Keith is right.

Fredrick wrote:

> I was not understanding what was being 'enriched' by what. Based on comments I think I am not the only one. Here is my informal definition of \$$\mathcal{X}\$$ a \$$\mathcal{V}\$$-enriched category. You start with a category \mathcal{X}\\).

No! _An enriched category is not, in general, a category_.

Don't be fooled by the terminology, or your eagerness to start talking about categories!

A category has a _set_ for any pair of objects, but our enriched categories have a mere _element of the set \$$\mathcal{V}\$$_ for any pair of objects!

Later we will look at more general enriched categories, which include both categories and the enriched categories we're studying now as special cases. However, even when we get to that situation, you must be careful. _An enriched category will not, in general, be a category_.

It might been wiser to call the things we're studying so far "enriched preorders". A plain old preorder has a _truth value_ for any pair of objects - that is, an element of \$$\textbf{Bool}\$$. We are "enriching" this idea - in the sense of making it more exciting - by replacing \$$\textbf{Bool}\$$ by an arbitrary symmetric monoidal preorder \$$\mathcal{V}\$$.

However, even the term "enriched preorder" could confuse you into thinking that every enriched preorder is a preorder. _It's not!_ After all, there's no reason that having an element of \$$\mathcal{V}\$$ gives you an element of \$$\textbf{Bool}\$$ in any interesting way.

I suddenly remember being confused by this issue myself, once upon a time. I was confused because there's a certain bunch of examples of enriched categories that _are_ categories... which we haven't talked about in this course yet. That's the reason for this terminology. But don't let the terminology fool you! [-X