Today I will finally define enriched profunctors. For this we need two ways to build enriched categories.

There are lots of ways to build new categories from old, and most work for \$$\mathcal{V}\$$-enriched categories too _if_ \$$\,\mathcal{V}\$$ is nice enough. We ran into two of these constructions for categories in [Lecture 52](https://forum.azimuthproject.org/discussion/2273/lecture-52-the-hom-functor/p1) when discussing the hom-functor

$\mathrm{hom}: \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathbf{Set} .$

To make sense of this we needed to show that we can take the 'product' of categories, and that every category has an 'opposite'. Now we're trying to understand \$$\mathcal{V}\$$-enriched profunctors \$$\Phi : \mathcal{X} \nrightarrow \mathcal{Y}\$$, which are really just \$$\mathcal{V}\$$-enriched functors

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

To make sense of this we need the same two constructions: the product and the opposite!

(This is no coincidence: soon we'll see that every \$$\mathcal{V}\$$-enriched category \$$\mathcal{X}\$$ has a hom-functor \$$\mathrm{hom} : \mathcal{X}^{\text{op}} \times \mathcal{X} \to \mathcal{V}\$$, which we can think of as a profunctor.)

So, first let's look at the product of enriched categories:

**Theorem.** Suppose \$$\mathcal{V}\$$ is a commutative monoidal poset. Then for any \$$\mathcal{V}\$$-enriched categories \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$, there is a \$$\mathcal{V}\$$-enriched category \$$\mathcal{X} \times \mathcal{Y}\$$ for which:

1. An object is a pair \$$(x,y) \in \mathrm{Ob}(\mathcal{X}) \times \mathrm{Ob}(\mathcal{Y}) \$$.

2. We define

$(\mathcal{X} \times \mathcal{Y})((x,y), \, (x',y')) = \mathcal{X}(x,x') \otimes \mathcal{Y}(y,y') .$

**Proof.** We just need to check axioms a) and b) of an enriched category (see [Lecture 29](https://forum.azimuthproject.org/discussion/2121/lecture-29-chapter-2-enriched-categories/p1)):

a) We need to check that for every object \$$(x,y) \$$ of \$$\mathcal{X} \times \mathcal{Y}\$$ we have

$I \le (\mathcal{X} \times \mathcal{Y})((x,y), \, (x,y)) .$

By item 2 this means we need to show

$I \le \mathcal{X}(x,x) \otimes \mathcal{Y}(y,y) .$

But since \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$ are enriched categories we know

$I \le \mathcal{X}(x,x) \text{ and } I \le \mathcal{Y}(y,y)$

and tensoring these two inequalities gives us what we need.

b) We need to check that for all objects \$$(x,y), (x',y'), (x'',y'') \$$ of \$$\mathcal{X} \times \mathcal{Y}\$$ we have

$(\mathcal{X} \times \mathcal{Y})((x,y), \, (x',y')) \otimes (\mathcal{X} \times \mathcal{Y})((x',y'), \, (x'',y'')) \le (\mathcal{X} \times \mathcal{Y})((x,y), \, (x'',y'')) .$

This looks scary, but long division did too at first! Just relax and follow the rules. To get anywhere we need to rewrite this using item 2:

$\mathcal{X}(x,x') \otimes \mathcal{Y}(y,y') \otimes \mathcal{X}(x',x'') \otimes \mathcal{Y}(y',y'') \le \mathcal{X}(x,x'') \otimes \mathcal{Y}(y,y'') . \qquad (\star)$

But since \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$ are enriched categories we know

$\mathcal{X}(x,x') \otimes \mathcal{X}(x',x'') \le \mathcal{X}(x,x'')$

and

$\mathcal{Y}(y,y') \otimes \mathcal{Y}(y',y'') \le \mathcal{Y}(y,y'') .$

Let's tensor these two inequalities and see if we get \$$(\star) \$$. Here's what we get:

$\mathcal{X}(x,x') \otimes \mathcal{X}(x',x'') \otimes \mathcal{Y}(y,y') \otimes \mathcal{Y}(y',y'') \le \mathcal{X}(x,x'') \otimes \mathcal{Y}(y,y'') .$

This is _almost_ \$$(\star) \$$, but not quite. To get \$$(\star) \$$ we need to switch two things in the middle of the left-hand side! But we can do that because \$$\mathcal{V}\$$ is a _commutative_ monoidal poset. \$$\qquad \blacksquare \$$

Next let's look at the opposite of an enriched category:

