Since the puzzles from [last time]( were quite substantial, let me copy the nice solutions provided by Anindya Bhattyacharyya and Matthew Doty here. They add up to the proof of this:

**Theorem.** Suppose \\(\mathcal{V}\\) is a commutative quantale. Then there is a category \\(\mathbf{Prof}_\mathcal{V}\\) whose objects are \\(\mathcal{V}\\)-enriched categories and whose morphisms are \\(\mathcal{V}\\)-enriched profunctors, where the composite of \\(\mathcal{V}\\)-enriched profunctors \\(\Phi \colon \mathcal{X} \nrightarrow \mathcal{Y} \\) and \\(\Psi \colon \mathcal{Y} \to \mathcal{Z}\\) is defined by

\[ (\Psi\Phi)(x,z) = \bigvee_{y \in \mathrm{Ob}(\mathcal{Y})} \Phi(x,y) \otimes \Psi(y,z).\]

We'll break the proof into lots of lemmas. In all that follows, \\(\mathcal{V}\\) is a commutative quantale and \\(\mathcal{W}, \mathcal{X}, \mathcal{Y}, \mathcal{Z}\\) are \\(\mathcal{V}\\)-enriched categories. Letters like \\(w,w',x,x',y,y',z,z'\\) will always stand for objects in the enriched categories with the same letters as their names - and when I write inequalities involving these I mean they're true _for all choices_ of these objects, unless I say otherwise.

First we check that the formula above really defines a \\(\mathcal{V}\\)-enriched profunctor!

**Lemma.** Suppose \\(\Phi \colon \mathcal{X} \nrightarrow \mathcal{Y} \\) and \\(\Psi \colon \mathcal{Y} \to \mathcal{Z}\\) are \\(\mathcal{V}\\)-enriched profunctors. Then the formula

\[ (\Psi\Phi)(x,z) = \bigvee_{y} \Phi(x,y) \otimes \Psi(y,z) \]

specifies a \\(\mathcal{V}\\)-enriched profunctor \\(\Psi\Phi \colon \mathcal{X} \to \mathcal{Y}\\).

**Proof.** We need to show that

\[ \Psi\Phi \colon \mathcal{X}^{\text{op}} \times \mathcal{Z} \to \mathcal{V} \]

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

\[ (\mathcal{X}^\text{op}\times \mathcal{Z})((x, z), (x', z')) \leq \mathcal{V}(\Psi\Phi)(x, z), (\Psi\Phi)(x', z')) \]

or in other words

\[ \mathcal{X}(x', x) \otimes \mathcal{Z}(z, z') \leq (\Psi\Phi)(x, z)\multimap(\Psi\Phi)(x', z') .\]

Using the definitions this becomes

\[ \mathcal{X}(x', x) \otimes \left(\bigvee_y \Phi(x, y)\otimes\Psi(y, z)\right) \otimes\mathcal{Z}(z, z') \leq \bigvee_y \Phi(x', y)\otimes\Psi(y, z') \]

or since \\(\otimes\\) distributes over \\(\bigvee\\) in a quantale,

\[ \bigvee_y \mathcal{X}(x', x) \otimes \Phi(x, y)\otimes\Psi(y, z) \otimes\mathcal{Z}(z, z') \leq \bigvee_y \Phi(x', y)\otimes\Psi(y, z'). \]

This is true because

\[ \mathcal{X}(x', x) \otimes \Phi(x, y) \le \Phi(x',y) \]


\[ \Psi(y, z) \otimes\mathcal{Z}(z, z') \le \Phi(y,z') \]

by the following theorem. \\( \qquad \blacksquare \\)

This is a very handy tool:

**Theorem.** The following are equivalent:

1. \\(\Phi : \mathcal{X} \nrightarrow \mathcal{Y}\\) is a \\(\mathcal{V}\\)-enriched profunctor.

2. \\(\mathcal{X}(x', x) \otimes \Phi(x, y) \otimes \mathcal{Y}(y, y') \leq \Phi(x',y')\\) for all \\(x,x',y,y'\\).

3. \\(\mathcal{X}(x', x) \otimes \Phi(x, y) \leq \Phi(x', y)\\) and \\(\Phi(x, y) \otimes \mathcal{Y}(y, y') \leq \Phi(x, y')\\) for all \\(x,x',y,y'\\).

**Proof.** Item 1 is true iff \\(\Phi : \mathcal{X}^\text{op} \times \mathcal{Y} \to \mathcal{V}\\) is a \\(\mathcal{V}\\)-functor, which by definition is true iff

\[ (\mathcal{X}^\text{op} \times \mathcal{Y})((x, y), (x', y')) \leq \mathcal{V}(\Phi(x, y), \Phi(x',y')) \]

for all \\(x,x',y,y'\\). Using the definitions, this is equivalent to

\[ \mathcal{X}(x', x) \otimes \mathcal{Y}(y, y') \leq \Phi(x, y) \multimap \Phi(x', y') .\]

By the definition of\\(\multimap\\) this in turn is equivalent to

\[ \Phi(x, y) \otimes \mathcal{X}(x', x) \otimes \mathcal{Y}(y, y') \leq \Phi(x', y') \]

Because \\(\otimes\\) is commutative this is the same as item 2. Thus, 1 is equivalent to 2.

