Proving the free group for $A \amalg B$ is equal to the coproduct of free groups for $A$ and $B$ Aluffi's "Algebra: Chapter 0" ex. II.5.8 suggests proving that $F(A \amalg B) = F(A) * F(B)$, where $F(S)$ is a free group for a set $S$.
I managed to prove that they are isomorphic by chasing the following diagram and using universal properties where needed (and some obvious morphisms are omitted to declutter):

(sorry for the screenshot, I couldn't figure out how to get the tikz-cd environment working here).
But isomorphism does not imply equality, so how do I prove that they are equal? Or is Aluffi a bit abusing the language here, and proving isomoprhism is sufficient?
 A: Yes, Aluffi is abusing the language and meaning $\cong$ instead of $=$.
At the same time, there is no real danger of confusion. Let me explain. In Aluffi's Algebra 0, there is never a "unique on the nose" definition of a coproduct of two groups. A coproduct $G * G'$ of two groups $G$ and $G'$ is defined in Exercise II.8.7, but the definition given there depends on presentations of $G$ and $G'$, and the result is independent on these presentations only if you treat it (or, rather, its cocone, with its maps from $G$ and from $G'$) up to isomorphism. So when he says $F\left(A \amalg B\right) = F\left(A\right) * F\left(B\right)$, the right hand side of this "equality" is only defined up to isomorphism, whence the equality must be understood as an isomorphism.
You can, of course, turn the definition of a coproduct in Exercise II.8.7 into a precise definition by picking the tautological presentations of $G$ and $G'$. (The tautological presentation of a group $G$ is the presentation which uses all elements of $G$ as generators and all entries of the multiplication table as relations.) Then, the equality $F\left(A \amalg B\right) = F\left(A\right) * F\left(B\right)$ does not hold on the nose, since the elements of the left hand side are equivalence classes of words over $\left(A \amalg B\right) \amalg \left(A \amalg B\right)^{-1}$ whereas the elements of the right hand side are equivalence classes of words over $F\left(A\right) \amalg F\left(B\right)$. So, again, you are looking for an isomorphism, not an equality in the literal sense.
A: I think that you did not proceed correctly with your proof. At the moment this exercise was presented in the book the existence of coproducts in the category Set has not been proven yet and you are assuming its existence implicitly by using it and its canonical arrows in a diagram.
I think that a diagram like the following is more accurate.

