The counit of an adjunction of the adjoint pair free and underlying functor I would like to know how the counit $\varepsilon$ for an adjunction $(F,U,\phi)$
$$\varepsilon:FUX\to X$$
works if $F$ is the free functor from $\mathbf{Set}$ to $\mathbf{Ab}$ and $U$ is the underlying functor from $\mathbf{Ab}$ to $\mathbf{Set}$?
I.e. what element in $X$ do I assign to the term $$x+y$$ say, if $$x,y\in X$$ by $$\varepsilon(x+y)$$?
 A: For an adjunction $L \dashv R$ we can always recover the counit (for an object $X$) $\varepsilon_X: LR(X) \to X$ by looking at what $Id_{R(X)}: R(X) \to R(X)$ corresponds to under the adjunction. That is, the adjunction is a natural ismorphism
$$
\operatorname{Hom}(F(X), Y) \cong \operatorname{Hom}(X, R(Y)),
$$
so in particular we have
$$
\operatorname{Hom}(LR(X), X) \cong \operatorname{Hom}(R(X), R(X)).
$$
Considering $Id_{R(X)} \in \operatorname{Hom}(R(X), R(X))$, this must correspond to some $LR(X) \to X$, which will be precisely $\varepsilon_X$.
So how does this apply to the adjunction $F: \mathbf{Ab} \rightleftarrows \mathbf{Set}: U$? A function $f: X \to U(G)$ (for a set $X$ and an abelian group $G$) corresponds to a group homomorphism $\tilde{f}: F(X) \to G$ by defining $\tilde{f}(x) = f(x)$ for all $x \in X$. Then since $F(X)$ is the free abelian group on the generator set $X$, this extends uniquely to a group homomorphism $\tilde{f}: F(X) \to G$. So in particular, for an abelian group $G$, $FU(G)$ is the free group generated by the set of elements in $G$. Then the identity function $Id_G: G \to G$ extends uniquely to a group homomorphism $FU(G) \to G$, and this is precisely $\varepsilon_G$.

Note that the entire story above applies dually to finding the unit $\epsilon_X: X \to RL(X)$ of an adjunction $L \dashv R$.
