Lecture 60 - Closed Monoidal Preorders

Okay, now I've rather carefully discussed one example of \(\mathcal{V}\)-enriched profunctors, and rather sloppily discussed another. Now it's time to build the general framework that can handle both these examples.

We can define \(\mathcal{V}\)-enriched categories whenever \(\mathcal{V}\) is a monoidal preorder: we did that way back in Lecture 29. We can also define \(\mathcal{V}\)-enriched functors whenever \(\mathcal{V}\) is a monoidal preorder: we did that in Lecture 31. But to define \(\mathcal{V}\)-enriched profunctors, we need \(\mathcal{V}\) to be a bit better. We can see why by comparing our examples.

Our first example involved \(\mathcal{V} = \textbf{Bool}\). A feasibility relation

[ \Phi : X \nrightarrow Y ]

between preorders is a monotone function

[ \Phi: X^{\text{op}} \times Y\to \mathbf{Bool} . ]

We shall see that a feasibility relation is the same as a \( \textbf{Bool}\)-enriched profunctor.

Our second example involved \(\mathcal{V} = \textbf{Cost}\). I said that a \( \textbf{Cost}\)-enriched profunctor

[ \Phi : X \nrightarrow Y ]

between \(\mathbf{Cost}\)-enriched categories is a \( \textbf{Cost}\)-enriched functor

[ \Phi: X^{\text{op}} \times Y \to \mathbf{Cost} ]

obeying some conditions. But I let you struggle to guess those conditions... without enough clues to make it easy!

To fit both our examples in a general framework, we start by considering an arbitrary monoidal preorder \(\mathcal{V}\). \(\mathcal{V}\)-enriched profunctors will go between \(\mathcal{V}\)-enriched categories. So, let \(\mathcal{X}\) and \(\mathcal{Y}\) be \(\mathcal{V}\)-enriched categories. We want to make this definition:

Tentative Definition. A \(\mathcal{V}\)-enriched profunctor

[ \Phi : \mathcal{X} \nrightarrow \mathcal{Y} ]

is a \(\mathcal{V}\)-enriched functor

[ \Phi: \mathcal{X}^{\text{op}} \times \mathcal{Y} \to \mathcal{V} .]

Notice that this handles our first example very well. But some questions appear in our second example - and indeed in general. For our tentative definition to make sense, we need three things:

1. We need \(\mathcal{V}\) to itself be a \(\mathcal{V}\)-enriched category.

2. We need any two \(\mathcal{V}\)-enriched category to have a 'product', which is again a \(\mathcal{V}\)-enriched category.

3. We need any \(\mathcal{V}\)-enriched category to have an 'opposite', which is again a \(\mathcal{V}\)-enriched category.

Items 2 and 3 work fine whenever \(\mathcal{V}\) is a commutative monoidal poset. We'll see why in Lecture 62.

Item 1 is trickier, and indeed it sounds rather scary. \(\mathcal{V}\) began life as a humble monoidal preorder. Now we're wanting it to be enriched in itself! Isn't that circular somehow?

Yes! But not in a bad way. Category theory often eats its own tail, like the mythical ourobous, and this is an example.

To get \(\mathcal{V}\) to become a \(\mathcal{V}\)-enriched category, we'll demand that it be 'closed'. For starters, let's assume it's a monoidal poset, just to avoid some technicalities.

Definition. A monoidal poset is closed if for all elements \(x,y \in \mathcal{V}\) there is an element \(x \multimap y \in \mathcal{V}\) such that

[ x \otimes a \le y \text{ if and only if } a \le x \multimap y ]

for all \(a \in \mathcal{V}\).

This will let us make \(\mathcal{V}\) into a \(\mathcal{V}\)-enriched category by setting \(\mathcal{V}(x,y) = x \multimap y \). But first let's try to understand this concept a bit!

We can check that our friend \(\mathbf{Bool}\) is closed. Remember, we are making it into a monoidal poset using 'and' as its binary operation: its full name is \( \lbrace \text{true},\text{false}\rbrace, \wedge, \text{true})\). Then we can take \( x \multimap y \) to be 'implication'. More precisely, we say \( x \multimap y = \text{true}\) iff \(x\) implies \(y\). Even more precisely, we define:

[ \text{true} \multimap \text{true} = \text{true} ] [ \text{true} \multimap \text{false} = \text{false} ] [ \text{false} \multimap \text{true} = \text{true} ] [ \text{false} \multimap \text{false} = \text{true} . ]

Puzzle 188. Show that with this definition of \(\multimap\) for \(\mathbf{Bool}\) we have

[ a \wedge x \le y \text{ if and only if } a \le x \multimap y ]

for all \(a,x,y \in \mathbf{Bool}\).

We can also check that our friend \(\mathbf{Cost}\) is closed! Remember, we are making it into a monoidal poset using \(+\) as its binary operation: its full name is \( [0,\infty], \ge, +, 0)\). Then we can define \( x \multimap y \) to be 'subtraction'. More precisely, we define \(x \multimap y\) to be \(y - x\) if \(y \ge x\), and \(0\) otherwise.

Puzzle 189. Show that with this definition of \(\multimap\) for \(\mathbf{Cost}\) we have

[ a + x \le y \text{ if and only if } a \le x \multimap y . ]

But beware. We have defined the ordering on \(\mathbf{Cost}\) to be the opposite of the usual ordering of numbers in \([0,\infty]\). So, \(\le\) above means the opposite of what you might expect!

Next, two more tricky puzzles. Next time I'll show you in general how a closed monoidal poset \(\mathcal{V}\) becomes a \(\mathcal{V}\)-enriched category. But to appreciate this, it may help to try some examples first:

Puzzle 190. What does it mean, exactly, to make \(\mathbf{Bool}\) into a \(\mathbf{Bool}\)-enriched category? Can you see how to do this by defining

[ \mathbf{Bool}(x,y) = x \multimap y ]

for all \(x,y \in \mathbf{Bool}\), where \(\multimap\) is defined to be 'implication' as above?

Puzzle 191. What does it mean, exactly, to make \(\mathbf{Cost}\) into a \(\mathbf{Cost}\)-enriched category? Can you see how to do this by defining

[ \mathbf{Cost}(x,y) = x \multimap y ]

for all \(x,y \in \mathbf{Cost}\), where \(\multimap\) is defined to be 'subtraction' as above?

Note: for Puzzle 190 you might be tempted to say "a \(\mathbf{Bool}\)-enriched category is just a preorder, so I'll use that fact here". However, you may learn more if you go back to the general definition of enriched category and use that! The reason is that we're trying to understand some general things by thinking about two examples.

Puzzle 192. The definition of 'closed' above is an example of a very important concept we keep seeing in this course. What is it? Restate the definition of closed monoidal poset in a more elegant, but equivalent, way using this concept.

To read other lectures go here.