Matthew wrote:

> If we go instead with

> \[ x \otimes a \le y \text{ if and only if } a \le x \multimap y \tag{✴'} \]

> Then your quick proof in [#6]( works, the theorem we proved holds without demanding commutativity, and \\((x\otimes -) \dashv (x\multimap -)\\) like you say.

This is an interesting issue. I was using the convention in Fong and Spivak's book (see Section 2.5.1), and I should stick with that to avoid confusion. It could be that in the noncommutative case the convention you mention here works better. I will avoid trouble by the simple expedient of assuming \\(\mathcal{V}\\) is commutative. But there's more to say...

The book mainly focuses on categories enriched over commutative monoidal posets - or symmetric monoidal preorders, which work more or less the same way. Things get more fussy if we drop the assumption of commutativity, or symmetry, and certain things just don't work at all.

For example, a \\(\mathcal{V}\\)-enriched category has an 'opposite' when \\(\mathcal{V}\\) is symmetric monoidal but not, I believe, when \\(\mathcal{V}\\) is merely monoidal! To define composition in the would-be opposite you need to switch two things in \\(\mathcal{V}\\past each other.

Similarly, we can take the product of two \\(\mathcal{V}\\)-enriched categories when \\(\mathcal{V}\\) is symmetric monoidal but not, I believe, when \\(\mathcal{V}\\) is merely monoidal. Again, to define composition we need to switch two things in \\(\mathcal{V}\\) past each other.

We will see these things soon.

For these two reasons it's hopeless (I believe) to develop the theory of enriched profunctors when \\(\mathcal{V}\\) is merely monoidal. Thus, I could have spared myself some grief by assuming \\(\mathcal{V}\\) is a commutative monoidal poset throughout this lecture. But in fact closed monoidal categories, and closed monoidal posets, are interesting even in the noncommutative case! So I couldn't resist getting into trouble.

I will rewrite the lecture to avoid this issue by assuming commutativity. The reason is that in the noncommutative case one must really distinguish between **left closed** and **right closed** monoidal posets. In one, the operation \\(a \otimes -\\) has a right adjoint for each \\(a\\). In the other, the operation \\( - \otimes a\\) has a right adjoint for each \\(a\\).

(When we want to distinguish these two, we do it by turning around the lollipop in \\(\multimap\\). I don't know how to do that in this TeX installation!)

**Puzzle.** You can think of a monoid as a 'discrete' monoidal poset, meaning one where \\(x \le y\\) iff \\(x = y\\). Show that a monoid which is _group_ gives a monoidal poset that is both left closed and right closed.

A monoidal poset (or monoidal category) that's both left and right closed is called **biclosed**.

You're raising the issue of whether it's _left closed_ or _right closed_ monoidal posets that become enriched over themselves... or both... or _neither_. You may have already settled this, but my instinct is that if one works, the other must too! The reason is that I don't believe the concept of "enriched category" has a left/right asymmetry built into it: I don't think there are "left" and "right" enriched categories.