What is the general case of QuickCheck's promote function?

Question

What is the general term for a functor with a structure resembling QuickCheck's promote function, i.e., a function of the form:

promote :: (a -> f b) -> f (a -> b)

(this is the inverse of flip $ fmap (flip ($)) :: f (a -> b) -> (a -> f b)). Are there even any functors with such an operation, other than (->) r and Id? (I'm sure there must be). Googling 'quickcheck promote' only turned up the QuickCheck documentation, which doesn't give promote in any more general context AFAICS; searching SO for 'quickcheck promote' produces no results.

Godden

Is sequenceA relevant? – Ethiopian

Let me see. Substituting into the type of sequenceA, we would get t = (->) a and f = f. So if (->) a had a Traversable instance, this function would exist for all a. I think Traversable ((->) a) requires (Bounded a, Enum a) of or the equivalent, though. – Godden

For what it's worth, the universe package family provides the requisite Traversable instance. – Gratulant

@Daniel Wagner, ah, given Finite, which I guess is essentially the same thing. – Godden

Answer 1

So far I found these ways of constructing an f with the promote morphism:

• f = Identity
• if f and g both have promote then the pair functor h t = (f t, g t) also does
• if f and g both have promote then the composition h t = f (g t) also does
• if f has the promote property and g is any contrafunctor then the functor h t = g t -> f t has the promote property

The last property can be generalized to profunctors g, but then f will be merely a profunctor, so it's probably not very useful, unless you only require profunctors.

Now, using these four constructions, we can find many examples of functors f for which promote exists:

f t = (t,t)

f t = (t, b -> t)

f t = (t -> a) -> t

f t = ((t,t) -> b) -> (t,t,t)

f t = ((t, t, c -> t, (t -> b) -> t) -> a) -> t

Also note that the promote property implies that f is pointed.

point :: t -> f t
point x = fmap (const x) (promote id)

Essentially the same question: Is this property of a functor stronger than a monad?

Reproduce

Could you add the actual instances you're suggesting? For example what does this instance for (t, t) look like? When implementing a -> (b, b) -> (a, a) -> (b, b), does it just apply the input function to both as and throw away 2/4 bs? – Fillister

@AsadSaeeduddin The instance for (t, t) requires us to implement promote:: (a -> (b, b)) -> (a -> b, a -> b), not a -> (b, b) -> (a, a) -> (b, b). Implementing promote:: (a -> (b, b)) -> (a -> b, a -> b) does not throw away anything; promote f = (fst . f, snd . f). Try to derive the general construction: if f and g have promote then h has promote where h a = (f a, g a). There is only one way to implement promote for h, and it does not throw away any information. – Reproduce

Answer 2

(<*>) :: Applicative f => f (a -> b) -> f a -> f b
(=<<) :: Monad m => (a -> m b) -> m a -> m b

Given that Monad is more powerful an interface than Applicative, this tell us that a -> f b can do more things than f (a -> b). This tells us that a function of type (a -> f b) -> f (a -> b) can't be injective. The domain is bigger than the codomain, in a handwavey manner. This means there's no way you can possibly preserve behavior of the function. It just doesn't work out across generic functors.

You can, of course, characterize functors in which that operation is injective. Identity and (->) a are certainly examples. I'm willing to bet there are more examples, but nothing jumps out at me immediately.

Mosera

Answer 3

Data.Distributive has

class Functor g => Distributive g where
  distribute :: Functor f => f (g a) -> g (f a)
  -- other non-critical methods

Renaming your variables, you get

promote :: (c -> g a) -> g (c -> a)

Using slightly invalid syntax for clarity,

promote :: ((c ->) (g a)) -> g ((c ->) a)

(c ->) is a Functor, so the type of promote is a special case of the type of distribute. Thus every Distributive functor supports your promote. I don't know if any support promote but not Distributive.

Omnipresent

Distributive is strictly stronger than promote. Every Distributive functor g is representable: that is, there exists a constant type b such that g t = b -> t . However, functors such as g t = (t -> a) -> t have promote but are not representable and thus fail to be Distributive. More generally, a functor g defined as g t = c t -> h t has promote if c is any contrafunctor and h has promote. This construction can define a Distributable` only if c is a constant (contra)functor. – Reproduce