Lecture 56 - Feasibility Relations

# Lecture 56 - Feasibility Relations

Let's start trying to understand enriched profunctors. We'll start with a very nice special case: 'feasibility relations'.

Definition. Suppose $$(X, \le_X)$$ and $$(Y, \le_Y)$$ are preorders. Then a feasibility relation from $$X$$ to $$Y$$ is a monotone function

[ \Phi : X^{\text{op}} \times Y \to \mathbf{Bool} .]

If $$\Phi$$ is a feasibility relation from $$X$$ to $$Y$$ we write $$\Phi : X\nrightarrow Y$$. If $$\Phi(x,y) = \text{true}$$, we say $$x$$ can be obtained given $$y$$.

The idea is that we use elements of $$X$$ to describe 'requirements' - things you want - and elements of $$Y$$ to describe 'resources' - things you have. A feasibility relation $$\Phi : X \nrightarrow Y$$ says when what you want can be obtained from what you have! And the fact that it's monotone makes a lot of sense.

In fact:

Theorem. A function $$\Phi : X^{\text{op}} \times Y \to \mathbf{Bool}$$ is a feasibility relation if and only if:

1. If $$\Phi(x,y) = \text{true}$$ and $$x' \le_X x$$ then $$\Phi(x',y) = \text{true}$$.

2. If $$\Phi(x,y) = \text{true}$$ and $$y \le_Y y'$$ then $$\Phi(x,y') = \text{true}$$.

Translating this into English, we see it makes perfect sense:

1. If what you want can be obtained from the resources you have, and then you change your mind and want less, you can still obtain what you want.

2. If what you want can be obtained from the resources you have, and then you acquire more resources, you can still obtain what you want.

But let's prove the theorem. This is mainly a nice review of various concepts.

Proof. First, remember that $$\textbf{Bool}$$ is the preorder with two elements $$\text{true}$$ and $$\text{false}$$, with

[ \text{false} \le \text{true} . ]

We can read $$\le$$ here as 'implies'.

Second, remember that $$X^{\text{op}}$$ is the opposite of the preorder $$X$$, with the definition of $$\le$$ turned around:

[ x \le_{X^{\text{op}}} x' \text{ if and only if } x' \le_X x ]

Third, remember that $$X^{\text{op}} \times Y$$ is the product of the preorders $$X^{\text{op}}$$ and $$Y$$. So, its elements are pairs $$(x,y)$$ with $$x \in X$$ and $$y \in Y$$, and we define a concept of $$\le$$ on these pairs by

[ (x,y) \le (x',y') \text{ if and only if } x' \le_X x \text{ and } y \le_Y y' .]

Note how I said $$x' \le x$$. This looks backwards, but it's not one of my usual typos! It works this way because of the $$\text{op}$$ in $$X^{\text{op}} \times Y$$.

Now we're ready to see exactly what a feasibility relation, that is a monotone function $$\Phi : X^{\text{op}} \times Y \to \mathbf{Bool}$$, really is. It's a function that obeys

[ (x,y) \le (x',y') \text{ implies } \Phi(x,y) \le \Phi(x',y') ]

or in other words

[ \text{ if } x' \le_X x \text{ and } y \le_Y y', \text{ then } \Phi(x,y) \text{ implies } \Phi(x',y') ]

Translating this into English to see what this means, it says:

If you can get what you want from the resources you have, and then you change your mind and want less, and also go out and get more resources, then you can still get what you want.

Here "less" really means "less than or equal to", and "more" really means "greater than or equal to" - English is not very good at saying these things quickly! So, the process of wanting less and getting more resources can always be broken into two steps:

1. wanting less but keeping the resources the same, and then

2. getting more resources but wanting the same thing.

So, this condition

[ \text{ if } x' \le_X x \text{ and } y \le_Y y', \text{ then } \Phi(x,y) \text{ implies } \Phi(x',y') ]

is equivalent to the combination of both these conditions:

1. If $$x' \le_X x$$, then $$\Phi(x,y)$$ implies $$\Phi(x',y)$$.

2. If $$y \le y'$$, then $$\Phi(x,y)$$ implies $$\Phi(x,y')$$.

But here's another equivalent way to say these two things:

1. If $$\Phi(x,y) = \text{true}$$ and $$x' \le_X x$$ then $$\Phi(x',y) = \text{true}$$.

2. If $$\Phi(x,y) = \text{true}$$ and $$y \le_X y'$$ then $$\Phi(x,y') = \text{true}$$.

Logic! Ain't it great? $$\qquad \blacksquare$$

Next time I'll show you how to draw pictures of feasibility relations, and look at some examples. We've already drawn pictures of preorders, or at least posets: they're called Hasse diagrams, and they look like a bunch of dots, one for each element of our poset $$X$$, and a bunch of arrows, enough so that $$x \le y$$ whenever there's a path of arrows leading from $$x$$ to $$y$$. So, to draw a feasibility relation $$\Phi : X \nrightarrow Y$$, we'll draw two Hasse diagrams and some extra arrows to say when $$\Phi(x,y) = \text{true}$$.

Finally, two puzzles:

Puzzle 169. I gave a verbal argument for how we can break up any inequality $$(x,y) \le (x',y')$$ in $$X^{\text{op}} \times Y$$ into two other inequalities. Can you write this out in a purely mathematical way?

Puzzle 170. What if we have a morphism in a product of categories $$\mathcal{C} \times \mathcal{D}$$. Can we always write it as a composite of two morphisms, copying the procedure in my verbal argument and Puzzle 169? How does this work?

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