Clarify relationship between Functor and Applicative

roconnor at roconnor at
Sat Feb 26 00:07:09 CET 2011

On Fri, 25 Feb 2011, Ross Paterson wrote:

> On Fri, Feb 25, 2011 at 05:34:18PM -0500, roconnor at 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 <>.
> 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                                      <>
``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.''

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