On 1/23/12 10:39 PM, Ryan Ingram wrote:
type m :-> n = (forall x. m x -> n x) class Functor f where fmap :: forall a b. (a -> b) -> f a -> f b -- Functor identity law: fmap id = id -- Functor composition law fmap (f . g) = fmap f . fmap gGiven Functors m and n, natural transformation f :: m :-> n, and g :: a -> b, how can I prove (f . fmap_m g) = (fmap_n g . f)?
That is the defining property of natural transformations. To prove it for polymorphic functions in Haskell you'll probably want to leverage parametricity.
I assume you don't know category theory, based on other emails in this thread. But the definition of a natural transformation is that it is a family of morphisms/functions { f_X :: M X -> N X | X an object/type } such that for all g :: a -> b we have that f_b . fmap_m g == fmap_n g . f_a
Thus, you can in principle define plenty of natural transformations which do not have the type f :: forall X. M X -> N X. The only requirement is that the family of morphisms obeys that equation. It's nice however that if a function has that type, then it is guaranteed to satisfy the equation (so long as it doesn't break the rules by playing with strictness or other things that make it so Hask isn't actually a category).
-- Live well, ~wren _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
