Clarify relationship between Functor and Applicative

[roconnor at theorem.ca]

Actually this means once you prove the identity applicative law

pure id <*> x = x

then it entails that

pure f <*> x = fmap f x

On Fri, 25 Feb 2011, roconnor at theorem.ca wrote:

> On Fri, 25 Feb 2011, Ross Paterson wrote:
>
>> On Fri, Feb 25, 2011 at 05:34:18PM -0500, roconnor at theorem.ca wrote:
>>> In the applicative documentation, it says for an Applicative functor f:
>>> 
>>> The Functor instance should satisfy
>>>
>>>       fmap f x = pure f <*> x
>>> 
>>> I think the documentation should be clarified that this does not
>>> need to be checked because it is a consequence of the other
>>> applicative laws.
>>> 
>>> See <http://hpaste.org/44315/applicative_implies_functor>.
>> 
>> Do you have a proof that Functor instances are uniquely determined?
>
> Suppose we have a functor f and another function
>
> foo :: (a -> b) -> f a -> f b
>
> Then as a consequence of the free theorem for foo,
> for any f :: a -> b and any g :: b -> c.
>
> foo (g . f) = fmap g . foo f
>
> In particular, if foo id = id, then
>
> foo g = foo (g . id) = fmap g . foo id = fmap g . id = fmap g
>
>

-- 
Russell O'Connor                                      <http://r6.ca/>
``All talk about `theft,''' the general counsel of the American Graphophone
Company wrote, ``is the merest claptrap, for there exists no property in
ideas musical, literary or artistic, except as defined by statute.''