# On Profunctors, Abruptly

02-15-2018 |

Disclaimer: This post reqiures a working understanding of functors in Haskell terms.

## Introduction (Theses)

If we’re comfortable with the notion of a functor in Haskell, we can intuitively infer what might be meant by a bifunctor. Consider the kind of a type that is a functor: * -> *. That is, it takes one type as an argument and returns a type. A bifunctor’s kind is one which takes two types and returns a type: * -> * -> *. Its definition looks like this:

class Bifunctor (b :: * -> * -> *) where
bimap :: (a -> b) -> (c -> d) -> f a c -> f b d

A bifunctor is useful for mapping over product types, such as (,):

instance Bifunctor (,) where
bimap f  g (x, y) = (f x, g y)

Now is a good time to take note of the direction of the arrows. We begin with an f a b and end up with a f c d, by way of two functions, a -> b, and c -> d. In both of their cases, the arrows go the same direction: forward.

1. The a -> b follows the same path as the f a _ -> f b _ in the bifunctor f.
2. Like the function before it, c -> d follows the direction of the functor as well f _ c -> f _ d.

## Antitheses

This direction-following is called covariance, and our Bifunctor is called a covariant functor. However, like all good things, category theory yields to the Hegelian Spirit and cannot achieve synthesis without both a thesis and an antithesisHere’s a Twitter thread from when I discovered Lambek and Lawvere’s explorations of Heraclitus’s and/or Hegel’s influence on mathematics.

. We have two antitheses two derive, so let us begin with a regression as a vehicle to the first.

### Antithesis: Category

In category theory, A category $$C$$ has an opposite category wherein the objects are the same, but the arrows are reversed. $$C$$’s opposite category would be denoted as $$C^{op}$$.

if this were C:

$\begin{xy} \xymatrix{ a \ar[r]^f \ar[d]_{g \circ f} & b \ar[dl]^g \\ c } \end{xy}$

Then this would be $$C^{op}$$:

$\begin{xy} \xymatrix{ a & b \ar[l]_{f^{op}} \\ c \ar[u]^{(g \circ f)^{op}} \ar[ur]_{g^{op}} } \end{xy}$

On to the next.

### Antithesis: Covariant

When we define a functor in a categorical sense rather than a Haskell sense, we define it as a mapping from one category $$C$$ to another $$D$$, whilst preserving structure: $$F : C \rightarrow D$$. To say a functor is covariant is to state that when a functor $$F$$, for example, represents morphisms on objects from $$C$$ in $$D$$, those morphisms’ arrows’ must go the same direction that they did in $$C$$. I.e., given $$F : C \rightarrow D$$ and our $$C$$ defined above, when we take $$f$$ into $$D$$ with $$F$$, the arrows must respect their initial directions, meaning the following statement must hold in $$D$$:

$F(f) : F(a) \rightarrow F(b)$

Okay, now that we have a more formal definition of a covariant functor, let’s use it a baseline to define its antithesis. Covariant’s antithesis is called contravariant, which, you guessed it, is a reversal of an arrow’s direction. That said, lets define a new functor $$F' : C \rightarrow D$$, but note that $$F'$$ is contravariant, meaning that the following holds in $$D$$.

$F'(f) : F'(b) \rightarrow F'(a)$

## Profunctors

Having finally met this post’s desiderata, I’ll present a Haskell definition of a profunctor, initially without comment.

class Profunctor (p :: * -> * -> *) where
dimap :: (a' -> a) -> (b -> b') -> p a b -> p a' b'

Take it in. Ruminate. What do you see?

Here’s a more formal definition as a hint:

$P : C^{op} \times C \rightarrow D$

A profunctor is a mapping from a category $$C$$ and its opposite $$C^{op}$$ to $$D$$. To put this in more practical terms, a profunctor is a bifunctor which is contravariant in one of its parameters. Let’s go back for now to the Haskell definition above for an illustration. A profunctor p starts with two functions, and a p a b, and ends up with a p a' b'. The functions we’re given map from a' -> a and b -> b', respectively. This tells us that p’s treatment of the first function, the a’s, will be contravariant. But how is this possible? How could we end up with a a' if that’s our starting point? Let’s draw it out!

1. Our first function, which we’ll call $$f$$, is a' -> a:

$\begin{xy} \xymatrix{ a' \ar[r]^f & a } \end{xy}$

2. And our next function, $$g$$, is from b -> b':

$\begin{xy} \xymatrix{ a' \ar[r]^f & a & b \ar[dll]^g \\ b' } \end{xy}$

3. Now, for the sake of our explanation, let’s say our third parameter p a b can be thought of as a more general way of writing the type of a function from a to b. We can think about it this way by replacing p with a concrete type constructor, the function type constructor (->), to be specific. In its case we’d write p a b as (->) a b or, in its more common form, a -> b. We’ll call this function $$h$$:

$\begin{xy} \xymatrix{ a' \ar[r]^f & a \ar[r]^h & b \ar[dll]^g \\ b' } \end{xy}$

4. There’s two things I’d like to stop and make note of here.

1. If we can call our third parameter, p a b a more general way of stating a -> b, then the same applies to dimap’s return type, p a' b'. It can be thought of as a' -> b'.
2. Let’s look again at the diagram and use what we know about category theory’s laws to discern what it would take to get a return type that was shaped like a' -> b'. It’s composition! To return our p a' b', or remember, our a' -> b', we just need to compose $$f$$, $$g$$, and $$h$$ in the right order: $$g \circ h \circ f$$:

$\begin{xy} \xymatrix{ a' \ar[r]^f \ar[d]_{g \circ h \circ f} & a \ar[r]^h & b \ar[dll]^g \\ b' } \end{xy}$

And with that, you can see what makes a profunctor tick!

## In Closing

This sums up the technical section of this this post, as its purpose was to explain what, exactly, a profunctor was, while simultaneously sharing a manner such that makes contravariance a tractable thing. It was through this intuition that I personally was able to get a handle on how to synthesize contravariance, and I hope it was helpful to anyone who comes across this post.

## In Closing Closing

Before I :wq on this one, though, I want to also make mention of the structure of the final technical section of this post. Drawing out the commutative diagrams one arrow at a time was huge in getting a grasp on these concepts for me, and it’s an approach I’ll reach for regularly from now on. If it’s not something you’ve tried before when taking on something dense in the land of category theory, I can certainly attest to its efficacy! Cheers!