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

[Matthew Brecknell]

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