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

[Chad Scherrer]

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
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