# [Haskell-cafe] Natural Transformations and fmap

wren ng thornton wren at freegeek.org
Fri Jan 27 06:30:10 CET 2012

```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 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)?

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

```