Haskell：关于hackage Control.Applicative文章中应用函子法的描述中的缺陷？：它说Applicative确定了Functor

Question

I think I found a flaw in the hackage article for Control.Applicative . As a description of the applicative functor laws, it says:

 class Functor f => Applicative f where 

A functor with application, providing operations to embed pure expressions ( pure ), and sequence computations and combine their results ( <*> ).

A minimal complete definition must include implementations of these functions satisfying the following laws:

identity

 pure id <*> v = v 

composition

 pure (.) <*> u <*> v <*> w = u <*> (v <*> w) 

homomorphism

 pure f <*> pure x = pure (fx) 

interchange

 u <*> pure y = pure ($ y) <*> u 

(note that this does not say anything about fmap) and it states that these laws determine the Functor instance:

As a consequence of these laws, the Functor instance for f will satisfy

 fmap fx = pure f <*> x 

I thought at first this was obviously wrong. That is, I guessed there must exist a type constructor t satisfying the following two conditions:

1. t is an instance of Applicative fulfilling the rules above, and
2. There are two different implementations of instance (Functor t) (ie there are two distinct functions fmap1, fmap2 :: (a->b) -> (t a->tb) , satisfying the functor laws).

If (and only if) the above is correct, the above statement must be rewritten to

The Functor instance for f must satisfy

 fmap fx = pure f <*> x 

As a consequence of these laws, this satisfies the Functor laws.

which is clearly correct, no matter whether my guess is right or not .

My question is : is my guess correct? Is there a t with the desired conditions?

---

What follows is the explanation of what I thought during trying to answer this question by myself.

If we were mere mathematicians uninterested in actual Haskell programming , we could easily answer this question affirmatively. In fact,

t = Identity
newtype Identity a = Identity {runIdentity :: a}

meets the requirements 1 and 2 above (in fact, almost anything will do). Indeed, Identity is trivially an instance of Functor and Applicative , as defined in Data.Functor.Identity . (This satisfies fmap f = (pure f <*>) .) To define another "implementation" of instance (Functor f) , Take, for each type a , two functions

transf_a, tinv_a :: a -> a

such that

tinv_a . transf_a = id

(This is, set-theoretically , easy). Now define

instance (Functor Identity) where
 fmap (f :: a->b) = Identity . transf_b . f . tinv_a . runIdentity

This satisfies the Functor laws and is a different implementation from the trivial one if there are

x :: a
f :: a -> b

such that

f x /= transf_b $ f $ tinv_a x

But it is not at all obvious whether we can do it in Haskell. Is something like:

class (Isom a) where
 transf, tinv :: a -> a

instance (Isom a) where
 transf = id
 tinv = id

specialized instance (Isom Bool) where
 transf = not
 tinv = not

possible in Haskell?

---

Edit

I forgot to write something very important. I recognize Control.Applicative as part of GHC base package, so I am also interested in whether the answer to my question changes with any GHC language extensions, such as FlexibleInstances or OverlappingInstances , which I don't understand yet.

Answer 1

Any type in Haskell can have at most one instance of Functor , so your guess is not correct: For no type t can there exist two different implementations of instance (Functor t) . See: http://article.gmane.org/gmane.comp.lang.haskell.libraries/15384

Answer 2

It is a property of the type a -> a that there are only two inhabitants of it, namely id :: a -> a (defined as id x = x ) and bottom :: a -> a defined as bottom = error "Infinite loop." .

If we restrict ourselves to only the first case as "reasonable", we come to an important mathematical theorem, which is that there is at most one function fmap of type forall a. forall b. (a -> b) -> fa -> fb forall a. forall b. (a -> b) -> fa -> fb forall a. forall b. (a -> b) -> fa -> fb satisfying fmap id = id and fmap f . fmap g = fmap (f . g) fmap f . fmap g = fmap (f . g) .

If we do not, then you are correct, we have one case where fmap = undefined and the other where fmap = (<*>) . pure fmap = (<*>) . pure . But that's a little cheesy.