# How is the free group on $S$ generators a cogroup?

According to nLab:

Cogroup objects in the category of groups are free groups, and to give a free group the structure of a cogroup object is the same a choosing a generating set. This is an old result of D.M. Kan’s.

What I understand: write $F$ for the free functor $\mathbf{Set} \rightarrow \mathbf{Grp}$. If $S$ is a set and $G$ is a group, then the set $\mathbf{Grp}(F(S),G)$ becomes a group in a natural way, since:

$$\mathbf{Grp}(F(S),G) \cong \mathbf{Set}(S,U(G)) \cong (UG)^S \cong U(G^S)$$

Hence we can induce a group structure on the set $\mathbf{Grp}(F(S),G)$ from the group $G^S$.

What I don't understand: how does this make $F(S)$ into a cogroup? In particular, what are the comultiplication, counit, and coinverse mappings?

You just need to follow the isomorphisms involved. You know what the group structure on $G^S$ is, so you know, for example, the multiplication function $$m : UG^S × UG^S → UG^S.$$ You know what the isomorphism $UG^S ≅ \mathrm{Hom}(FS, G)$ is, so you can calculate $$m' : \mathrm{Hom}(FS, G) × \mathrm{Hom}(FS, G) → \mathrm{Hom}(FS, G),$$ and from there $$m'' : \mathrm{Hom}(FS ⊔ FS, G) → \mathrm{Hom}(FS, G).$$ This is natural in $G$, so by Yoneda lemma it comes from a morphism $$m''' : FS → FS ⊔ FS.$$ Now just remembering the "by Yoneda lemma" part won't do you any good, but Yoneda lemma or at least its proof tells you exactly how to calculate $m'''$ from $m''$.
• Is this correct? The homomorphism $m''' : FS \rightarrow FS \sqcup FS$ is described by the corresponding function $S \rightarrow FS \sqcup FS$, which seems to be the map $s \mapsto \eta_0(s)\eta_1(s)$, where $\eta_0$ is the "left" inclusion of $S$ into $FS \sqcup FS$ and $\eta_1$ is the "right" inclusion. Is this correct? I can't see what else it would be... – goblin Jun 6 '15 at 17:10