Dear Dan, you posted an entry on the Comonad.Reader blog about free monoids in Haskell. I commented about FM being a monad, and meanwhile verified that indeed every monoid is an FM-algebra. The proofs are analogous to proving that the continuation (_ -> r) -> r for fixed r is a monad and that r is an algebra for it, respectively. Moreover, the FM-algebra stucture is a monoid homomorphism w.r.t. the monoid instance you gave for (FM a). I'd like to ask what happens when one omits the Monoid constraint. That is, what are the elements of newtype Y a = Y { runY :: forall b. (a -> b) -> b } (Y a) is like (Control.ForFree.Yoneda Identity a), that is, elements of (Y a) should be natural transformations from the reader functor ((->) a) to the identity functor. If that is true, then the Yoneda lemma tells us that (Y a) is isomorphic to (Identity a), but at least operationally that seems not to be the case: u = runY (return undefined) has different semantics than u' = runY (Y undefined) as can be seen by applying both to const (). So return does not map ⊥ to (Y ⊥). I wonder whether one can exhibit other elements not equal to any return(x). Of course (Identity a) is actually the lifted a, since ⊥≠I dentiyt ⊥. Yet this does not serve as an explanation for u vs. u', as both u and u' are of the form (Y _), but one function evaluates its argument while the other does not. Olaf