What category is the universal property of the Free Group a diagram in? wikipedia says that the free group is defined by a universal property:

The free group $F_S$ is the universal group generated by the set $S$. This can be formalized by the following universal property: given any function $f$ from $S$ to a group $G$, there exists a unique homomorphism $φ: F_S → G$ making the following diagram commute (where the unnamed mapping denotes the inclusion from $S$ into $F_S$):




My question is, in what category is this a diagram? Is it in Grp or Set? Either way I'm confused, because $S$ is not a group, suggesting it's in Set, but the uniqueness of $\phi$ only holds for homomorphisms, not general functions, suggesting this is in Grp.
 A: As the definition mentions, $f$ and the unnamed inclusions are just functions while $\varphi$ is a group homomorphism. Hence the diagram is not in $\mathbf{Grp}$, nor actually in $\mathbf{Set}$ (in the sense that the diagram in $\mathbf{Set}$ would not force $\varphi$ to be a group homomorphism).
The construction gives in fact a functor from $\mathbf{Set}$ to $\mathbf{Grp}$ assigning to each set $S$ the free group $F_S$, and to each function $g:S\to T$ the morphism $\varphi_g:F_S \to F_T$ associated to the map $f=\iota_T\circ g:S\to F_T$ by the universal property (where $\iota_T:T\to F_T$ is the inclusion).
A: I often think of the free group over $S$ to be the initial object in the category of groups with $S$ specified points, or more formally the category of groups $(G, *)$ along with a specified function from $S$ to $G$, where morphisms consist of a group homomorphism that makes the functions from $S$ agree.
This definition captures what the diagram is trying to convey: it is showing initiality of $F_S$ where the morphism consists of the whole wedge coming out of $S$.
As other answers have said, this construction results in a functor from $\mathbf{Set}$ to $\mathbf{Grp}$, which is left adjoint to the forgetful functor, but I don't believe this context (while fascinating, and points to many interesting generalizations) is necessary to understand the free group.
A: As you say, $S$ is a set, so this is a diagram in $\text{Set}$. The fact that we force $\varphi$ to be a homomorphism of groups is extra structure that isn't captured by the diagram alone.
You might consider this unsatisfying, so alternatively we can explicitly name the forgetful functor $U : \text{Grp} \to \text{Set}$ from groups to sets, which is being implicitly applied to $G$ here, and regard $f$ as a morphism $f : S \to U(G)$ in $\text{Set}$, then talk about the universal property in terms of the adjunction
$$\text{Hom}_{\text{Grp}}(F(S), G) \cong \text{Hom}_{\text{Set}}(S, U(G)).$$
A: Though the other answers tell how it could be interpreted either in ${\bf Set}$ or in ${\bf Grp}$, using the adjoint functors $U$ (implicitly) or $F$, a third construction exists answering properly this question:
Take the disjoint union of categories ${\bf Set}$ (draw it to the left) and ${\bf Grp}$ (draw to the right), and for any set $S$ and group $G$, add the functions $S\to U(G)$ as (so called hetero-) morphisms $S\to G$.
All occuring compositions are function compositions.
This construction is known as the cograph (or collage) of the profunctor
$$U^*:{\bf Set}^{op}\times{\bf Grp}\to{\bf Set}\ (S,G)\mapsto \hom_{\bf Set}(S,UG)\,.$$
The left adjoint, $F$ of $U$, can be alternatively described by reflections on the subcategory ${\bf Grp}$.
