Defining unit and counit for the adjunction between $\mathbf{Grp}$ and $\mathbf{Ab}$

Question

So I'm trying to write the proof for the following example of adjunction as training for an upcoming exam and I can't for the love of God figure this out. Here are the details.

I have the following functors $F \colon \mathbf{Grp} \rightarrow \mathbf{Ab}$ (free functor) and $U \colon \mathbf{Ab} \rightarrow \mathbf{Grp}$ (inclusion functor).

I have the unit and counit written out too (which I hope I wrote right):

• $\eta \colon \mathrm{id}_{\mathbf{Grp}} \Rightarrow U \circ F$ aka $\eta_A : A \rightarrow (U \circ F)(A)$,

• $\epsilon \colon F \circ U \Rightarrow \mathrm{id}_{\mathbf{Ab}}$ aka $\epsilon_{A_{\mathrm{ab}}} \colon F(A) \rightarrow A_{\mathrm{ab}}$,

• plus the triangular diagrams (I didn't include them here).

I'm trying to use the triangular identities to write the proof, but I'm stuck at the point where I need to define the unit and counit in order to proceed with the rest of the proof. If I were working with list, I'd write something like $\eta_{A} (a) = [a],$ but it's not list I'm working with.

Could someone explain it to me how I should proceed?

Answer 1

Basics on abelianization

Let us recall how abelianization works:

• For every group $G$, its abelianization $\newcommand{\ab}{\mathrm{ab}} G^{\ab}$ is the quotient $G / [G, G]$ where $[G, G]$ is the commutator subgroup of $G$.
• We have for every group $G$ the homomorphism of groups $$ \newcommand{\longto}{\longrightarrow} p_G \colon G \longto G^{\ab} \,, \quad g \longmapsto [g] \,, $$ which is the canonical homomorphism from $G$ to $G^{\ab}$.
• The group $G^{\ab}$ together with the homomorphism $p_G$ satisfies the following universal property: given an abelian group $A$, every homomorphism of groups $φ$ from $G$ to $A$ factors uniquely through $p_G$. More explicitly, there exists for every homomorphism of groups $φ$ from $G$ to $A$ a unique homomorphism of groups $ψ$ from $G^{\ab}$ to $A$ with $φ = ψ ∘ p_G$.
• The abelianization construction $(-)^{\ab}$ is functorial: for every homomorphism $φ$ from $G$ to $H$ there exists a unique homomorphism of groups $φ^{\ab}$ from $G^{\ab}$ to $H^{\ab}$ that makes the following diagram commute: $$ \require{AMScd} \begin{CD} G @> φ >> H \\ @V p_G VV @VV p_H V \\ G^{\ab} @>> φ^{\ab} > H^{\ab} \end{CD} \tag{$\ast$} $$ (As a formula, $p_H ∘ φ = φ^{\ab} ∘ p_G$.) The homomorphism $φ^{\ab}$ is explicitly given by $$ φ^{\ab} \colon G^{\ab} \longto H^{\ab} \,, \quad [g] \longmapsto [φ(g)] \,. $$

Let us check in the following that the abelianization functor $F := (-)^{\ab}$ from $\newcommand{\Grp}{\mathbf{Grp}} \Grp$ to $\newcommand{\Ab}{\mathbf{Ab}} \Ab$ is left-adjoint to the forgetful functor $U$ from $\Ab$ to $\Grp$.

Via natural bijection of $\newcommand{\Hom}{\operatorname{Hom}} \Hom$-sets

The universal property of the abelianization gives us for every group $G$ and every abelian group $A$ a bijection $$ Φ_{G, A} \colon \Hom_{\Ab}(G^{\ab}, A) \longto \Hom_{\Grp}(G, A) \,, \quad ψ \longmapsto ψ ∘ p_G \,. $$ In other words, a bijection $$ Φ_{G, A} \colon \Hom_{\Ab}(F(G), A) \longto \Hom_{\Grp}(G, U(A)) \,. $$ This bijection is natural in both $G$ and $A$, and therefore gives an adjunction $$ F ⊣ U \,. $$

So how do the unit $η$ and counit $ε$ of this adjunction look like? Let us recall the general construction:

• Suppose we have an adjunction $F ⊣ U$ between two functors $\newcommand{\cat}{\mathscr} F \colon \cat{C} \to \cat{D}$ and $U \colon \cat{D} \to \cat{C}$, given by bijections $$ Φ_{X, Y} \colon \Hom_{\cat{D}}(F(X), Y) \longto \Hom_{\cat{C}}(X, U(Y)) \,. $$ The unit $η$ and counit $ε$ are then natural transformations $$ η \colon \mathrm{Id}_{\cat{C}} \Longrightarrow U F \,, \quad ε \colon F U \Longrightarrow \mathrm{Id}_{\cat{D}}. $$
• For every object $X$ of $\cat{C}$, the component $η_X$ is a morphism from $X$ to $UF(X)$. Under the bijection $$ Φ_{X, F(X)} \colon \Hom_{\cat{D}}(F(X), F(X)) \longto \Hom_{\cat{C}}(X, UF(X)) \,, $$ the morphism $η_X$ corresponds to $\newcommand{\id}{\mathrm{id}} \id_{F(X)}$.
• For every object $Y$ of $\cat{D}$, the component $ε_Y$ is a morphism from $FU(Y)$ to $Y$. Under the bijection $$ Φ_{U(Y), Y} \colon \Hom_{\cat{D}}(FU(Y), Y) \longto \Hom_{\cat{C}}(U(Y), U(Y)) \,, $$ the morphism $ε_Y$ corresponds to $\id_{U(Y)}$.

