Categorial definition of free products? If $X$ and $Y$ are objects of a concrete category $\mathcal{C}$, is there an accepted definition of "free product of $X$ and $Y$," generalizing the  in the special case where $\mathcal{C}$ is the concrete category of groups? See also, free product of groups.
Edit. In $\mathbf{Grp},$ the free product is just the coproduct, but I see no reason why this should be the "right" definition of "free product" general. Free products ought to have a special relationship to free objects. If we can find that relationship, then we can either prove that coproducts always enjoy that relationship, or else, find a case where they do not.
 A: Yes of course, even for any category: it is called the coproduct :)
A: Most notions of "freeness" start with a functor $G:\mathcal C\to \mathcal D$ and finding an adjoint $F:\mathcal D\to \mathcal C$ with $\operatorname{Hom}_{\mathcal C}(FD,C)\equiv \operatorname{Hom}_{\mathcal D}(D,GC)$. For example, with $\mathcal C$ the category of groups, and $\mathcal D$ the category of sets, and $G$ the forgetful functor sending a group to the underlying set.
Then the diagonal functor $G:\mathcal C\to \mathcal C^2$ with $C\mapsto (C,C)$ has as its corresponding $F$ the coproduct.
A: For the question "what's the link between a free product and a free object?", the answer is given in the wikipedia page, before "contents":

The point is that a disjoint union of groups is not a group but it is a groupoid. A groupoid  G has a universal group  U(G)  and the universal group of a disjoint union of groups is the free (= coproduct) of the groups.

Remarks:


*

*It seems that here $U(-)$ denotes the free functor left adjoint to the forgetful functor $ \mathbf{Grp}\longrightarrow \mathbf{Grpoid}$

*The disjoint union is the in $\mathbf{Grpoid}$
To me, the OP's question is obviously: is there a distinct categorical notion of free product (distinct from coproduct in general but such that for groups they coincide, exactly as coproduct and products in other examples)?
