Lecture 30 - Preorders as Enriched Categories

# Lecture 30 - Preorders as Enriched Categories

Okay, last time I dumped the definition of "enriched category" on you. Now let's figure out what it does! To get going we need to pick a monoidal preorder $$\mathcal{V}$$. Then:

Definition. A $$\mathcal{V}$$-enriched category, say $$\mathcal{X}$$, consists of:

1. a set of objects, $$\mathrm{Ob}(\mathcal{X})$$, and

2. an element $$\mathcal{X}(x,y) \in \mathcal{V}$$ for any pair of objects $$x,y$$.

such that:

a) $$I\leq\mathcal{X}(x,x)$$ for any object $$x$$, and

b) $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$ for any objects $$x,y,z$$.

We often say "$$\mathcal{V}$$-category" instead of $$\mathcal{V}$$-enriched category, just to save time. Life is short.

To understand this definition, we need examples! The simplest interesting example is $$\mathcal{V} = \mathbf{Bool}$$. This is the set of Booleans, or "truth values":

[ \mathbf{Bool} = \{ \texttt{true}, \texttt{false} \} . ]

It's a poset where $$x \le y$$ means "$$x$$ implies $$y$$". Namely, $$\texttt{false} \le \texttt{true}$$ and everything is less than or equal to itself, and that's all. We make it monoidal by taking $$x \otimes y$$ to mean "$$x$$ and $$y$$", so the unit $$I$$ is $$\texttt{true}$$. Let's see what this gives.

Theorem. A $$\mathbf{Bool}$$-category is a preorder.

Proof. Let's start with a $$\mathbf{Bool}$$-category and see what it actually amounts to. A $$\mathbf{Bool}$$-enriched category, say $$\mathcal{X}$$, consists of:

1. a set of objects, $$\mathrm{Ob}(\mathcal{X})$$ and

2. an element $$\mathcal{X}(x,y) \in \mathbf{Bool}$$ for any two objects $$x,y$$

So, we get a truth value for each pair $$x,y$$. Let's make up a relation $$\le$$ on $$\mathrm{Ob}(\mathcal{X})$$ such that $$x \le y$$ is true when $$\mathcal{X}(x,y) = \texttt{true}$$, and false when $$\mathcal{X}(x,y) = \texttt{false}$$.

Next:

a) $$I\leq\mathcal{X}(x,x)$$ for any object $$x$$.

In our example this means

[ \texttt{true} \textrm{ implies that } x \le x . ]

In other words, $$x \le x$$ is true. So this is just the reflexive law for $$\le$$!

b) $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$ for any objects $$x,y,z$$.

In our example this means

[ x \le y \textrm{ and } y \le z \textrm{ implies } x \le z .]

So this is just the transitive law for $$\le$$!

It's easy to turn this argument around and get the converse: any preorder gives a $$\mathbf{Bool}$$-category $$\mathcal{X}$$ if we define $$\mathcal{X}(x,y) = \texttt{true}$$ when $$x \le y$$ and $$\mathcal{X}(x,y) = \texttt{false}$$ otherwise. $$\qquad \blacksquare$$

Yee-hah! We've managed to define preorders in a new, more abstract, more confusing way!

But of course that's not our goal. That was just a warmup. A much cooler example occurs when we take $$\mathcal{V} = \mathbf{Cost}$$. This is the set of costs, or non-negative real numbers including infinity:

$$\mathbf{Cost} = [0,\infty] .$$

It's a poset where the order is defined in the opposite of the usual way: we use $$\ge$$ instead of $$\le$$. It becomes a monoidal using addition, with $$0$$ as the unit. In case you're nervous about infinity, this means that $$x + \infty = \infty = \infty + x$$ and $$x \le \infty$$ for all $$x$$.

Puzzle 88. Show that these choices actually makes $$\mathbf{Cost}$$ into a symmetric monoidal poset.

Puzzle 89. Figure out exactly what a $$\mathbf{Cost}$$-category is, starting with the definition above. Again, what matters is not your final answer so much as your process of deducing it!

Hint: given a $$\mathbf{Cost}$$-category $$\mathcal{X}$$, write $$\mathcal{X}(x,y)$$ as $$d(x,y)$$ and call it the distance between the objects $$x,y \in \mathrm{Ob}(\mathcal{X})$$. The axioms of an enriched category then say interesting things about "distance".

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© 2018 John Baez