I am currently on the road, so please excuse my poor formatting.

This theorem does not fix \\(\mathcal{V}\\). I have been reading your blog article from a few years ago and it’s references to acquaint myself with enriched categories, Simon. I see now that not fixing \\(\mathcal{V}\\) is unorthodox. I am sorry, I didn’t intend to defy convention.

Consider some \\(\mathcal{X}\\) enriched in some \\(\mathcal{V}\\). Since I am not fixing a \\(\mathcal{V}\\), I can’t talk about a particular preorder \\(\le\\) when comparing \\(\mathcal{X}(a,b)\\) and \\(\mathcal{X}(c,d)\\). Because of this, I thought it would make since to clarify the order associated with \\(\mathcal{X}\\) as \\(\le_\mathcal{X}\\).

A ‘monoidal preorder over a discrete poset’ is a monoidal preorder where the underlying preorder is a discrete poset. That is not a great definition. I will attempt to clarify.

In a discrete poset, \\(a \le b\\) is true if and only if \\(a = b\\).

Every monoid can be made into a monoidal preorder over a discrete poset. Just construct the trivial poset over the domain of the monoid.

A simple example would be \\(\mathbb{Z}/(2)\\). Since discrete posets are equivalence relations, we can represent them with their equivalence classes. In this case the equivalence class is {{0},{1}}.

I will try to go into how to construct an enriched category for the discretely ordered \\(\mathbb{Z}/(2)\\) group or any other discretely ordered group later.