Universal Property of the Free Abelian Group on $S$.

Question

This problem is from Dummit and Foote, 6.3.11.

Problem: Let $S$ be a set. The group with presentation $(S,R)$ with $R = \{[s,t] \mid s,t \in S\}$ is called the free-abelian group on $S$ denoted $A(S)$. So specifically $A(S) = F(S)/ \langle \langle R \rangle \rangle$. Show that if $G$ is an abelian group and $\varphi:S \rightarrow G$ is a set map then there is a unique homomorphism $\psi:A(S) \rightarrow G$ such that $\psi|_S = \varphi$. Deduce that if $|S| = n$ that $A(S) \cong \mathbb{Z}^n$.

Here's my attempt at satisfying the universal property.

Attempt: Let $\varphi:S \rightarrow G$ be a set map. We will extend $\varphi$ to a map on $F(S)$ by specifying what it does to words, so for all $s \in S$ set $\varphi(s^{-1}) = \varphi(s)^{-1}$ and $\varphi(s_1 \dots s_n) = \varphi(s_1) \dots \varphi(s_n)$. We can then define a map $\phi:F(S) \rightarrow G$ by $\phi(s_1\dots s_n) = \varphi(s_1) \dots \varphi(s_n)$. To define a homomorphism from $A(S)$ into $G$ it suffices to have a map that sends the relations in $R$ to the identity; in this case commutators of $S$ need to be sent to $1$ in $G$. But we've already specified enough for this to be true, let $s,t \in S$ then $$ \phi([s,t]) = \phi(sts^{-1}t^{-1}) = \varphi(s)\varphi(t)\varphi(s^{-1})\varphi(t^{-1}) = \varphi(s)\varphi(s^{-1})\varphi(t)\varphi(t^{-1}) = 1_G, $$ where swapping the terms followed from the fact that $G$ is an abelian group. Therefore we have defined a homomorphism from $A(S) \rightarrow G$ since any commutator of $S$ is sent to $1$ by $\phi$. Note that by construction we have that $\phi|_S = \varphi$ as required, and it remains to show that $\phi$ is the unique such homomorphism. Let $\gamma:A(S) \rightarrow G$ be another homomorphism satisfying $\gamma|_S = \varphi$, then it follows that $$ \gamma(s_1 \dots s_n) = \gamma(s_1) \dots \gamma(s_n) = \varphi(s_1) \dots \varphi(s_n) = \phi(s_1 \dots s_n), $$ and so $\varphi$ is unique. So $A(S)$ satisfies the universal property.

I think this is ok, but I'm having a harder time with the deduction that $A(S) \cong \mathbb{Z}^n$. I think I can define a set mapping $\varphi:S \rightarrow \mathbb{Z}^n$ by $\varphi(s_i) = (0, \dots , 0, 1, 0 , \dots ,0)$ i.e the $n$-tuple with zeros in all entries but the $i$'th. And this would extend by the universal property to a homomorphism from $A(S)$ into $\mathbb{Z}^n$ and all I would need is to show the function is bijective. I suppose succinctly my questions are as follows.

1. Is the proof of the universal property alright?
2. Is my approach alright for deducing that $A(S) \cong \mathbb{Z}^n$?
3. Could anyone enlighten me as to why this feels so much like a vector space? I know that it's a broad ask but lots of the free group theory (in particular extending to a homomorphism uniquely by specifying the images $\varphi(s_i)$) seems very analogous to vector space theory and I'd like to tie them together in my mind if possible.

Thanks in advance for the help.