Free Product of Groups with Presentations 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!
 A: The Universal Property of the free product $G*H$ of groups $G$ and $H$ is that there exist homomorphisms $i_G:G \to G*H$ and $i_H:H \to G*H$ such that, for any group $K$ and any homomorphisms $\tau_G:G \to K$ and $\tau_H:H \to K$, there is a unique homomorphism $\phi:G*H \to K$ with $\phi i_G=\tau_G$ and $\phi i_H=\tau_H$.
It is not hard to prove uniqueness of $G*H$ up to isomorphism directly from this definition.
To prove existence, let $G = \langle A \mid R \rangle$ and $H = \langle B \mid S \rangle$ be presentations of $G$ and $H$ , and let $P = \langle A \cup B \mid R \cup S \rangle$. Then by basic properties of group presentations, the identity maps on $A$ and on $B$ induce homomorphisms $i_G:G \to P$ and $i_H:H \to P$.
Also, given $K,\tau_G,\tau_H$ as above, define $\psi:F(A \cup B) \to K$ by $\psi(a) = \tau_G(a)$ and $\psi(b) = \tau_H(b)$ for $a \in A,b \in B$. Then, since $\tau_G$ and $\tau_H$ are homomorphisms of $G$ and 0f $H$, $\psi$ maps all elements of $R \cup S$ to the identity, and so $\phi$ induces a homomorphism $\phi:P \to H$ with $\phi i_G=\tau_G$ and $\phi i_H=\tau_H$. Hence $P \cong G*H$.
A: 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.
A: I'll try to give another proof of the fact $G_i=\langle X_i|R_i\rangle \Rightarrow *G_i=\langle  \sqcup X_i|  \sqcup R_i\rangle$. Please let me know if it is coorect.
We have the following diagramm (I'm sorry for this bad picture but I could not type it here)

where $\pi_i:G_i\to *G_i$ are monomorphisms and $\phi_i:F(X_i)\to \frac{F(X_i)}{\langle \langle R_i\rangle\rangle}$ are the projections.
Then from universal property of free products we have that there exists $\phi: *F(X_i)\to *G_i$ with $$\phi\circ j_i=\pi_i\circ \phi_i$$
Then $\phi$ is onto since: every $\phi_i$ maps $X_i$ onto a set of generators of $G_i$ so $\phi$ has to map every $X_i$ to a set of generators of $G_i$ hence $im\phi=*G_i$.
It is also: $ker\phi=\langle \langle \sqcup R_i\rangle\rangle$ since $j_i,\pi_i$ are $1-1$.
Finally we have an isomorphism $\bar{\phi}:\frac{*F(X_i)}{ker\phi}\to *G_i$
