This question is inspired by the feedback given to my previous question
Scalaz provides a wrapper class Kleisli[M[_], A, B] for a function A => M[B].
Kleisli[M[_], A, B] is a semigroup if M[_] is a semigroup. Suppose that M[_] is a functor. Is it correct that Kleisli is a functor too ? What if M[_] is an applicative or monad ?
The fully unapplied Kleisli isn't much of anything—it has kind (* -> *) -> * -> * -> *, and I don't know of any meaningful type classes for that kind.
If you have a monad for a type constructor F[_], though, then Kleisli[F, ?, ?] is an Arrow (which is a type class for things of kind * -> * -> *).
Similarly, if F[_] has a functor instance, then Kleisli[F, A, ?] has a functor instance for any specific A. In fact the same thing works for applicative functors and monads—have an instance of any of them for F[_] and you have an instance for Kleisli[F, A, ?] as well.
Lastly, if you have a semigroup (or monoid) for F[B], then Kleisli[F, A, B] is a semigroup (or monoid) for any A.