Puzzle 5: mu f = let x = f x in x
One cheating strategy that I like: https://www.haskell.org/hoogle/?hoogle=forall+a.+%28a+-%3E+a%29+-%3E+a
Any intuitions available for the connection between the \\(\mu \\), adjoints and recursion? The high level picture isn't yet super clear to me.
Puzzle 6: presumably the y combinator, although I still don't really understand the relation between the untyped lambda expression for the y combinator \\(\lambda f (\lambda x (f (x x)) (\lambda x (f (x x)) \\) and the Haskell version.