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2 Addressed comment to question; edited body

At least if one takes labeled graphs (LGrphs) and labeled digraphs the functor you suggest, say D, is right adjoint to the forgetful functor which I'll call U. There is a canonical natural transformation UD -> Id_{LGrphs} which just collapses the doubled edges to the ones they came from. For a labeled digraph Q and a labeled graph G the bijection on hom-sets then sends a map UQ -> G to the unique lift Q -> DG which sends each directed edge of Q to the corresponding edge of the appropriate direction which lifts the assignment coming from UQ -> G. I guess when you just define morphisms in terms of adjacency or directed adjacency and edges are indistinguishable then it is still fine?

My guess without thinking much for a free digraph construction, i.e.

And the comment to your question indicates that there can't be a left adjointto the forgetful functor would be to do something like sending G to the coproduct (disjoint union) of all possible ways of directing G. But it isn't obvious to me (at least right now) if this actually defines , so no free digraph just a functorcofree one.

1

At least if one takes labeled graphs (LGrphs) and labeled digraphs the functor you suggest, say D, is right adjoint to the forgetful functor which I'll call U. There is a canonical natural transformation UD -> Id_{LGrphs} which just collapses the doubled edges to the ones they came from. For a labeled digraph Q and a labeled graph G the bijection on hom-sets then sends a map UQ -> G to the unique lift Q -> DG which sends each directed edge of Q to the corresponding edge of the appropriate direction which lifts the assignment coming from UQ -> G. I guess when you just define morphisms in terms of adjacency or directed adjacency and edges are indistinguishable then it is still fine?

My guess without thinking much for a free digraph construction, i.e. a left adjoint to the forgetful functor would be to do something like sending G to the coproduct (disjoint union) of all possible ways of directing G. But it isn't obvious to me (at least right now) if this actually defines a functor.