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

Chad Scherrer said:

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

It seems like it should be obvious, but I haven't had any luck with it yet. Any help is greatly appreciated.

Have a look at this post and it's follow-ups: http://www.haskell.org/pipermail/haskell-cafe/2006-December/020041.html