Given fmap id = id, fmap (f . g) = fmap f . fmap g follows from the free theorem for fmap.
This was published as an aside in a paper a long time back, but I forget where. -Edward Kmett On Mon, Jan 4, 2010 at 5:14 PM, Paul Brauner <paul.brau...@loria.fr> 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 :) > > Paul > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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