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 > yet. > Any help is greatly appreciated.
Have a look at this post and it's follow-ups: http://www.haskell.org/pipermail/haskell-cafe/2006-December/020041.html _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