**Theorem.** Suppose \$$\mathcal{V}\$$ is a monoidal poset. Then for any \$$\mathcal{V}\$$-enriched category \$$\mathcal{X}\$$ there is a \$$\mathcal{V}\$$-enriched category \$$\mathcal{X}^{\text{op}}\$$, called the **opposite** of \$$\mathcal{X}\$$, for which:

1. The objects of \$$\mathcal{X}^{\text{op}}\$$ are the objects of \$$\mathcal{X}\$$.

2. We define

$\mathcal{X}^{\text{op}}(x,x') = \mathcal{X}(x',x) .$

**Proof.** Again we need to check [axioms a) and b) of an enriched category](https://forum.azimuthproject.org/discussion/2121/lecture-29-chapter-2-enriched-categories/p1).

a) We need to check that for every object \$$x \$$ of \$$\mathcal{X}^{\text{op}}\$$ we have

$I \le \mathcal{X}^{\text{op}}(x,x) .$

Using the definitions, this just says that for every object \$$x \$$ of \$$\mathcal{X}\$$ we have

$I \le \mathcal{X}(x,x) .$

This is true because \$$\mathcal{X}\$$ is an enriched category.

b) We also need to check that for all objects \$$x,x',x'' \$$ of \$$\mathcal{X}^{\text{op}} \$$ we have

$\mathcal{X}^{\text{op}} (x,x') \otimes \mathcal{X}^{\text{op}}(x',x'') \le \mathcal{X}^{\text{op}}(x,x'') .$

Using the definitions, this just says that for all objects \$$x,x',x'' \$$ of \$$\mathcal{X}\$$ we have

$\mathcal{X}(x',x) \otimes \mathcal{X}(x'',x') \le \mathcal{X}(x'',x) .$

We can prove this as follows:

$\mathcal{X}(x',x) \otimes \mathcal{X}(x'',x') = \mathcal{X}(x'',x') \otimes \mathcal{X}(x',x) \le \mathcal{X}(x'',x) .$

since \$$\mathcal{V}\$$ is commutative and \$$\mathcal{X}\$$ is an enriched category. \$$\qquad \blacksquare \$$

Now we are ready to state the definition of an enriched profunctor! I gave a tentative definition back in [Lecture 60](https://forum.azimuthproject.org/discussion/2287/lecture-60-chapter-4-closed-monoidal-posets/p1), but we didn't really know what it meant, nor under which conditions it made sense. Now we do!

**Definition.** Suppose \$$\mathcal{V}\$$ is a closed commutative monoidal poset and \$$\mathcal{X},\mathcal{Y}\$$ are \$$\mathcal{V}\$$-enriched categories. Then a \$$\mathcal{V}\$$-enriched profunctor

$\Phi : \mathcal{X} \nrightarrow \mathcal{Y}$

is a \$$\mathcal{V}\$$-enriched functor

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

There are lot of adjectives here: "closed commutative monoidal poset". They're all there for a reason. Luckily we've seen our friends \$$\mathbf{Bool}\$$ and \$$\mathbf{Cost}\$$ have all these nice properties - and so do many other examples, like the power set \$$P(X)\$$ of any set \$$X\$$.

Alas, if we want to _compose_ \$$\mathcal{V}\$$-enriched profunctors we need \$$\mathcal{V}\$$ to be even _nicer!_ From our work with feasibility relations we saw that composing profunctors is done using a kind of 'matrix multiplication'. For this to work, \$$\mathcal{V}\$$ needs to be a 'quantale'. So, next time I'll talk about quantales. Luckily all the examples I just listed are quantales!

Here's a puzzle to keep you happy until next time. It's important:

**Puzzle 197.** Suppose \$$\mathcal{V}\$$ is a closed commutative monoidal poset and \$$\mathcal{X}\$$ is any \$$\mathcal{V}\$$-enriched category. Show that there is a \$$\mathcal{V}\$$-enriched functor, the **hom functor**

$\mathrm{hom} \colon \mathcal{X}^{\text{op}} \times \mathcal{X} \to \mathcal{V}$

defined on any object \$$(x,x') \$$ of \$$\mathcal{X}^{\text{op}} \times \mathcal{X} \$$ by

$\mathrm{hom}(x,x') = \mathcal{X}(x,x') .$

If you forget the definition of enriched functor, you can find it in [Lecture 32](https://forum.azimuthproject.org/discussion/2169/lecture-32-chapter-2-enriched-functors/p1).

**[To read other lectures go here.](http://www.azimuthproject.org/azimuth/show/Applied+Category+Theory#Chapter_4)**