Michael Hong wrote:

>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?

I think we're free to choose between,
\\[
x \sim y \mapsto x \leq y,
\\]

and,

\\[
x \sim y \mapsto x = y,
\\]

Either way, the operation is destructive.