The free group $F(S)$ is the group given by a set $S$ with the universal property: For every group $G$ and map $f: S \to G$ there is a unique homomorphism $\phi: F(S) \to G$, such that $\phi \circ i = f$ where $i: S \to F(S)$ is the inclusion.

If we ignore that $S$ is a set and $F(S)$ and $G$ are groups, $F(S)$ could be the colimit of the diagram with exactly one object.

Is this true? Is the universal property generally a colimit of a diagram with one object? And if it's true how can one precise this given that $S, F(S)$ and $G$ are not in the same category?


The category you look for is the cograph of the profunctor $U^*:{\bf Set}^{op}\times{\bf Grp}\to{\bf Set}$ induced by the forgetful functor $U:{\bf Grp}\to{\bf Set}$, i.e. $U^*(S,G)=\hom_{\bf Set}(S,UG)$.
It basically connects the categories ${\bf Set}$ and ${\bf Grp}$ with further morphisms from sets to groups: an arrow $S\to G$ will be just a function $S\to UG$ in this case.

And, the free group $F(S)$ is the reflection of $S$ in the subcategory ${\bf Grp}$ of the cograph of $U^*$.

Finally, the colimit (or limit) of any single object $S$ in any category is $S$ itself (or any isomorphic).

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