Subject: Re: Clarify relationship between Functor and Applicative Newsgroups: gmane.comp.lang.haskell.libraries Date: Friday 25th February 2011 23:07:09 UTC (over 6 years ago) On Fri, 25 Feb 2011, Ross Paterson wrote: > On Fri, Feb 25, 2011 at 05:34:18PM -0500, [email protected] 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.'' |
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