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