
1 Feb
2007
1 Feb
'07
7:18 p.m.
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. Thanks, Chad