On Tue, Jan 5, 2010 at 7:14 AM, 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 :)
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.)