Direct limit and amalgamation (Serre's "Trees") At the very beginning of Serre's Trees, it's taken that the groups $G_i$ (indexed over some set $I$, with no additional specifications) are equipped with homomorphisms $f_{ij}:G_i\to G_j$, collected in $F_{ij}$ defined for all $i,j\in I$. From this, he constructs the direct limit $G$ with maps $f_i:G_i\to G$, satisfying $f_j\circ f_{ij}=f_i$ and a universal property. Then, he gives the example $I=[2]$, with the only specifications on the groups $G_1$ and $G_2$ being the existence of homomorphisms $f_i:A\to G_i$ for some other group $A$, and claims that $G$ in this setting is the amalgamated product $G_1*_AG_2$.
My question is exactly what implicit information about $F_{12}$ and $F_{21}$ is given in that example. Here is my best guess. Say $G_1$ is generated by (fixed) $S$, and $T\subset A$ has the set bijection $\phi:T\overset{\sim}{\to}S$ where $\phi=f_1|_T$. Then $f_T=f_2\circ\phi^{-1}$, which extends to all of $G_1$, and $F_{12}=\{f_T:T\subset A,f_1|_T:T\overset{\sim}{\to}S\}$. $F_{21}$ arises analogously. I think this is equivalent to the notion of amalgamation that I've seen elsewhere of handling $f_1(x)f_2(x)^{-1}$, but wanted to be certain by phrasing it explicitly in the notation Serre introduces for the direct limit. Thanks in advance.
 A: Okay, just looked at the book. So for each pair of indices $i$ and $j$; we have a family $F_{ij}$ of morphisms. Far be it for me to correct Serre, so let me instead point out that most people would not call this construction a "direct limit", but rather a colimit. "Direct limit" is usually restricted for the case where $I$ is a directed set (partially ordered set in which every finite subset has an upper bound), and where for each $i$ and $j$, if $i\nleq j$ then $F_{ij}$ is empty; if $i\leq j$ then $F_{ij}=\{f_{ij}\}$ is a singleton; $f_{ii}=\mathrm{Id}_{G_i}$; and the morphisms are requied to satisfy $f_{ik} = f_{jk}\circ f_{ij}$ whenever $i\leq j\leq k$.
I don't think you are reporting the example correctly: the index set is not "$[2]$" as you report (by which I suspect you mean $\{1,2\}$). Here is what he says, explicitly:

Example. Take three groups, $A$, $G_1$, and $G_2$, and two homomorphisms, $f_1\colon A\to G_1$ and $f_2\colon A\to G_2$. One says that the corresponding group $G$ obtained by amalgating $A$ in $G_1$ and $G_2$ by means of $f_1$ and $f_2$, we denote it by $G_1*_A G_2$. One can have $G=\{1\}$ even though $f_1$ and $f_2$ are non-trivial.

So, here your index set has three elements: $1$, $2$, and then a silent index used for $A$. I would denote it by $G_0=A$. Then you have $F_{01}=\{f_1\colon A\to G_1\}$; $F_{02}=\{f_2\colon A\to G_2\}$; and $F_{10}=F_{20}=F_{12}=F_{21}=\varnothing$. In other words, you entire system consists of exactly three groups and exactly two maps, and nothing else. Then you take the colimit/direct limit of this system, and you get the free product of $G_1$ and $G_2$ amalgamated over $A$.
Note that $f_1$ and $f_2$ are not assumed to be embeddings; that's why the amalgamated product could be trivial.
Your $S$ and $T$ are misguided. If $G_1$ were generated by $S$ and you had a bijection from a subset of $A$ onto $S$ with the restriction of a homomorphism, then the homomorphism would be onto.
The "classic" situation is when $f_1$ and $f_2$ are embeddings, so that $A$ can be identified with a "common" subgroup of $G_1$ and $G_2$; in that case, $G_1*_A G_2$ is defined as $G_1*G_2/N$, where $G_1*G_2$ is the free product, and $N$ is the smallest normal subgroup of $G_1*G_2$ containing all elements of the form $f_1(a)f_2(a)^{-1}$; it is then necessary to prove that this results in a group that contains isomorphic copies of $G_1$ and $G_2$, which intersect precisely "at" $A$. In the more general situation here, where $f_1$ and $f_2$ are not required to be monomorphisms/injective, this is what is usually called a pushout of $f_1$ and $f_2$; that is, you have a diagram
$$\require{AMScd}
\begin{CD}
A@>f_1>>G_1\\
@Vf_2VV\\
G_2
\end{CD}$$
and then you seek a universal group $G$ and maps $g_1,g_2$ that fits into the diagram as follows:
$$\begin{CD}
A@>f_1>>G_1\\
@Vf_2VV@VVg_1V\\
G_2@>g_2>>G
\end{CD}$$
so that $g_1f_1=g_2f_2$ and satisfies the relevant universal property relative to that condition.
