex) and a map $s : E \to E$ (reversing the direction of an edge) satisfying $$i \circ s = f$$ $$f \circ s = i$$ $$s \circ s = 1_E$$ This definition makes sense in any category: we can replace sets $V,E$ by objects in the category and maps $i,f,s$ by morphisms in the category. In other words, a graph object in a category $\mathcal{C}$ as a functor $\mathcal{G}\to \mathcal{C}$, where $\mathcal{G}$ is the category with two objects $V,E$ and three nonidentity morphisms $i,f,s$, with composition defined by the identities above.

For example we can apply that definition in the category of groupoids (technically this is really a $2$-category, so we should probably talk about $2$-functors from $\mathcal{G}$, i.e. replace equality of morphisms in the three conditions above by isomorphism of 1-morphisms). Any groupoid can be thought of as a set with a group attached to each element. With this in mind, I believe that if we unpack the definitions we find that a graph object in the category of groupoids is exactly a graph of groups, but without the constraint that the morphisms from edge groups to vertex groups have to be monomorphisms (there are also some subtleties with half-loops, i.e. edges $e\in E$ such that $s(e)=e$, but these are usually excluded in the theory of graphs of groups, at least in the references I know).

Question: Has this point of view been studied before? Does the theory of graphs of groups (in particular the definition of the fundamental group) extend to this slightly more general setting, without the monomorphism condition and with half-loops allowed? If so, does this approach makes some proofs or definitions more elegant?

Secondary question: Is this notion of "graph object" interesting in any other category?

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    $\begingroup$ My feeling is that graphs of groups can be viewed as a kind of homotopy colimit in the category of groups. For instance, the HNN extension corresponds to a kind of mapping cyclinder and the amalgamated free product of a homotopy pushout. The graph of spaces approach to these by Scott and Wall is via homotopy colimits $\endgroup$
    – Benjamin Steinberg
    Commented Sep 24, 2022 at 17:39
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    $\begingroup$ A graph of groups is a diagram of a certain shape in the 2-category of group(oids) and its fundamental group(oid) is given by taking the homotopy colimit. We can of course talk about homotopy colimits of diagrams in a lot of generality. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 24, 2022 at 17:57
  • $\begingroup$ I've never heard of homotopy colimits; that seems interesting. Could you elaborate on how the fundamental groupoid of a graph of groups can be interpreted as a homotopy colimit? $\endgroup$
    – Antoine Labelle
    Commented Sep 24, 2022 at 23:13
  • $\begingroup$ Isn't a graph of groups exactly a graph object in the category of groups with injective maps? I'm not sure why you think you need groupoids... $\endgroup$
    – HJRW
    Commented Oct 5, 2022 at 7:54
  • $\begingroup$ @HJRW No, graph objects in the category of groups would have a group structure on their vertex set and edge set, that's not what graph of groups are. We need groupoids since they can be interpreted (up to equivalence) as sets with a group attached to each element. $\endgroup$
    – Antoine Labelle
    Commented Oct 5, 2022 at 12:58

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Just to give a small answer to your first question: if you drop the monomorphism condition, you quickly run into some trouble. One of the early exercises in Serre's book Trees is to show that a certain direct limit of nontrivial groups is trivial. I forget the exact example, but think something like $C_2 \amalg_{\mathbb{Z}} G$ where $G$ is simple, $\mathbb{Z}$ maps injectively into $G$ and surjectively onto the cyclic group of order two.

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  • $\begingroup$ Why is that particularly problematic? For example, the tensor product of nontrivial rings can sometimes be trivial, but that doesn't mean it's not useful to define and study the tensor product in full generality. $\endgroup$
    – Antoine Labelle
    Commented Nov 29, 2022 at 3:55
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    $\begingroup$ @AntoineLabelle: What you're talking about is arbitrary pushouts of groups. As you say, they can be useful -- eg every quotient is a pushout, and of course we like to study quotients of groups. But the most basic result of the theory of graphs of groups -- specifically, the fact that the vertex groups embed -- fails for arbitrary pushouts. So we need a name for this "stronger" theory, and the name for that is graphs of groups. $\endgroup$
    – HJRW
    Commented Nov 29, 2022 at 11:50
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    $\begingroup$ ... (cont'd). What is true is that every pushout canonically defines a graph of groups -- just replace every vertex and edge group with its image in the pushout. But, as in Rylee's example, the vertex groups might become trivial when you do this. $\endgroup$
    – HJRW
    Commented Nov 29, 2022 at 11:51

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