In our specific example, this means the following:

• For every group $G$, the component $η_G$ is a homomorphism of groups from $G$ to $UF(G) = G^{\ab}$. It is the image of $\id_{G^{\ab}}$ under the bijection $$ Φ_{G, G^{\ab}} \colon \Hom_{\Ab}(G^{\ab}, G^{\ab}) \longto \Hom_{\Grp}(G, G^{\ab}) \,, \quad ψ \longmapsto ψ ∘ p_G \,. $$ Therefore, $η_G = p_G$ is the canonical homomorphism from $G$ to $G^{\ab}$. In formulas, we have $$ η_A \colon A \longto A^{\ab} \,, \quad a \longmapsto [a] \,. $$

• For every abelian group $A$, the component $ε_A$ is a homomorphism of groups from $FU(A) = A^{\ab}$ to $A$. Under the bijection $$ Φ_{A, A} \colon \Hom_{\Ab}(A^{\ab}, A) \longto \Hom_{\Grp}(A, A) \,, \quad ψ \longmapsto ψ ∘ p_A \,. $$ the homomorphism $ε_A$ corresponds to $\id_A$. In other words, we have $ε_A ∘ p_A = \id_A$. The homomorphism $ε_A$ is therefore given by $$ ε_A \colon A^{\ab} \longto A \,, \quad [a] \longmapsto a \,. $$

Via triangle identities

We can also use the universal property of the abelianization to first construct $η$ and $ε$, and then check that the triangle identities are satisfied:

• We already have for every group $G$ a special homomorphism from $G$ to $G^{\ab}$, namely the canonical homomorphism $p_G$. So let $η_G := p_G$, i.e., $$ η_G \colon G \longto G^{\ab} \,, \quad g \longmapsto [g] \,. $$ We need to check that these homomorphisms assemble into a natural transformation $η$ from $\mathrm{Id}_{\Grp}$ to $U ∘ F$. To this end, we need to check that for every homomorphism of groups $φ \colon G \to H$ the following diagram commutes. $$ \require{AMScd} \begin{CD} G @> φ >> H \\ @V η_G VV @VV η_H V \\ G^{\ab} @>> φ^{\ab} > H^{\ab} \end{CD} \tag{$\ast$} $$ But this is precisely the commutative diagram $(\ast)$ from the first section of this post.

• We need for every abelian group $A$ a homomorphism $ε_A$ from $A^{\ab}$ to $A$. By the universal property of the abelianization $A^{\ab}$, the desired homomorphism $ε_A \colon A^{\ab} \to A$ corresponds to a homomorphism $A \to A$. There is an obvious homomorphism $A \to A$, namely the identity homomorphism, so we choose $ε_A$ as the corresponding homomorphism from $A^{\ab}$ to $A$. More explicitly, $$ ε_A \colon A^{\ab} \longto A \,, \quad [a] \longmapsto a \,. $$ We need to check that the resulting transformation $ε$ from $UF$ to $\mathrm{Id}_{\Ab}$ is natural. In other words, we need to check that for every homomorphism of abelian groups $ψ \colon A \to B$ the following diagram commutes: $$ \require{AMScd} \begin{CD} A^{\ab} @> ψ^{\ab} >> B^{\ab} \\ @V ε_A VV @VV ε_B V \\ A @>> ψ > B \end{CD} \tag{$\ast$} $$ This holds true because $$ ε_B( ψ^{\ab}( [a] ) ) = ε_B( [ψ(a)] ) = ψ(a) = ψ( ε_A( [a] ) ) \,. $$

Now that we have constructed the natural transformations $η$ and $ε$, we check that they satisfy the two triangle identities $$ εF ∘ Fη = \id_F \quad\text{and}\quad Uε ∘ ηU = \id_U \,. $$

• For every group $G$ we have \begin{align*} (εF ∘ Fη)_G( [g] ) &= ε_{G^{\ab}}( (η_G)^{\ab}( [g] ) ) \\ &= ε_{G^{\ab}}( [η_G(g)] ) \\ &= ε_{G^{\ab}}( [[g]] ) \\ &= [g] \\ &= \id_{G^{\ab}}( [g] ) \\ &= (\id_F)_G( [g] ) \end{align*} for every element $g$ of $G$, therefore $(εF ∘ Fη)_G = (\id_F)_G$ and thus $εF ∘ Fη = \id_F$. This shows the first triangle identity.

• For every abelian group $A$ we have \begin{align*} (Uε ∘ ηU)_A( a ) = ε_A( η_A( a ) ) = ε_A( [a] ) = a = \id_A( a ) = (\id_U)_A( a ) \end{align*} for every element $a$ of $A$, therefore $(Uε ∘ ηU)_A = (\id_U)_A$, and thus $Uε ∘ ηU = \id_U$. This shows the second triangle identity.

We can also retrieve the natural bijection $$ Φ_{G, A} \colon \Hom_{\Ab}(G^{\ab}, A) \to \Hom_{Grp}(G, A) $$ via $η$ and $ε$: the bijection $Φ_{G, A}$ is given by $U(-) ∘ η_G = (-) ∘ η_G$, and its inverse is given by $ε_A ∘ F(-) = ε_A ∘ (-)^{\ab}$.