Re: [Haskell-cafe] (a -> [b]) vs. [a -> b]

Chad Scherrer wrote:

Are (a -> [b]) and [a -> b] isomorphic? I'm trying to construct a function

f :: (a -> [b]) -> [a -> b]

that is the (at least one-sided) inverse of

f' :: [a -> b] -> a -> [b] f' gs x = map ($ x) gs

Anything better than this? f g = [\x -> g x !! n | n <- [0..]] -Yitz