Let $F_n$ be the free group on $n$ generators. Of course, every finitely-presentable group $G$ is a finite colimit of copies of $F_n$, where $n$ is allowed to vary. But is $G$ a finite colimit of copies of $F_2$?

Of course, because $F_{2n}$ is a finite coproduct of copies of $F_2$, we have that any finitely-presentable group $G$ is a finite colimit of finite colimits of copies of $F_2$ -- a "2-fold" finite colimit of copies of $F_2$. But I'm curious about the 1-fold case.

To make the question a bit more concrete, let's unwind what it means to be a finite colimit of copies of $F_2$:

Let $G$ be a group. Then $G$ is a finite colimit of copies of copies of $F_2$ if and only if $G$ admits a presentation of the following description:

There are $2n$ generators coming in pairs $x_1,y_1, \dots x_n, y_n$;

There is a finite set of generating relations, each of the form $w(x_i,y_i)=v(x_j,y_j)$, where $w,v$ are group words and $1 \leq i \leq j \leq n$.

So for example, $x_1y_1^2x_1^{-1} = y_2^{-1}x_2$ is a permissible generating relation (with $i=1,j=2$) but $x_1 x_2 = x_3$ is not a permissible generating relation because only 2 different subscripts are allowed to appear in a permissible generating relation. So my question is:

**Question:** Let $G$ be a finitely-presented group.

Is $G$ a finite colimit of copies of $F_2$?

Equivalently, does $G$ admit a presentation of the above form?

**Edit:**

The form of the presentation can be constrained even further, to look like this:

- There is a finite set of generating relations, coming in pairs each of the form $x_i = w(x_j,y_j)$, $y_i=v(x_j,y_j)$, where $w,v$ are group words and $1 \leq i, j \leq n$.

Other variations are possible too; I'm not sure what the most convenient description to work with might be.