Proving that disjoint unions of presentations are coproducts of groups I'm working through Aluffi's Algrebra: Chapter 0 and I need some assistance with an excercise.

Aluffi, Ex. II.8.7
Let $(A|R)$, resp. $(A'|R')$, be a presentation for a group $G$ in Grp, resp. $G'$; we may assume that $A, A'$ are disjoint. Prove that the group $G \ast G'$ presented by
  \begin{equation}
(A \cup A'|R \cup R')
\end{equation}
  satisfies the universal property for the coproduct of $G$ and $G'$ in Grp. (Use the universal properties of both free groups and quotients to construct natural homomorphisms $G \to G \ast G', G' \to G \ast G'.$)

What I have done so far:
First off, we have natural set-functions $A,A' \to F(A \cup A')$. Hence we have unique homomorphisms $F(A),F(A') \to F(A \cup A')$ whose restrictions to $A,A'$ are equal to the functions $A,A' \to F(A \cup A')$, by the universal property of free groups.
By combining these homomorphisms with the natural projection $F(A \cup A') \to F(A \cup A')/(R \cup R')$, we get homomorphisms $F(A),F(A') \to F(A \cup A')/(R \cup R')$.
Since $a(R) = b(R) \Rightarrow a(R \cup R') = b(R \cup R')$, we get a natural homomorphism $\pi_1:G \cong F(A)/(R) \to F(A \cup A')/(R \cup R')$, by the universal property of quotients. Similarly, we get a natural homomorphism $\pi_2:G' \cong F(A')/(R') \to F(A \cup A')/(R \cup R') = G \ast G'$.
If I'm not mistaken, these homomorphisms simply map $a(R) \mapsto a(R \cup R')$, for all $a(R) \in G$, and similarly for elements of $G'$. If we have a group $H$ and homomorphisms $f:G \to H, g:G' \to H$, then the only function $\varphi:G \ast G' \to H$ s.t. $\varphi \circ \pi_1 = f, \varphi \circ \pi_2 = g$, must map $a(R \cup R') \mapsto f(a(R))$ for all $a \in F(A)$, and similarly for all $b \in F(A'), b(R \cup R') \mapsto g(b(R'))$. All that is left, is to show that this defines a function $\psi$.
So, we need to show that $ab^{-1} \in (R \cup R') \Rightarrow \psi (ab^{-1}(R \cup R')) = e_H$, for any $a,b \in F(A \cup A')$.
This is what I can't do. How do I go about showing this, or am I going in a wrong direction?
Edit: Note that (R) stands for the normal closure of R in F(A).
One answer:
Note that $(R \cup R') := \langle \{ghg^{-1} \mid g \in F(A \cup A'), h \in R \cup R'\} \rangle$. Now, for any $h \in R \cup R', g \in F(A \cup A')$, we have 
\begin{align}
\psi (ghg^{-1}(R \cup R'))  & = \psi (g(R \cup R')) \psi (h(R \cup R')) \psi (g^{-1}(R \cup R'))
\\ & =  \psi (g(R \cup R')) \psi (g^{-1}(R \cup R'))
\\ & =  \psi (gg^{-1}(R \cup R'))
\\ & = e_H.
\end{align}
Hence, for any $a \in (R \cup R')$, we have $\psi (a) = e_H$, which completes the proof.
 A: I guess by $(R)$ you mean the normal closure $\langle R^{F(A)} \rangle$ of $R$ in $F(A)$?
There is a unique homomorphism $\psi:F(A \cup A') \to H$ with $\psi(a) = f(a)$ for $a \in A$ and $\psi(a) = g(a)$ for $a \in A'$. Since all elements of $R$ lie in $\ker(f)$ and all elements of $R'$ lie in $\ker(g)$, we have $R \cup R' \subset \ker(\psi)$ and hence, since $\ker(\psi) \unlhd F(A \cup A')$, we have $(R \cup R') \le \ker(\psi)$.
So $\psi$ induces a well-defined map $\varphi:G*H = F(A \cup A')/(R \cup R') \to H$, and $\varphi$ satisfies $\varphi \circ \pi_1 = f$ and $\varphi \circ \pi_2 = g$, as required.
A: Often it is much more convenient to work with representable functors.
The universal property of the group $(A|R)$ is the following: If $H$ is a group, there is a natural isomorphism (i.e. bijection)
$$\hom((A|R),H) \cong \{f \in \hom(A,|H|) : f(R)=\{1\}\},$$
where by $f(R)$ I mean the set of all $f(r)$, $r \in R$, and $f(r)$ is defined by extension of $f$ on the set of words. It follows for test groups $H$:
$$\hom((A \sqcup A'|R \sqcup R'),H)$$
$$ \cong \{g \in \hom(A \sqcup A',|H|) : g(R \sqcup R')=\{1\}\}$$
$$= \{g \in \hom(A \sqcup A',|H|) : g(R)=\{1\} \wedge g(R')=\{1\}\}$$
$$\cong \{(f,f') \in \hom(A,|H|) \times \hom(A',|H|) : f(R)=\{1\} \wedge f'(R')=\{1\}\}$$
$$ \cong \hom((A|R),H) \times \hom((A'|R'),H).$$
These isomorphisms are clearly natural in $H$. Thus, by definition of a coproduct, $(A \sqcup A'|R \sqcup R')$ is a coproduct of $(A|R)$ and $(A'|R')$.
