I have a question about the forgetful functor \\(\mathbf{Preord} \stackrel{R_2}{\longrightarrow} \mathbf{Poset} \\). How would this functor forget 2-cycles? Since we are keeping the objects, it seems to me there can be a choice of which arrow to forget but don't think it can be mapped to the other arrow coming back since source and target won't match. Does the whole cycle just get forgotten ie mapped into an identity?

Upon further pondering, I have another question. When going from \\(\mathbf{Poset} \stackrel{R_2}{\longrightarrow} \mathbf{Set} \\), how exactly does the functor forget the arrows? Where do they get mapped to? In order to preserve source and target, all connected objects need to get mapped to one object?

Upon further pondering, I have another question. When going from \\(\mathbf{Poset} \stackrel{R_2}{\longrightarrow} \mathbf{Set} \\), how exactly does the functor forget the arrows? Where do they get mapped to? In order to preserve source and target, all connected objects need to get mapped to one object?