A \(\mathcal{V}\) functor must have \(\mathcal{V}\) categories for domain and codomain, so the \(\mathcal{V}\) of a \(\mathcal{V}\) profunctor must be considered as enriched in itself.
Matrix algorithm for computing distance matrix of profunctor: \(d_X * M_\phi * d_Y\)
A \(\mathcal{V}\) profunctor, denoted \(\mathcal{X}\overset{\phi}{\nrightarrow} \mathcal{Y}\) - where \(\mathcal{V}=(V,\leq,I,\otimes)\) is a (unital commutative) quantale, and \(\mathcal{X},\mathcal{Y}\) are \(\mathcal{V}\) categories.
A \(\mathcal{V}\) functor \(\mathcal{X}^{op}\times Y \xrightarrow{\phi} \mathcal{V}\)
Bool-profunctors and their interpretation as bridges
Let’s consider Bool-profunctors. Recall a preorder (Bool-category) can be drawn as a Hasse diagram.
A Bool-profunctor \(X \overset{\phi}{\nrightarrow} Y\) can look like this:
With bridges coming from the profunctor, one can now use both paths to get from points in \(X\) to points in \(Y\).
There is a path from N to e, so \(\phi(N,e)=\)\(true\) but \(\phi(W,d)=\)\(false\).
We could put a box around both preorders and define a new preorder, called the collage.
Cost profunctors and their interpretation as bridges.
Cost categories are Lawvere metric spaces and can be represented by weighted graphs
This is a depiction of a Cost-profunctor
The distance from a point in the left to a point in the right will run through the left, across a bridge, then through through the right.
\(\phi(B,x)=\)\(11\),\(\phi(A,z)=\)\(20\),\(\phi(C,y)=\)\(17\)
Is it true that a Bool-profunctor is exactly the same as a feasibility relation?
Monotone maps are Bool-functors between Bool-categories, so the definitions line up
Fill out the Cost-matrix associated with Example 4.13 NOCARD
\(\phi\) | x | y | z |
---|---|---|---|
A | 17 | 21 | 20 |
B | 11 | 15 | 14 |
C | 14 | 18 | 17 |
D | 12 | 9 | 15 |
Show that a \(\mathcal{V}\) profunctor is the same as a function \(Ob(\mathcal{X})\times Ob(\mathcal{Y}) \xrightarrow{\phi} V\) such that, \(\forall x,x' \in \mathcal{X}, y,y' \in \mathcal{Y}\), the following holds in \(\mathcal{V}\):
\(\mathcal{X}(x',x)\otimes \phi(x,y) \otimes \mathcal{Y}(y,y') \leq \phi(x',y')\)
A \(\mathcal{V}\) profunctor must be a function satisfying the following constraint, according to the \(\mathcal{V}\) functor definition:
\(Z((x,y),(x',y')) \leq\) \(\mathcal{V}(\phi((x,y)),\phi((x',y')))\)
where \(Z = \mathcal{X}^{op}\times \mathcal{Y}\)
Unpacking the definition of a product \(\mathcal{V}\) category, we obtain
\(\mathcal{X}^{op}(x,x') \otimes \mathcal{Y}(y,y') \leq \mathcal{V}(\phi((x,y)),\phi((x',y')))\)
And applying opposite category definition: \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y') \leq \mathcal{V}(\phi((x,y)),\phi((x',y')))\)
Noting the definition of \(\multimap\) for a \(\mathcal{V}\) category enriched in itself:
\(\mathcal{V}(v,w)=v\multimap w\), so now we have: \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y') \leq \phi((x,y)) \multimap \phi((x',y'))\)
From the constraint of a hom-element of a symmetric monoidal preorder \(\mathcal{V}\), i.e. \(a \leq (v \multimap w)\) iff \((a \otimes v) \leq w\), we see that the first case pattern matches with:
\(a \mapsto\) \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y')\)
\(v \mapsto\) \(\phi((x,y))\)
\(w \mapsto\) \(\phi((x',y'))\)
So using the iff we can rewrite as \((a \otimes v) \leq w\), and use the commutativity of \(\otimes\) to obtain the desired expression.