Can two non-functors compose to a functor?

7
votes

We can have two types f, g :: * -> * such that they're not monads, but their composition is. For example for an arbitrary fixed s:

f a := s -> a
g a := (s, a)

g a isn't a monad (unless we restrict s to a monoid), but f (g a) is the state monad s -> (s, a). (Unlike functors and applicative functors, even if both f and g were monads, their composition might not be.)

Is there a similar example for functors or applicative functors? That is that the composition of f and g is a a functor (or an applicative functor), even though

  1. one of f and g isn't an (applicative) functor and the other is, or
  2. neither of them is an (applicative) functor,
2
haskellcompositionfunctorapplicativecategory-theory
The ContT transformer makes Functors and Applicatives without requiring that the transformed structure is a Functor, but isn't built by composition. hackage.haskell.org/package/transformers-0.4.1.0/docs/… Similarly, the free applicative Ap adds Applicative behavior to any Functor, without requiring that it be Applicative, but again isn't built by composition. hackage.haskell.org/package/free-4.9/docs/… – Cirdec
perhaps not an answer but whenever F and G are adjoint you get a monad in the composition. – Philip JF

2 Answers

8
votes

This is not a (covariant) functor

f x = x -> r

but f . f is the "continuation" functor (also a monad):

f (f x) = (x -> r) -> r

This is probably not the best example because f is a contravariant functor.

7
votes

Let g :: *->*. Then Const A . g is a functor for any A, in fact isomorphic to Const A.