TITLE: Defining a coproduct in $\mathsf{Grp}$ using group presentations
QUESTION [5 upvotes]: I've encountered this exercise in Aluffi's Algebra: Chapter 0. It might be helpful to say that the book doesn't introduce functors at this stage,, and that the book defines a presentation of a group $G$ as a pair
$(A\mid \Psi)$, where $\Psi$ is a subset of $F(A)$ such that $G \cong F(A)/R$, where $R$ is the smallest normal subgroup of $F(A)$ containing $\Psi$.
So, here is the exercise.

Let $(A\mid \Psi)$, resp., $(A'\mid \Psi')$, be a presentation for a group $G$, resp. $G'$; we may assume that $A,A'$ are disjoint. Prove that the group $G\ast G'$ presented by $(A\cup A'\mid \Psi\cup\Psi')$ satisfies the universal property for the coproduct of $G$ and $G'$ in $\mathsf{Grp}$.
(Use the universal properties of both free groups and quotients to construct natural homomorphisms $G\rightarrow G\ast G'$ and $G'\rightarrow G\ast G'$.

First of all, why would we assume that $A,A'$ are disjoint? If they are then their union is their disjoint union as well. But can any 2 groups be presented with $A\cap A' = \emptyset$?
Secondly, I still can't prove it even for disjoint $A$ and $A'$. So, will be thankful for any tips.

REPLY [6 votes]: Let $G\ast G'=(A \cup A' \mid \Psi \cup \Psi')$.
By the universal property for free groups, for the set map $f: A \to G\ast G'$ where $f(a)=a$, there exists a unique group homomorphism $\bar f: F(A) \to G\ast G'$ defined on the generating set $A$ by $\bar f(a)=f(a)=a$ and making the relevant diagram commute. Similarly, for $g: A' \to G\ast G'$, we have a homomorphism $\bar g: F(A') \to G\ast G'$.
Next, $R \subset \ker \bar f$ and so by the universal property for quotients there exists a unique homomorphism $\tilde f: G=F(A)/R \to G\ast G'$ defined by $\tilde f(aR)=\bar f(a)=f(a)=a$ and making the relevant diagram commute. Similarly, there exists a hom $\tilde g: G' = F(A')/R' \to G \ast G'$.
So, we actually have our natural maps $G \hookrightarrow G\ast G'$ and $G' \hookrightarrow G \ast G'$.
Now, for any group $Z$ and any pair of homomorphisms $\varphi: G \to Z$ and $\psi: G' \to Z$, we know these maps are determined by where they map the generators in $A$ and $A'$, respectively. So, there exists a unique map $\sigma: G\ast G' \to Z$ by mapping $a \in A$ to $\sigma(a)=\varphi(a)$ and $a' \in A'$ to $\sigma(a')=\psi(a')$.