Proving that there is a functor $F: Grp \rightarrow Ab$ s.t $F(G)=G_{ab}$- How to deal with the quotient

Question

Let $G$ be a group. $f$ a group hom. $f:G\rightarrow H $

I define:
$F(G)=G_{ab}$
and
$F$ : $G$ $\rightarrow$ $G_{ab}$ a homomorphism, where $G_{ab} : = G/[G,G]$ is the abelianization of group $G$. such that

$F(f):=f_{ab}$ where $f_{ab}([r])=[f(r)]$

I assume F is well-defined on objects, because the abelianization is an ableian group

I think the checks that I have to perform are a part from F mapping identity to the identity and the composition, are :

1. that $f_{ab}$ is a group hom: (provided 2) has been shown as observed in the comments)

$f_{ab}([r][s])=f_{ab}([rs])=[f(rs)]=[f(r)f(s)]=f_{ab}([r])f_{ab}([s])$

2. that $f_{ab}$ maps $G_{ab}$ to $H_{ab}$:

   Let $[r] \in G_{ab}$ —> what does this imply? I know that being in [G,G] means being a product of commutators, that is if $g \in [G,G]$, then $g=[g_1,h_1][g_2,h_2],….[g_n,h_n]$ for some finite $n$ but then I am not sure what is the form of the elements in the quotient. Quotienting should be just a regrouping, but still the quotient is something that always comes back to hunt me.

$f_{ab}([r])=[f(r)]$, I have to prove that this is in $H_{ab}$. I know that since f is a group hom
$f([G,G])\subset [H,H]$, but I don't know how to use it here

I know that $G/[G,G]=\{a[G,G]: a \in G\}$, not sure if it helps

3. that $f_{ab}$ does not depend on the representative of the class, i.e if $[r] = [s] \implies f_{ab}([r])=f_{ab}([s])$

$f_{ab}([r])=[f(r)]$… I am not sure how to continue here

4. $F(id_G)=id_{F(G)}$

   Let $[r]\in G_{ab}$

$F(id_G)([r])=(id_G)_{ab}([r])=[id_G(r)]=[r]….?$

5. $F(f\circ g)=F(f)\circ F(g)$

Anything else that I am missing to prove? How do I complete the missing pieces?. At least with points 2 and 3, I would appreciate a detailed explanation, I've been the whole day on this

Many thanks

Answer 1

I'm not sure what you mean by (2). If you drop your notation for the cosets, we define $f^{ab}:G^{ab} \rightarrow H^{ab}$ by $f^{ab}(g[G,G]) = f(g)[H,H]$, which is clearly an element of the abelianization of $H$.

I think a more illuminating way to see what's going on is to see where the map $f^{ab}$ comes from. First, you have the factor isomorphism theorem.

Theorem: Let $f:G \rightarrow H$ be a group homomorphism and $N \unlhd G$. Then $f$ factors as $f = \psi \pi$ with $\pi:G \rightarrow G/N$ and a well-defined $\psi:G/N \rightarrow H$ if and only if $N \subseteq \ker f$. If this is the case, then $\psi$ is unique and acts by $\psi(gN) = f(g)$.

With this you can prove the universal property of the abelianization.

Theorem: For any group $G$, its commutator subgroup is normal, and $G/[G,G]$ is abelian. Moreover, if $\varphi:G \rightarrow A$ is any group homomorphism into an abelian group, then $\varphi$ factors through the abelianization of $G$ by $\varphi = \psi \pi$ for $\pi:G \rightarrow G/[G,G]$ and a unique group homomorphism $\psi:G/[G,G] \rightarrow A$.

This says whenever we have a group homomorphism $\varphi$ from $G$ into an abelian group, it will factor through the abelianization of $G$.

Now let $\varphi:G \rightarrow H$ be a group homomorphism, i.e a morphism of $\mathbf{Group}$. Denote $\pi:H \rightarrow H/[H,H]$ for the natural projection onto the abelianization. Then we get a composite $\pi \varphi:G \rightarrow H/[H,H]$, and this is still a group homomorphism. However, it is a homomorphism into an abelian group, and so by the universal property of the abelianization, there is a unique group homomorphism $\psi:G/[G,G] \rightarrow H/[H,H]$ for which $\varphi = \psi \pi_G$, where $\pi_G$ is projection onto the abelianization of $G$.

In particular, the functor $F$ from $\mathbf{Group}$ to $\mathbf{Ab}$ groups sending $G$ to $G^{ab}$ and a morphism $f:G \rightarrow H$ to the unique map $\psi:G^{ab} \rightarrow H^{ab}$ which is defined by the universal property of the abelianization. In this way the map $f^{ab}$ is already well-defined from the factor isomorphism theorem, and its definition is more intuitive.

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Edit: As FShrike mentioned, your attempt at well-defined is not quite there. Here is how you might show it manually:

Suppose $[r] = [s]$. This means that $s^{-1}r \in [G,G]$. In turn, $f(s^{-1}r) \in [H,H]$. Because $f$ is a group homomorphism, $f(s)^{-1}f(r) \in [H,H]$ which implies that $[f(s)] = [f(r)]$, and so $f^{ab}([r]) = f^{ab}([s])$ and the map is well-defined.

In this one uses that if $f$ is a homomorphism $f([a,b]) = [f(a),f(b)]$, which extends to finite products of commutators. You should review the universal property of the abelianization as FShrike mentioned.