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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