Universal Mapping Property of Free Abelian Groups Let S be a set and $F=F_S$ the free group on S. Let $F'$ be the commutator subgroup of $F$. Set $A=A_S = F/F'$, and call it the free Abelian group on $S$. Prove the universal mapping property of the free Abelian group: for any function $f:S \rightarrow G$, where G is an Abelian group, there exists a unique group homomorphism $\varphi:A \rightarrow G$ so that the diagram

$S \xrightarrow{f} G$,
$S \xrightarrow{a \mapsto [a]} A$,
$A \xrightarrow{\varphi} G$

commutes.

We did a proof of this for free groups in class but I am not sure how to apply it to this. I think part of my issue is that I am having a hard time figuring out what the elements of A look like. Will $A=\{[s](a^{-1}b^{-1}ab) \vert a,b,s \in S\}$? If so, then could I just send $\varphi(w)$ to $f(w)$, with $w \in F$?
There is nothing special about the freeness. Given a group $G$ and a morphism $f:G \to A$ with $A$ abelian, $f$ must factor through the abelianization of $G$, which is $G / [G,G]$.

To be precise: This means that there exists a unique $\tilde f$ with $\tilde f \circ \pi = f$ where $\pi$ is the quotient map.
First of all free abelian group (in your sense) is obviously commutative because u kill commutator. Now if we have any abelian group $G$ and function $f:S\rightarrow G$ then we have unique morphism $f_*:F\rightarrow G$. Now we know that $[F,F]\subset ker(f_*)$ so we have unique factorization $f_*=gh$ where $h:F\rightarrow A$ is canonical projection and $g:A\rightarrow G$ unique morphism u looking for.
Secondly one can easily proof that $A_S=\bigoplus_S\mathbb Z\subset \prod_S\mathbb Z$. This is subgroup of product generated by sequences witch are not $0$ only on one coordinate.