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:

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

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

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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