Lecture 59 - Cost-Enriched Profunctors

# Lecture 59 - Cost-Enriched Profunctors

We've been looking at feasibility relations, as our first example of enriched profunctors. Now let's look at another example. This combines many ideas we've discussed - but don't worry, I'll review them, and if you forget some definitions just click on the links to earlier lectures!

Remember, $$\mathbf{Bool} = \lbrace \text{true}, \text{false} \rbrace$$ is the preorder that we use to answer true-or-false questions like

• can we get from here to there?

while $$\mathbf{Cost} = [0,\infty]$$ is the preorder that we use to answer quantitative questions like

• how much does it cost to get from here to there?

or

• how far is it from here to there?

In $$\textbf{Cost}$$ we use $$\infty$$ to mean it's impossible to get from here to there: it plays the same role that $$\text{false}$$ does in $$\textbf{Bool}$$. And remember, the ordering in $$\textbf{Cost}$$ is the opposite of the usual order of numbers! This is good, because it means we have

[ \infty \le x \text{ for all } x \in \mathbf{Cost} ]

just as we have

[ \text{false} \le x \text{ for all } x \in \mathbf{Bool} .]

Now, $$\mathbf{Bool}$$ and $$\mathbf{Cost}$$ are monoidal preorders, which are just what we've been using to define enriched categories! This let us define

and

We can draw preorders using graphs, like these:

An edge from $$x$$ to $$y$$ means $$x \le y$$, and we can derive other inequalities from these. Similarly, we can draw Lawvere metric spaces using $$\mathbf{Cost}$$-weighted graphs, like these:

The distance from $$x$$ to $$y$$ is the length of the shortest directed path from $$x$$ to $$y$$, or $$\infty$$ if no path exists.

All this is old stuff; now we're thinking about enriched profunctors between enriched categories.

A $$\mathbf{Bool}$$-enriched profunctor between $$\mathbf{Bool}$$-enriched categories also called a feasibility relation between preorders, and we can draw one like this:

What's a $$\mathbf{Cost}$$-enriched profunctor between $$\mathbf{Cost}$$-enriched categories? It should be no surprise that we can draw one like this:

You can think of $$C$$ and $$D$$ as countries with toll roads between the different cities; then an enriched profunctor $$\Phi : C \nrightarrow D$$ gives us the cost of getting from any city $$c \in C$$ to any city $$d \in D$$. This cost is $$\Phi(c,d) \in \mathbf{Cost}$$.

But to specify $$\Phi$$, it's enough to specify costs of flights from some cities in $$C$$ to some cities in $$D$$. That's why we just need to draw a few blue dashed edges labelled with costs. We can use this to work out the cost of going from any city $$c \in C$$ to any city $$d \in D$$. I hope you can guess how!

Puzzle 182. What's $$\Phi(E,a)$$?

Puzzle 183. What's $$\Phi(W,c)$$?

Puzzle 184. What's $$\Phi(E,c)$$?

Here's a much more challenging puzzle:

Puzzle 185. In general, a $$\mathbf{Cost}$$-enriched profunctor $$\Phi : C \nrightarrow D$$ is defined to be a $$\mathbf{Cost}$$-enriched functor

[ \Phi : C^{\text{op}} \times D \to \mathbf{Cost} ]

This is a function that assigns to any $$c \in C$$ and $$d \in D$$ a cost $$\Phi(c,d)$$. However, for this to be a $$\mathbf{Cost}$$-enriched functor we need to make $$\mathbf{Cost}$$ into a $$\mathbf{Cost}$$-enriched category! We do this by saying that $$\mathbf{Cost}(x,y)$$ equals $$y - x$$ if $$y \ge x$$, and $$0$$ otherwise. We must also make $$C^{\text{op}} \times D$$ into a $$\mathbf{Cost}$$-enriched category, which I'll let you figure out to do. Then $$\Phi$$ must obey some rules to be a $$\mathbf{Cost}$$-enriched functor. What are these rules? What do they mean concretely in terms of trips between cities?

And here are some easier ones:

Puzzle 186. Are the graphs we used above to describe the preorders $$A$$ and $$B$$ Hasse diagrams? Why or why not?

Puzzle 187. I said that $$\infty$$ plays the same role in $$\textbf{Cost}$$ that $$\text{false}$$ does in $$\textbf{Bool}$$. What exactly is this role?

By the way, people often say $$\mathcal{V}$$-category to mean $$\mathcal{V}$$-enriched category, and $$\mathcal{V}$$-functor to mean $$\mathcal{V}$$-enriched functor, and $$\mathcal{V}$$-profunctor to mean $$\mathcal{V}$$-enriched profunctor. This helps you talk faster and do more math per hour.

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