I am trying to show that every finitely generated free group is a cogroup object in the category of groups. (Note I believe that this is also true for non-finite free groups, but that is probably much harder to prove - I think it is theorem of Kan's)
I know that the coproduct in the category of groups is just the free product. So then is it just basically show that the free product satisfies the usual diagrams for co-associativity, co-identity and co-inverse?
I guess I am confused by the map $C \to C \coprod C$ (can't get my head around co-multiplication yet).
This was OK in the category of abelian groups - as the coproduct is the direct product and the there is an obvious map $c \mapsto (c,c)$. But (and maybe I am struggling with the notion of a free product here), if I tried something similar - take an element from a finitely generated free group (say on $n$ generators) and then take the coproduct with itself - what does that give? Wouldn't it just leave a single element? (As all adjacent elements belong to the same set the reduced word cancels everything out?)
Not looking for an answer, I guess just some clarity - I am sure I am not understand something properly.
(I guess one could also do this by showing that $\mbox{Hom}(\quad , G)$ take a group structure?)
Edit: Actually is $C \ast C = C$? ($\ast$ is the free product) (No - as Theo points out!)