# [Haskell-cafe] Re: Automatic Recognition of Functors

apfelmus at quantentunnel.de apfelmus at quantentunnel.de
Thu Mar 1 10:31:42 EST 2007

```Walter Potter wrote:
> Folks,
>
> Given f:: a -> b it is very natural to lift f to P f :: P a -> P b where
> P is the power set functor. Or L f :: [a] -> [b].
>
> We are modeling structures using repeated application of the power
> functor, via repeated application of [ ].
>
> It would be very nice if Haskell would recognize this lifting. That is,
> if f :: a -> b then one automatically has f :: [a] -> [b]
> without using fMap.
>
> We can do something similar with classes in the following way:
>
> Given
>
> (+.) :: a -> a -> a
>
> (+.) w [ ] = w
> (+.) [ ] w = w
> (+.) (a:as) (b:bs) = (a+b) :(as + bs)
>
> Now given
>
> (+.) x y = x+y
>
> One can compute
> [[1,2],[3,4]] +. [ [2,3],[1,2,.3]].
>
> I know I'm asking for a bit more here. I might need to use  fMap f : [
> a] -> [ b].
> But I can't seem to get by with
> fMap f [[1,2],[3,4]] when f :: Int -> Int
>
> We often need to lift functions to higher power maps.
>
> It would be nice to have a way to do this with ease.

You could try to overload the specific f you want to lift, but I guess
that you have arbitrary f that need to be lifted.

By introducing explicit functor composition, you can reduce multiple
liftings to a single one:

newtype Comp f g a = Comp { unComp :: f (g a)) } deriving (Show,Eq)

instance (Functor f, Functor g) => Functor (Comp f g) where
fmap f = Comp . fmap (fmap f) . unComp

> fmap (+1) \$ Comp [[1,2],[3,4]]
Comp {unComp = [[2,3],[4,5]]}

Of course, this shifts the problem because now, you have to lift into a
stack of 'Comp's like 'Comp (Comp f g) h'. But it may be useful if you
are working with abstract types anyway.

Regards,
apfelmus

```