# Lecture 32 - Enriched Functors

Whenever we study some kind of mathematical gadget, we should also study *maps between* gadgets of this kind. For example:

- When we study sets, we should also study functions.
- When we study preorders, we should also study monotone functions.
- When we study monoids, we should also study monoid homomorphisms.
- When we study monoidal preorders, we should also study monoidal monotones.

The reason is that mathematical structures don't live in isolation - or at least, they don't want to! They like to live in communities: they like to talk to each other.

Now we're studying enriched categories. So we should also study maps between enriched categories. These are called "enriched functors".

To get going, we need to choose a monoidal preorder \(\mathcal{V}\). I defined \(\mathcal{V}\)-categories at the end of Lecture 30. And now I'll define maps between these:

**Definition.** Let \(\mathcal{X}\) and \(\mathcal{Y}\) be \(\mathcal{V}\)-categories. A **\(\mathcal{V}\)-functor from \(\mathcal{X}\) to \(\mathcal{Y}\)**, denoted \(F\colon\mathcal{X}\to\mathcal{Y}\), is a function

[ F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y}) ]

such that

[ \mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x')) ]

for all \(x,x' \in\mathrm{Ob}(\mathcal{X})\).

Remember, both \(\mathcal{X}(x,x')\) and \(\mathcal{Y}(F(x),F(x'))\) are elements of \(\mathcal{V}\), which is a monoidal preorder. So, it makes sense to say that one is less than or equal to the other.

As always when meeting a new definition, it's good to look at examples. What's \(\mathbf{Bool}\)-functor between \(\mathbf{Bool}\)-categories?

We've seen a \(\mathbf{Bool}\)-category \(\mathcal{X}\) is just a preorder, where we say \(x \le x'\) iff \(\mathcal{X}(x,y) = \texttt{true}\). So, the obvious guess is that a \(\mathbf{Bool}\)-functor is just a monotone function.

Why is this obvious? Well, \(\mathbf{Bool}\)-functors go between \(\mathbf{Bool}\)-categories, which are preorders, and monotone functions also go between preorders... so life would be simple if these were the same thing!

And indeed it's true! Suppose we have a \(\mathbf{Bool}\)-functor \(F\colon\mathcal{X}\to\mathcal{Y}\). This is a function

[ F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y}) ]

such that

[ \mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x')). ]

But what does this last inequality mean? Remember, both \(\mathcal{X}(x,x') \) and \( \mathcal{Y}(F(x),F(x'))\) are elements of \( \mathbf{Bool} = \{\texttt{true}, \texttt{false} \} \). In \( \mathbf{Bool} \), the symbol \(\leq\) means "implies". When we make \(\mathcal{X} \) into a preorder, \(\mathcal{X}(x,x') = \texttt{true}\) means that \(x \le x'\). Similarly, \( \mathcal{Y}(F(x),F(x')) = \texttt{true} \) means that \(F(x) \le F(x')\). So, the inequality just says

[ x \le x' \textrm{ implies } F(x) \le F(x') . ]

This says that \(F\) is a monotone function!

Why don't you try one now? We've seen a \(\mathbf{Cost}\)-category is a Lawvere metric space.

**Puzzle 92.** What's a \(\mathbf{Cost}\)-functor?

It should be some kind of nice map between Lawvere metric spaces; this can help you *guess* the answer, but then you should check your answer.

**To read other lectures go here.**