Haskell Functor implied law

Question

Typeclassopedia says:

\"A similar argument also shows that any Functor instance satisfying the first law (fmap id = id) will automatically satisfy the second law as well

Answer 1

Using seq, we can write an instance which satisfies the first rule but not the second one.

data Foo a = Foo a
    deriving Show

instance Functor Foo where
    fmap f (Foo x) = f `seq` Foo (f x)

We can verify that this satisfies the first law:

fmap id (Foo x)
= id `seq` Foo (id x)
= Foo (id x)
= Foo x

However, it breaks the second law:

> fmap (const 42 . undefined) $ Foo 3
Foo 42
> fmap (const 42) . fmap undefined $ Foo 3
*** Exception: Prelude.undefined

That said, we usually ignore such pathological cases.