3-Functor-3-Laws

1. Although it is not possible for a Functor instance to satisfy the first Functor law but not the second (excluding undefined), the reverse is possible. Give an example of a (bogus) Functor instance which satisfies the second law but not the first.

instance Functor Maybe where
  fmap :: (a -> b) -> Maybe a -> Maybe b
  fmap _ _ = Nothing

fmap id (Just x) = Nothing /= Just x = id (Just x)

fmap (f . g) _ = Nothing
(fmap f . fmap g) _ = Nothing

2. Which laws are violated by the evil Functor instance for list shown above: both laws, or the first law alone? Give specific counterexamples.

-- Evil Functor instance
instance Functor [] where
  fmap :: (a -> b) -> [a] -> [b]
  fmap _ [] = []
  fmap g (x:xs) = g x : g x : fmap g xs

Both laws. Here is failure for first law.

fmap id [()] = fmap id (():[])
  = id () : id () : fmap id []
  = [(), ()]
  /= id [()] = [()]

And here is failure for second law.

fmap (id . id) [()] = fmap id [()] = [(), ()]

(fmap id . fmap id) [()] = fmap id [(), ()]
  = id () : id () : fmap id [()] = [(), (), (), ()]