Next note that \\(I \leq \mathcal{X}(x, x)\\) and \\(I \leq \mathcal{Y}(y, y)\\) by the definition of a \\(\mathcal{V}\\)-category. Thus, given 2 we can set \\(y' = y\\) to get

\[ \mathcal{X}(x', x) \otimes \Phi(x, y) = \mathcal{X}(x', x) \otimes \Phi(x, y) \otimes I \leq \mathcal{X}(x', x) \otimes \Phi(x, y) \otimes \mathcal{Y}(y, y) \leq \Phi(x', y) \]

or we can set \\(x' = x\\) to get

\[ \Phi(x, y) \otimes \mathcal{Y}(y, y') = I \otimes \Phi(x, y) \otimes \mathcal{Y}(y, y') \leq \mathcal{X}(x, x) \otimes \Phi(x, y) \otimes \mathcal{Y}(y, y') \leq \Phi(x, y') \]

Thus, 2 implies 3. On the other hand, using 3 we can show 2:

\[ \mathcal{X}(x', x) \otimes \Phi(x, y) \otimes \mathcal{Y}(y, y') \leq \Phi(x', y) \otimes \mathcal{Y}(y, y') \leq \Phi(a, b) \]

Thus, all three conditions are equivalent. \\( \qquad \blacksquare \\)

Next we check that composition is associative:

**Lemma.** Suppose \\(\Theta \colon \mathcal{W} \nrightarrow \mathcal{X}, \Phi \colon \mathcal{X} \nrightarrow \mathcal{Y} \\) and \\(\Psi \colon \mathcal{Y} \to \mathcal{Z}\\) are \\(\mathcal{V}\\)-enriched profunctors. Then

\[ (\Psi \Phi) \Theta = \Psi (\Phi \Theta) .\]

**Proof.** We need to show

\[ \bigvee_{x} \Theta(x,y) \otimes \left( \bigvee_{y} \Phi(x,y) \otimes \Psi(y,z) \right) = \bigvee_{y} \left( \bigvee_{x} \Theta(x,y) \otimes \Phi(x,y) \right) \otimes \Psi(y,z) .\]

Because \\(\otimes\\) distributes over \\(\bigvee\\), it's enough to show

\[ \bigvee_{x} \bigvee_{y} \Theta(x,y) \otimes \left( \Phi(x,y) \otimes \Psi(y,z) \right) = \bigvee_{y} \bigvee_{x} \Theta(x,y) \otimes \Phi(x,y) \otimes \Psi(y,z) .\]

Since \\(\otimes\\) is associative, and we always have \\(\bigvee_{x} \bigvee_{y} = \bigvee_{y} \bigvee_{x} \\) when the joins in question exist, this is true. \\( \qquad \blacksquare \\)

Next we show our would-be category \\(\mathbf{Prof}_\mathcal{V}\\) has identity morphisms.

**Lemma.** The \\(\mathcal{V}\\)-enriched functor \\( \mathrm{hom} \colon \mathcal{X}^{\text{op}} \times \mathcal{X} \to \mathcal{V} \\), defined by

\[ \mathrm{hom}(x,x') = \mathcal{X}(x,x') ,\]

corresponds to a \\(\mathcal{V}\\)-enriched profunctor

\[ 1\_{\mathcal{X}} \colon \mathcal{X} \nrightarrow \mathcal{X} \]

that serves as an identity for composition.

**Proof.** We need to check the left and right unit laws, but they are very similar so we'll only do one:

\[ \Phi 1_{\mathcal{X}} = \Phi \]

for any \\(\mathcal{V}\\)-enriched profunctor \\(\Phi \colon \mathcal{X} \to \mathcal{Y}\\). This amounts to proving

\[ \bigvee_{x'} \mathcal{X}(x,x') \otimes \Phi(x',y) = \Phi(x,y) \]

First we'll show

\[ \Phi(x,y) \le \bigvee_{x'} \mathcal{X}(x,x') \otimes \Phi(x',y) \]

and then we'll show the reverse inequality. Since \\(\mathcal{V}\\) is a poset this will mean the two sides are equal.

For the inequality above it's enough to find one choice of \\(x'\\) that makes

\[ \Phi(x,y) \le \mathcal{X}(x,x') \otimes \Phi(x',y) . \]

The obvious guess is \\(x' = x\\); then we need

\[ \Phi(x,y) \le \mathcal{X}(x,x) \otimes \Phi(x,y) . \]

But the [definition of enriched category]( says that \\( I \le \mathcal{X}(x,x)\\), so

\[ \Phi(x,y) = I \otimes \mathcal{\Phi}(x,y) \le \mathcal{X}(x,x) \otimes \Phi(x,y) \]

as desired. Next we show the reverse inequality:

\[ \bigvee_{x'} \mathcal{X}(x,x') \otimes \Phi(x',y) \le \Phi(x,y) .\]

For this it's enough to prove that for _all_ \\(x'\\) we have

\[ \mathcal{X}(x,x') \otimes \Phi(x',y) \le \Phi(x,y) .\]

We've seen this in the theorem above, so we're done. \\( \qquad \blacksquare \\)

Whew - that was a good workout! #:-S

**[To read other lectures go here.](**