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Re: [Haskell-cafe] lawless instances of Functor

 Subject: Re: [Haskell-cafe] lawless instances of Functor Derek Elkins Tue, 5 Jan 2010 08:01:46 +0900
 ```On Tue, Jan 5, 2010 at 7:14 AM, Paul Brauner <[email protected]> wrote: > Hi, > > I'm trying to get a deep feeling of Functors (and then pointed Functors, > Applicative Functors, etc.). To this end, I try to find lawless > instances of Functor that satisfy one law but not the other. > > I've found one instance that satisfies fmap (f.g) = fmap f . fmap g > but not fmap id = id: > > data Foo a = A | B > > instance Functor Foo where >  fmap f A = B >  fmap f B = B > > -- violates law 1 > fmap id A = B > > -- respects law 2 > fmap (f . g) A = (fmap f . fmap g) A = B > fmap (f . g) B = (fmap f . fmap g) B = B > > But I can't come up with an example that satifies law 1 and not law 2. > I'm beginning to think this isn't possible but I didn't read anything > saying so, neither do I manage to prove it. > > I'm sure someone knows :) Ignoring bottoms the free theorem for fmap can be written: If h . p = q . g then fmap h . fmap p = fmap q . fmap g Setting p = id gives h . id = h = q . g && fmap h . fmap id = fmap q . fmap g Using fmap id = id and h = q . g we get, fmap h . fmap id = fmap h . id = fmap h = fmap (q . g) = fmap q . fmap g So without doing funky stuff involving bottoms and/or seq, I believe that fmap id = id implies the other functor law (in this case, not in the case of the general categorical notion of functor.) _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe ```