# 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?

## 1 Answer

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 $m''$.

Of course there's another method which is much faster, but offers less insight, and isn't guaranteed to work -- guessing :)

• 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
• @goblin: Well, if that's what you get by the calculation described above, then yes. But sure, even without computation, I'd say it's the most reasonable guess, and pairing it with the other reasonable guess (coinverse = letterwise inverse), you can see that the cogroup axioms will be satisfied. Of course, you should also check that this really is the right choice for the coinverse morphism. It's the same thing as above, only simpler since it's unary. Counit on the other hand's no effort at all, since Grp has a zero object. I can also fill in some of the details if you want me to. – user54748 Jun 6 '15 at 20:21