Abelianization is left adjoint to the forgetful functor

Question

I am trying to show that the map $ab:Grp \rightarrow AbGrp$ is left adjoint to the forgetful functor.

I have seen in a similar question the approach is as follows:

Let $f:G\rightarrow H$ be a group homomorphism where $H$ is an abelian group and $G$ is just a group. Then there exists a unique $r:G/[G,G]\rightarrow H$ described by $r(g+[G,G]) = f(g)$. I think this is what is meant by

"$f$ factors uniquelty through $G/[G,G]$."

My question is firstly, is this correct? And secondly, why does this mean we have an adjunction?

Answer 1

Yes, your interpretation is correct (but why did you write the group operation additively in $g+[G,G]$?)

The forgetful functor $U\colon\mathbf{AbGrp}\to\mathbf{Grp}$ is just the identity on objects and morphisms, so for any group $G$ and abelian group $A$, $ab\colon\mathbf{Grp}\to\mathbf{AbGrp}$ is the left-adjoint of $U$ if and only if there is a (natural in $G,A$) bijection $$ \operatorname{Hom}_\mathbf{AbGrp}(ab(G),A)\cong\operatorname{Hom}_\mathbf{Grp}(G,A). $$

Every group homomorphism $G\to A$ factors uniquely through $G/[G,G]$, so we identify to $f\colon G\to A$ in $\operatorname{Hom}_\mathbf{Grp}(G,A)$ with the induced $\bar{f}\colon ab(G)\to A$. Conversely, for any homomorphism $g\colon ab(G)\to A$, we have a homomorphism $f\colon G\to A$ by $G\to ab(G)\to A$. These are clearly inverse of each other so we have a bijection.

Moreover, this is natural in $G,A$: if $H$ is any group and $B$ any abelian group, with $f'\colon H\to G$, $g'\colon A\to B$ are any morphisms (of groups and abelian groups respectively), then we have $$\require{AMScd} \begin{CD} \operatorname{Hom}_\mathbf{AbGrp}(ab(G),A) @>{ab(f')\circ-\circ g'}>> \operatorname{Hom}_\mathbf{AbGrp}(ab(H),B);\\ @| @| \\ \operatorname{Hom}_\mathbf{Grp}(G,A) @>{f'\circ-\circ g'}>> \operatorname{Hom}_\mathbf{Grp}(H,B); \end{CD} $$ commutes, where $ab(f')$ is the unique map $ab(G)\to ab(H)$ arising from $G\xrightarrow[f']{} H\to ab(H)$ factoring uniquely through $ab(G)$.

Answer 2

Yes, it is correct.

Similar formulation:

For every $f\in\mathbf{Grp}(G,U(H))$ there is a unique $r\in\mathbf{AbGrp}(G/[G,G],H)$ such that $f=U(r)\circ\eta$ where $U$ denotes the forgetful function $\mathbf{AbGrp}\to\mathbf{Grp}$ and $\eta:G\to U(G/[G,G])$ is the arrow in $\mathbf{Ab}$ that is prescribed by $g\mapsto g[G,G]$.

This implies for every pair $G,H$ the existence of a bijection $\mathbf{Grp}(G,U(H))\to\mathbf{AbGrp}(G/[G,G],H)$ that can be shown to be natural in $G$ and $H$. So we are dealing with an adjunction of the forgetful functor and the functor that you denote as $\mathsf{ab}$.

If it is not clear to you that this is an adjunction then can you please explain what your understanding of an adjunction is?