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](https://forum.azimuthproject.org/discussion/2121/lecture-29-chapter-2-enriched-categories/p1). We can also define \\(\mathcal{V}\\)-enriched functors whenever \\(\mathcal{V}\\) is a monoidal preorder: we did that in [Lecture 31](https://forum.azimuthproject.org/discussion/2169/lecture-32-chapter-2-enriched-functors/p1). 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](https://forum.azimuthproject.org/discussion/2292/lecture-62-chapter-4-constructing-enriched-categories/p1).

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](https://en.wikipedia.org/wiki/Ouroboros), 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]\\)](https://forum.azimuthproject.org/discussion/2128/lecture-31-chapter-2-lawvere-metric-spaces/p1). 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](https://forum.azimuthproject.org/discussion/2121/lecture-29-chapter-2-enriched-categories/p1) 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.](http://www.azimuthproject.org/azimuth/show/Applied+Category+Theory#Chapter_4)**

We can define \\(\mathcal{V}\\)-enriched categories whenever \\(\mathcal{V}\\) is a monoidal preorder: we did that way back in [Lecture 29](https://forum.azimuthproject.org/discussion/2121/lecture-29-chapter-2-enriched-categories/p1). We can also define \\(\mathcal{V}\\)-enriched functors whenever \\(\mathcal{V}\\) is a monoidal preorder: we did that in [Lecture 31](https://forum.azimuthproject.org/discussion/2169/lecture-32-chapter-2-enriched-functors/p1). 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](https://forum.azimuthproject.org/discussion/2292/lecture-62-chapter-4-constructing-enriched-categories/p1).

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](https://en.wikipedia.org/wiki/Ouroboros), 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]\\)](https://forum.azimuthproject.org/discussion/2128/lecture-31-chapter-2-lawvere-metric-spaces/p1). 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](https://forum.azimuthproject.org/discussion/2121/lecture-29-chapter-2-enriched-categories/p1) 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.](http://www.azimuthproject.org/azimuth/show/Applied+Category+Theory#Chapter_4)**