Let $F=F(A)$ be a free group, and let $f:A→G$ be a set-function from the set $A$ to an abelian group $G$. Show that $f$ induces a unique homomorphism $F/[F,F]→G$, where $[F,F]$ is the commutator subgroup of $F$. Conclude that $F/[F,F]≅F_{ab}(A)$.
I currently worked out the following:
We need to define a proper homomorphism $\tilde{f} : F/[F,F] \to G$. By the universal property of free group, we have a unique homomorphism $\varphi : F \to G$ induced from $f$. Now observe that for $g,h \in A$, $$ \varphi(g)\varphi(h)\varphi(g)^{-1}\varphi(h)^{-1} = \varphi(ghg^{-1}h^{-1}) = e $$ as $G$ is commutative, so we know that $\varphi$ vanish on $[F,F]$. Now we just define $$ \tilde{f} : F/[F,F] \to G \quad \text{ by } \quad \tilde{f}(x[F,F]) = \varphi(x). $$ It is a fast check that $\tilde{f}$ is the required homomorphism.
Now It remains to show $F/[F,F]≅F_{ab}(A)$. My textbook told me to show via universal property, so I have the diagram
Can I conclude by saying that for every $f$ there is a unique $\tilde{f}$ such that the diagram formed by $\pi \circ j, f$ and $f$ commutes?