[Haskell-cafe] Natural Transformations and fmap

[Ryan Ingram]

I've been playing around with the relationship between monoids and monads
(see
http://www.jonmsterling.com/posts/2012-01-12-unifying-monoids-and-monads-with-polymorphic-kinds.htmland
http://blog.sigfpe.com/2008/11/from-monoids-to-monads.html), and I put
together my own implementation which I'm quite happy with, that you can see
at http://hpaste.org/56903 ; relying only on the extensions RankNTypes,
TypeOperators, NoImplicitPrelude, ScopedTypeVariables;

At the end of that paste, I prove the three Haskell monad laws from the
functor laws and "monoid"-ish versions of the monad laws, but my proofs all
rely on a property of natural transformations that I'm not sure how to
prove; given

    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 g

Given 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)?  Is there some more
fundamental law of natural transformations that I'm not aware of that I
need to use?  Is it possible to write a natural transformation in Haskell
that violates this law?

  -- ryan
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