[Haskell-cafe] (a -> [b]) vs. [a -> b]
haskell at brecknell.org
Thu Feb 1 20:02:42 EST 2007
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
> Any help is greatly appreciated.
Have a look at this post and it's follow-ups:
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