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
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