Let $A$ and $B$ be groups and $A*B$ the free product. Let $N=\langle\langle A \rangle\rangle$ be the smallest normal subgroup of $A*B$ containing $A$. Show that $A*B/N \cong B$ without utilizing a specific construction of the free product. That is, formulate a proof without explicitly examining the elements of $A*B$.
For a homework assignment I turned in last week, I was asked to show that $A*B/N \cong B$, which I did by showing that every element $Na_1b_1\cdots a_nb_n \in A*B/N$ is of the form $Nb$ for some $b\in B$ and then by defining $\phi:A*B/N \to B$ as $\phi(Na_1b_1\cdots a_nb_n)=\phi(Nb)=b$, which can fairly easily be shown to be an isomorphism. While this sufficed for the assignment, I don't particularly like this proof.
I'm interested in knowing if it can be shown that $A*B/N \cong B$ without reference to the elements or specific construction of $A*B$. That is, does $A*B/N \cong B$ follow from the universal property of the free product or related theorems? All of my attempts so far have been fruitless.
Let me give a general categorical version of this statement and then prove it.
Proof: By definition of $p$, $pi$ is the zero map $A\to B$, and $p0$ is also the zero map $A\to B$, so $pi=p0$. Now suppose $q:A*B\to C$ is such that $qi=q0$. Let $f=qj:B\to C$ where $j:B\to A*B$ is the inclusion. Then $$fpi=f0=0=q0=qi$$ (here I abuse notation and write $0$ for any zero map) and $$fpj=f1_B=f=qj.$$ Since $A*B$ is a coproduct with inclusions $i$ and $j$, this implies $fp=q$, so $q$ factors through $p$. Moreover, the factorization is unique, since if $f'pj=qj$ then $f'=qj$ (since $pj=1_B$).
In the case of groups, the statement that $p$ is the coequalizer of $i$ and $0$ says exactly that a map on $A*B$ factors (uniquely) through $p$ iff $A$ is contained in its kernel. This is exactly the universal property of the quotient of $A*B$ by the normal subgroup generated by $A$, so this says $B\cong A*B/N$.