The group coequalizer of doubling and zero on the integers is not its underlying-set coequalizer

Reduction modulo $2$ coequalizes $f$ and $g$. If a group homomorphism $h:\mathbb{Z}\to H$ satisfies $hf=hg$, then $h(2n)$ is the identity for every $n$, so $h$ kills $2\mathbb{Z}$ and factors uniquely through $\mathbb{Z}\mathrm{/}2\mathbb{Z}$. Thus this is the group coequalizer by [L1]–[L2].

In $\mathbf{S}\mathbf{e}\mathbf{t}$, the generated equivalence relation identifies every even integer with $0$, because the generating pairs are $(2n,0)$. No odd integer occurs in such a pair, so each odd integer remains a singleton equivalence class. The quotient map coequalizes $f$ and $g$, and every function $u$ with $uf=ug$ is constant on the even class and therefore factors uniquely through this quotient, proving its Set universal property.

The set coequalizer is therefore infinite, while the underlying set of the group coequalizer has the residue classes of $0$ and $1$. The underlying-set functor, including the one in the free-group adjunction of [L3], does not preserve this coequalizer.