# 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.

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

$$\text{Hom}_{\text{Grp}}(F(S), G) \cong \text{Hom}_{\text{Set}}(S, U(G)).$$

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).

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.

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 $${\bf Grp}$$.