Profunctor

Definition

A profunctor between categories $C$ and $D$ is defined to be a functor $D^{\text{op}}\times C\to \text{Set}$, and is denoted $C\nrightarrow D$.

In Haskell, this is approximated as

  -- lmap (f . g) = lmap f . lmap g
  -- rmap (f . g) = rmap g . rmap f
  -- lmap id = id
  -- rmap id = id
  class Profunctor p where
    lmap :: (a -> a') -> p a b -> p a' b
    rmap :: (b -> b') -> p a b' -> p a b

Strong

  class Profunctor p => Strong p where
    strong :: p a b -> p (a, c) (b, c)

In Haskell, every Functor is strong with respect to (,).

  newtype Forget' r a b = Forget' { runForget :: a -> r }

  instance Profunctor (Forget' r) where
    dimap f _ (Forget' p) = Forget' (p . f)

Mapping

  -- Laws:
  --   map' . rmap f ≡ rmap (fmap f) . map'
  --   map' . map' ≡ dimap Compose getCompose . map'
  --   dimap Identity runIdentity . map' ≡ id

  class (Traversing p, Closed p) => Mapping p where
    map' :: Functor f => p a b -> p (f a) (f b)
    roam :: ((a -> b) -> s -> t) -> p a b -> p s t

Two observations:

• map' is a generalisation of fmap. Try setting p ~ (->) .

• roam very closely resembles over, from lens:

      type ASetter s t a b = (a -> Identity b) -> s -> Identity t
      --                   ≡ (a -> b) -> s -> t
      over :: ASetter s t a b -> (a -> b) -> s -> t
  

Closed

  -- Laws:
  --   lmap (. f) . closed ≡ rmap (. f) . closed
  --   closed . closed ≡ dimap uncurry curry . closed
  --   dimap const ($()) . closed ≡ id
  class Profunctor p => Closed p where
    closed :: p a b -> p (x -> a) (x -> y)

The first law is very interesting. The base Profunctor laws do not define any relation between lmap and rmap. Closed, however,

(quiver) \begin{tikzcd} {p\;a\;b} \\ {p\;(x \to a)\;(x \to b)} \\ {p\;a\;b} \arrow["{\text{closed}}", from=1-1, to=2-1] \arrow["{\text{lmap}\;(\circ f)}"', from=2-1, to=3-1] \arrow["{\text{rmap}\;(\circ f)}", shift left=3, from=2-1, to=3-1] \end{tikzcd} ParseError: No such environment: tikzcd at position 7: \begin{̲t̲i̲k̲z̲c̲d̲}̲ {p\;a\;b} \\ …

Inbox

• For a profunctor p a b, a must only appear positive in p's definition, and b can only appear negative.

• I'm currently re-experiencing the feeling of trying to gather an intuition for Monad and Applicative. "Yes, I see what Strong does. Show me a Profunctor that is not Strong! What exactly does it mean to have this power‽"

References

• [cite:@hu2013love]