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I am new for abstract algebra, and I am now reading the book "Algebra: Chapter 0" by myself. I meet the following problem:

(Chapter II 7.12 in the book) Let $F=F(A)$ be the free group of a set $A$, and let $f: A\to G$ be a set function from $A$ to an abelian group $G$. Denote $[F,F]$ be the commutator subgroup of $F$, prove that $f$ induces a unique homomorphism $F/[F,F]\to G$, and prove that $F/[F,F]\cong F^{ab}(A)$ where $F^{ab}(A)$ is the free abelian group of $A$.

I come up with a proof idea. First put $(f,G)$ into the category $\mathcal{F}^{A}$ where the objects are all $(f,G)$, since we know that $F$ is the initial object in this category, we have a unique homomorphism $\varphi: F\to G$ induced by $f$. We also know that the commutator subgroup is normal, therefore, if $[F,F]\subseteq ker\varphi$, there exists a unique group homomorphism $\psi: F/[F,F]\to G$. Since $G$ is abelian, we have $\varphi(aba^{-1}b^{-1})=\varphi(a)\varphi(b)\varphi(a)^{-1}\varphi(b)^{-1}=\varphi(a)\varphi(a)^{-1}\varphi(b)\varphi(b)^{-1}=e$, which indicates that $[F,F]\subseteq ker\varphi$. Therefore, there is a unique group homomorphism $\psi: F/[F,F]\to G$ induced by $\varphi$, so it is a unique homomorphism induced by $f$.

To prove $F/[F,F]\cong F^{ab}(A)$, since $G$ is abelian, and also $F/[F,F]$ is abelian, we can put them into the category $\mathcal{F}_{Ab}^{A}$ where the objects are all $(f,G)$ with G abelian, since we have proved that there is a unique homomorphism from $F/[F,F]$ to any abelian group $G$ induced by $f$, $F/[F,F]$ is an initial object in $\mathcal{F}_{Ab}^{A}$, which means $F/[F,F]\cong F^{ab}(A)$.

Could someone help me check is this proof is correct? Thank you!

cbyh
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    A free group is not an initial object in the category of groups (unless it is the trivial group). Unless you meant to talk about the coslice catery at $F$? – tomasz Aug 17 '20 at 16:24
  • Related : https://math.stackexchange.com/questions/666155/abelianization-of-free-group-is-the-free-abelian-group – Arnaud D. Aug 17 '20 at 16:37
  • Oh yes, sorry about that, it is not Grp but the category $F^{A}$ where each element is $(f,G)$, $G$ is a group and $f$ is a function from $A$ to $G$. I have corrected it, but I guess the other statements are still true. – cbyh Aug 17 '20 at 16:39

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