How can I understand “(.) . (.)”?

Question

I believe I understand fmap . fmap for Functors, but on functions it\'s hurting my head for months now.

I\'ve seen that you can just apply the definitio

Answer 1

You can also use your understanding of fmap . fmap.

If you have two Functors foo and bar, then

fmap . fmap :: (a -> b)  ->  foo (bar a)    ->   foo (bar b)

fmap . fmap takes a function and produces an induced function for the composition of the two Functors.

Now, for any type t, (->) t is a Functor, and the fmap for that Functor is (.).

So (.) . (.) is fmap . fmap for the case where the two Functors are (->) s and (->) t, and thus

(.) . (.) :: (a -> b) -> ((->) s) ((->) t a) -> ((->) s) ((->) t b)
          =  (a -> b) -> (s -> (t -> a))     -> (s -> (t -> b))
          =  (a -> b) -> (s ->  t -> a )     -> (s ->  t -> b )

it "composes" a function f :: a -> b with a function of two arguments, g :: s -> t -> a,

((.) . (.)) f g = \x y -> f (g x y)

That view also makes it clear that, and how, the pattern extends to functions taking more arguments,

(.)             :: (a -> b) -> (s ->           a) -> (s ->           b)
(.) . (.)       :: (a -> b) -> (s -> t ->      a) -> (s -> t ->      b)
(.) . (.) . (.) :: (a -> b) -> (s -> t -> u -> a) -> (s -> t -> u -> b)

etc.