Why is a -> a not a functor?

问题

Specifically referring to https://bartoszmilewski.com/2015/04/07/natural-transformations/

Author says "This is not a functor".

I can define fmap :: (a -> b) -> (a -> a) -> (b -> b) as fmap f aa = id, which seems to adhere to the functor laws.

I don't mean why it's not explicitly part of the Functor typeclass in X language, I just mean why it wouldn't be acknowledged as a functor.

回答1:

In the context of Haskell, I think you're talking about newtype Endo a = Endo (a -> a) (using a newtype to get the required * -> * kind).

Indeed we could define

instance Functor Endo where
    fmap _ _ = Endo id

But one of the Functor laws is fmap id = id, i.e. fmapping with id has to be the same as doing nothing. Your suggested definition violates this rule:

fmap id (Endo toUpper)

should result in Endo toUpper, but your code makes it Endo id. One of those transforms 'a' to 'A', the other turns 'a' into 'a'.

来源：https://stackoverflow.com/questions/48853854/why-is-a-a-not-a-functor