Aluffi II.8.7 suggests proving that, given groups $G, G'$ admitting representations $(A \mid R), (A' \mid R')$ respectively, where $A, A'$ are disjoint, their coproduct in is $(A \cup A' \mid R \cup R')$. Unlike a couple of other questions, I'm trying to do this without looking at the specific construction of the free groups, using just the corresponding universal properties.
First, I've already proven earlier that $F(A \amalg A') \cong F(A) * F(A')$, so I'll be using them interchangeably. I'll also be using $\langle R \rangle$ to denote the smallest normal subgroup generated by relations $R$.
Then, consider the following diagram.
$j_1$ exists by the definition of coproduct, $\iota_1$ exists by the definition of the quotient group, and similarly for $j_1', \iota_1'$. There also exists the morphism $\iota$, since $\frac{F(A \amalg A')}{\langle R \cup R' \rangle}$ is just a quotient group. Next, consider the morphism $\iota j_1$. For arbitrary $r \in R \subset F(A)$, $\iota$ must$^1$ map $r$ to the identity element, since $r \in R \subset R \cup R'$. Thus $\langle R \rangle \subseteq \ker \iota$, so by the universal property of quotients there exists $j_2$ such that $\iota j_1 = j_2 \iota_1$. Similarly for $j_2'$.
Now, for an arbitrary group $H$ and $\varphi : G \rightarrow H, \varphi : G' \rightarrow H$ we can consider the morphisms $\varphi \iota_1, \varphi' \iota_2$. By the universal property for coproducts there exists an unique $\sigma : F(A) * F(B) \rightarrow H$ such that $\sigma j_1 = \varphi \iota_1$, $\sigma j_1' = \varphi' \iota_1'$.
Similarly to the previous argument, $\forall r \in R: r \in \ker \sigma$, and similarly for $\forall r' \in R'$, so $\langle R \cup R' \rangle \subset \ker \sigma$, so by the universal property for quotients there exists an unique $\eta$ such that $\sigma = \eta \iota$, which, combined with the previous step, implies $\eta \iota j_1 = \varphi \iota_1$, $\eta \iota j_1' = \varphi' \iota_1'$. But we have proven earlier that $\iota j_1 = j_2 \iota_1$, so we get $\eta j_2 \iota_1 = \varphi \iota_1$. Since $\iota_1$ is injective, it has a right-inverse, multiplying by which yields $\eta j_2 = \varphi$. Similarly $\eta j_2' = \varphi'$.
Uniqueness of $\eta$ follows from the uniqueness of $\sigma$ and $\iota$.
Does this sound reasonable or have I missed something?
$^1$ I don't exactly like this part, since $r$ does not necessarily belong as is to $F(A) * F(A')$, but I managed to persuade myself that it's OK since otherwise $\langle R \cup R' \rangle$ wouldn't make sense, so it's sort of notational abuse (or convenience). Is that correct?
Best Answer
Seems almost correct, although admittedly it's been a bit hard to read. Careful though, the morphisms $\iota_i$ are epimorphisms (in this context, surjective) but not injective. Note that this is what you use: epis are right-cancellative.
As for the foot note, there's no need to be concerned. A slight abuse of notation is going on, if you want, think of the union of the relations as the union of the image of each relation set via the canonical inclusion of $F(A), F(A')$ to the coproduct. Techically speaking, when you identify $F(A) \ast F(A')$ with $F(A \sqcup A')$ these also carry to a different set, to which we will give the same name.
Here's the same argument, a bit more organized (if you can, draw a diagram as you read). I will note $A_1, A_2$ for the generator sets, $R_1,R_2$ the relations, $G_i = \langle A_i | R_i \rangle$ the quotients and $\pi_i : F(A_i) \to G_i$ the projections.
Let
$$ f_1 : \frac{F(A_1)}{\langle R_1 \rangle} \to H, \quad f_2 : \frac{F(A_2)}{\langle R_2 \rangle} \to H $$
be two morphisms. They induce a pair of morphisms $f_i\pi_i : F(A_i) \to H$. Thus we have a unique morphism from the coproduct,
$$ f : F(A_1) \ast F(A_2) \to H $$
such that $f\iota_i = f_i\pi_i$, where $\iota_i : F(A_i) \hookrightarrow F(A_1) \ast F(A_2)$ are the canonical inclusions.
Now if $r \in R_1$ is a relation, then
$$ f(r) = f(\iota_1(r)) = f_1\pi_1(r) = f_1(1) = 1. $$
By the same argument applied to $R_2$, we see that $\langle R_1 \cup R_2 \rangle \subset \ker f$. Thus there exists a unique morphism
$$ \hat{f} : \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle} \to H $$
such that $\hat{f}\pi = f$ with $\pi : F(A_1) \ast F(A_2) \rightarrow \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2\rangle}$ the projection. We also see that $R_i \subset \ker \pi\iota_i$ and so each $\iota_i$ induces an arrow
$$ \hat{\iota_i} : \frac{F(A_i)}{\langle R_i\rangle} \to \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle}. $$
so that $\pi\iota_i = \hat{\iota}_i\pi_i$ Now, since
$$ \hat{f}\hat{\iota_i}\pi_i = \hat{f}\pi\iota_i = f\iota_i = f_i\pi_i $$
and each $\pi_i$ is epi, we get
$$ \hat{f}\hat{\iota}_i = f_i. $$
Finally, the arrow $\hat{f}$ is unique in this sense: if we have another arrow $g$ such that $g\hat{\iota}_i = f_i$ for $i= 1,2$, then applying $\pi_i$ we get
$$ f_i\pi_i = g\hat{\iota}_i\pi_i = g\pi \iota_i $$
for both $i$. However, $f$ is the unique morphism such that $f \iota_i =f_i \pi_i$ and so $g\pi = f$. But $\hat{f}$ was the only morphism so that $\hat{f} \pi = f$, hence it must be $f = g$.
This finally concludes the proof that $\frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle}$ toghether with the arrows $\hat{\iota}_i : F(A_i)/\langle R_i \rangle \to \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle}$ are "a" coproduct for the objects $F(A_1)/\langle R_1\rangle$ and $F(A_1)/\langle R_2\rangle$. Since all coproducts are isomorphic, we get
$$ F(A_1)/\langle R_1 \rangle \ast F(A_1)/\langle R_2 \rangle\simeq \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle}. $$
Now simply identify $F(A_1) \ast F(A_2) \simeq F(A_1 \sqcup A_2)$ and note that the (group generated by the) relations carry to what we have (once again) noted $\langle R_1 \cup R_2\rangle$, to obtain
$$ F(A_1)/\langle R_1 \rangle \ast F(A_1)/\langle R_2 \rangle\simeq \frac{F(A_1 \sqcup A_2)}{\langle R_1 \cup R_2 \rangle} $$
as desired.