Error in characterization of free theorem for functors

[dfoxfranke]

Chapter 6 in the section "Polymorphism" claims that it is impossible to write a function of type (a -> b) -> f a -> f b that doesn't satisfy forall f g. fmap f . fmap g = fmap (f . g). But this is false:

data F a = A a | B a

bad_fmap f (A x) = B (f x)
bad_fmap f (B x) = A (f x)

A result you do get from the free theorem for fmap is that fmap id = id <==> forall f g. fmap f . fmap g = fmap (f . g); that is, if your fmap satisfies one of the two functors laws then it necessarily satisfies the other one too. The above example satisfies neither.