Lecture 64  The Category of Enriched Profunctors
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) ]
and
[ \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:

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

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

\(\mathcal{X}(x', x) \otimes \Phi(x, y) \leq \Phi(x', y)\) 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 wouldbe 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
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