如何仅基于Distributive特性编写Representable实例？及超类关联疑问

Great questions—let's unpack these about Haskell's Distributive and Representable functors, since their relationship is one of those subtle but important bits of category theory applied to Haskell.

1. 如何仅利用Distributive特性编写Representable实例？ #

First, let's ground ourselves: Representable functors require two core methods (tabulate :: (x -> a) -> f a and index :: f a -> x -> a) plus laws that these form an isomorphism. Distributive only gives us distribute :: Functor g => g (f a) -> f (g a).

Here's the thing: you can't write a full Representable instance using only the Distributive instance alone—you need one extra critical piece: a representing type x (the type that f is isomorphic to (->) x for). But once you have that type, you can use Distributive's distribute to define the required Representable methods.

Let's walk through a concrete example with Identity (which is both Distributive and Representable):

-- First, we know the representing type is ()
instance Representable Identity where
  type Rep Identity = ()
  -- Use distribute with Identity to turn a (() -> a) into Identity a
  tabulate f = runIdentity $ distribute (Identity f)
  -- Use distribute with fmap const to turn Identity a into (() -> a)
  index fa = runIdentity . distribute (fmap const fa)

Generalizing this pattern: for any distributive functor f with representing type x, you can define:

instance Representable f where
  type Rep f = x
  tabulate f = runIdentity $ distribute (Identity f)
  index fa = runIdentity . distribute (fmap const fa)

The key takeaway: Distributive gives you the machinery to connect f to reader-like behavior, but you still need to explicitly provide the representing type that makes f isomorphic to (->) x.

2. 若已为复杂自定义类型Foo实现Distributive实例，能否仅依托该实例的特性来编写Foo的Representable实例？若无法实现，为何Distributive是Representable的超类？ #

Short answer: No, you can't. #

Here's why: Distributive is a strictly broader class than Representable. Every Representable functor is Distributive, but not every Distributive functor is Representable.

For example, consider the continuation functor Cont r a = (a -> r) -> r. It has a valid Distributive instance:

instance Distributive (Cont r) where
  distribute gCont = \k -> fmap (\c -> c k) gCont

But Cont r is not representable—there's no type x such that (a -> r) -> r ≅ x -> a for all a. The continuation functor's structure is fundamentally different from a reader functor, even though it's distributive.

Your Foo type could be similar: having a Distributive instance only means it's a right adjoint functor (a category theory property), but Representable requires the stronger condition that it's isomorphic to a reader functor (->) x. That isomorphism isn't guaranteed by Distributive alone.

Why is Distributive a superclass of Representable? #

Because every Representable functor is automatically Distributive—you can derive the Distributive instance directly from the Representable methods:

instance Representable f => Distributive f where
  distribute gfa = tabulate $ \x -> fmap (`index` x) gfa

This works because representable functors are just reader functors in disguise, and reader functors are trivially distributive (since distribute for (->) x just maps each value in the functor to its application to x).

Making Distributive a superclass enforces that only valid distributive functors can be marked as representable, and it gives you the Distributive methods for free once you've written your Representable instance.

内容的提问来源于stack exchange，提问作者David Banas