TITLE: Free Product of Groups with Presentations
QUESTION [6 upvotes]: There is a highly believable theorem:
Let $A, B$ be disjoint sets of generators and let $F(A), F(B)$ be the corresponding free groups.  Let $R_1 \subset F(A)$, $R_2 \subset F(B)$ be sets of relations and consider the quotient groups $\langle A | R_1 \rangle$ and $\langle B | R_2 \rangle$.
Then $\langle A \cup B | R_1 \cup R_2 \rangle \simeq \langle A | R_1 \rangle \ast \langle B | R_2 \rangle$, where $\ast$ is the free product.
I can supply a horrible proof using normal closures and elements and such, but surely there is a reasonable diagram-chase proof using the universal properties of free groups and the free product.  However, I have been unable to construct it; I think there is probably some intermediary but basic categorical statement needed about presentations.
There is a hint given for the problem (it is coming from Lee, Topological Manifolds, problem 9-4) that is:
If $f_1 : G_1 \rightarrow H_1$ and $f_2 : G_2 \rightarrow H_2$ are homomorphisms and for $j = 1,2$ we have $i_j, i_j'$ the injections of $G_j, H_j$ into $G_1 \ast G_2$ and $H_1 \ast H_2$ respectively, there is a unique homomorphism $f_1 \ast f_2 : G_1 \ast G_2 \rightarrow H_1 \ast H_2$ making the square $(i_j, f_j, i_j', f_1 \ast f_2)$ commute for $j = 1,2$.
This is almost trivial to prove with the UMP for free products, but I can't wrangle it into a proof of the top statement.  Also, anything legitimately trivial like $F(A) \ast F(B) \simeq F(A \cup B)$ or $F(F(S)) \simeq F(S)$ is fair game.
I can get a whole lot of arrows, but I am having trouble conjuring up any arrows with $\langle A | R_1 \rangle \ast \langle B | R_2 \rangle$ as their domain.  We also know that the free product is the coproduct in $\textbf{Grp}$ but admittedly I haven't tried exploiting this yet.
Can anyone help with a hint?  Thanks a lot!

REPLY [3 votes]: This an old question that recently got bumped, but there's another nice proof that uses the fact that colimits commute with one another and left adjoints, so I thought I'd add it for potential future visitors.
Let $G_i =\langle X_i\mid R_i\rangle$ for $1\le i \le n$ be groups given by presentation. Another way of saying this is that we have the coequalizer diagram
$$
F(R_i)\rightrightarrows F(X_i)\to G_i,
$$
with the top map of the fork being induced by $R_i\subseteq F(X_i)$, and the bottom map being the zero map.
Now the coproduct in the category of groups is the free product, $\newcommand\bigast{\mathop{{\Large *}}}\bigast$, so taking the coproduct of these coequalizers, since colimits commute, we get the coequalizer
$$
\bigast_{i=1}^n F(R_i) \rightrightarrows \bigast_{i=1}^n F(X_i)\to \bigast_{i=1}^n G_i,
$$
but the free group functor is left adjoint to the forgetful functor, so it also commutes with colimits, so we get a coequalizer
$$
F\left(\coprod_{i=1}^n R_i\right) \rightrightarrows F\left(\coprod_{i=1}^n X_i\right) \to \bigast_{i=1}^n G_i.
$$
Translating this back into a statement about group presentations, this says that
$\bigast_{i=1}^n G_i$ is given by the presentation $\left\langle \coprod_{i=1}^n X_i\mid \coprod_{i=1}^n R_i \right\rangle$, as desired.