Difference between free monads and fixpoints of functors?

Question

I was reading http://www.haskellforall.com/2013/06/from-zero-to-cooperative-threads-in-33.html where an abstract syntax tree is derived as the free monad of a functor repres

Answer 1

There is a deep and 'simple' connection.

It's a consequence of adjoint functor theorem, left adjoints preserve initial objects: L 0 ≅ 0.

Categorically, Free f is a functor from a category to its F-algebras (Free f is left adjoint to a forgetful functor going the other way 'round). Working in Hask our initial algebra is Void

Free f Void ≅ 0

and the initial algebra in the category of F-algebras is Fix f: Free f Void ≅ Fix f

import Data.Void
import Control.Monad.Free

free2fix :: Functor f => Free f Void -> Fix f
free2fix (Pure void) = absurd void
free2fix (Free body) = Fix (free2fix <$> body)

fixToFree :: Functor f => Fix f -> Free f Void
fixToFree (Fix body) = Free (fixToFree <$> body)

Similarly, right adjoints (Cofree f, a functor from Hask to the category of F-coalgebras) preserves final objects: R 1 ≅ 1.

In Hask this is unit: () and the final object of F-coalgebras is also Fix f (they coincide in Hask) so we get: Cofree f () ≅ Fix f

import Control.Comonad.Cofree

cofree2fix :: Functor f => Cofree f () -> Fix f
cofree2fix (() :< body) = Fix (cofree2fix <$> body)

fixToCofree :: Functor f => Fix f -> Cofree f ()
fixToCofree (Fix body) = () :< (fixToCofree <$> body)

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Just look how similar the definitions are!

newtype Fix f 
  = Fix (f (Fix f))

Fix f is Free f with no variables.

data Free f a 
  = Pure a 
  | Free (f (Free f a))

Fix f is Cofree f with dummy values.

data Cofree f a 
  = a <: f (Cofree f a)