There are lots of ways to build new categories from old, and most of these 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)**