# [Haskell-cafe] Natural Transformations and fmap

Ryan Ingram ryani.spam at gmail.com
Tue Jan 24 06:06:52 CET 2012

```On Mon, Jan 23, 2012 at 8:05 PM, Daniel Fischer <

> On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote:
> > 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)?
>
> Unless I'm utterly confused, that's (part of) the definition of a natural
> transformation (for non-category-theorists).
>

Alright, let's pretend I know nothing about natural transformations and
just have the type declaration

type m :-> n = (forall x. m x -> n x)

And I have
f :: M :-> N
g :: A -> B
instance Functor M -- with proofs of functor laws
instance Functor N -- with proofs of functor laws

How can I prove
fmap g. f :: M A -> N B
=
f . fmap g :: M A -> N B

I assume I need to make some sort of appeal to the parametricity of M :-> N.

> > 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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