Show there is a surjective homomorphism from $\mathbb{Z}\ast\mathbb{Z}$ onto $C_2\ast C_3$

Show there is a surjective homomorphism from $\mathbb{Z}\ast\mathbb{Z}$ onto $C_2\ast C_3$, where $\ast$ denotes the coproduct in the category $\mathsf{Grp}$.

Note the exercise in this book (Algebra Chapter 0, Aluffi) is meant, or at least hinted at, to be done using the universal property of the coproduct, ie: coproduct of $A$ and $B$ in $\mathsf{Grp}$ is initial in the category $\mathsf{Grp}^{A,B}$, instead of using the definition of the free product.

By property of the coproduct, for any object $A$ and two homomorphisms $f_1,f_2:\mathbb{Z}\to A$, and two homomorphisms $i_1,i_2:\mathbb{Z}\to\mathbb{Z}\ast\mathbb{Z}$, there exists a unique homomorphism $\sigma:\mathbb{Z}\ast\mathbb{Z}\to A$ such that the diagram commutes, ie: $f_1=\sigma i_1$ and $f_2=\sigma i_2$. Now let $A=C_2\ast C_3$. Now what?

Very soon after the definition of coproducts (by the universal property), there should be a theorem (or remark) saying that the coproduct is functorial. In particular, any homomorphisms $A\to B$ and $C\to D$ induce a homomorphism $A*C\to B*D$. In your situation, you undoubtedly know some surjections $\mathbb Z\to C_2$ and $\mathbb Z\to C_3$; they induce a homomorphism $\mathbb Z*\mathbb Z\to C_2*C_3$. All that remains is to check that this is surjective, which you can do either by invoking general information about coproducts of epimorphisms or by observing that $C_2*C_3$ is generated by the set consisting of (the images under the canonical injections of) the generator of $C_2$ and the generator of $C_3$.

Under exercise 3.7, Aluffi hints at what the elements in $\mathbb Z \ast \mathbb Z$ look like. With this in mind:

Let $1_a$ and $1_b$ be the generators of $\mathbb Z \ast \mathbb Z$. Let $\alpha$ and $\beta$ be the generators of $C_2 \ast C_3$.

The elements of $\mathbb Z \ast \mathbb Z$ look like $1_a^{i_1} 1_b^{j_1}\cdots 1_a^{i_k} 1_b^{j_k}$ where $i_m, j_n \in \mathbb Z$.

The elements of $C_2 \ast C_3$ look like $\alpha^{i_1} \beta^{j_1}\cdots \alpha^{i_k} \beta^{j_k}$ where $0\le i_m \le 1$ and $0 \le j_n \le 2$.

The map $\mathbb Z \ast \mathbb Z \to C_2 \ast C_3$ defined by $1_a \mapsto \alpha$ and $1_b \mapsto \beta$ is a homomorphism and for any $\alpha^{i_1} \beta^{j_1}\cdots \alpha^{i_k} \beta^{j_k}$ in $C_2 \ast C_3$ $$1_a^{i_1} 1_b^{j_1}\cdots 1_a^{i_k} 1_b^{j_k} \mapsto \alpha^{i_1} \beta^{j_1}\cdots \alpha^{i_k} \beta^{j_k}.$$

Assume $f$ and $g$ are distinct, but $f \circ u_Z = g \circ u_Z$. Then $f \circ u_Z \circ i_Z = g\circ u_Z \circ i_Z$. But by the universal property of $u_Z$, $u_Z \circ i_Z = i_3 \circ x\ mod\ 3$. Substitution yields: $f\circ i_3 \circ x\ mod\ 3 = g\circ i_3 \circ x\ mod\ 3$. $x\ mod\ 3$ is an epimorphism so we can right cancel it to yield $f\circ i_3 = g\circ i_3$ and similarly that $f\circ i_2 = g\circ i_2$.

$f$ is the universal arrow making the coproduct diagram for $C_2*C_3$ and $(A, f\circ i_2 , f\circ i_3)$ commute so it is unique. But because $g \circ i_2 = f\circ i_2$ and $g\circ i_3 = f\circ i_3$ the arrow $g$ is universal for the same diagram, implying $f = g$.