Difference between function composition operator (.) and fmap (<$>)

Question

Currently reading through this article (which is pretty brilliant btw) and have a pretty simple question:

If I combine two functions like (+3) and (+2) with <$>, it seems to give me a new function that adds 5 to whatever is passed to it. If I do the same with the function composition operator, i.e. (+3) . (+2), would it not do the same thing? If that is true, is there a relationship here between these two operators such that they do the same thing in this simple case?

Is this even an intelligent question?

Answer 1

To find information about the Functor instance for functions, match up the types to find the relevant instance:

fmap :: (a -> b) -> f a -> f b

Then here a ~ Int, b ~ Int and f ~ (->) Int.

You can see all of the Functor instances that come with GHC here. (->) is just an infix type operator with two type parameters. We usually see it applied as Int -> Int, but this is equivalent to (->) Int Int. There is a Functor instance for the (partially applied) type (->) r (for any type r::*).

Looking at the ((->) r) instance for Functor, we see that fmap = (.), so there is no practical difference between (+3) . (+2) and fmap (+3) (+2) (same as (+3) <$> (+2).