# Kleisli Categories and Free Monads

Given a monad **m** one can construct the *Kleisli category* `Control.Arrow.Kleisli m`

, in this post we’ll explore what happens when we start with a free monad.

```
module KleisliCategoriesAndFreeMonads where
import Prelude hiding (id, (.))
import Control.Category (Category (id, (.)))
import Control.Monad (ap, join)
import qualified Control.Monad.Free as Free
```

## Kleisli contstruction

```
newtype Kleisli m a b = Kleisli { runKleisli :: a -> m b }
instance Monad m => Category (Kleisli m) where
id = Kleisli return
Kleisli f . Kleisli g = Kleisli (\x -> f =<< g x)
```

The Keisli composition is so useful, it has it’s own operator defined in `Control.Monad`

:

## Free monads

Free monad construction gives you a monad out of a functor. It will satisfy all the monad laws: associativity, and unitality axioms, and it is a universal construction.

```
data Free f a
= Return a
| Free (f (Free f a))
instance Functor f => Functor (Free f) where
fmap f (Return a) = Return (f a)
fmap f (Free ff) = Free (fmap f <$> ff)
instance Functor f => Applicative (Free f) where
pure = return
(<*>) = ap
instance Functor f => Monad (Free f) where
return = Return
Return a >>= f = f a
Free ff >>= f = Free (fmap (>>= f) ff)
```

The universal property of a free monad can be expressed with a class, which I borrowed from free-algebras package (I don’t include all details that are not important for this blog post; The details where described in this post):

```
class FreeAlgebra1 (m :: (* -> *) -> * -> *) where
-- | Natural transformation that embeds generators into `m`.
liftFree :: Functor f => f a -> m f a
-- | The freeness property.
foldNatFree
:: forall d f a .
( Monad d
, Functor f
)
=> (forall x. f x -> d x)
-- ^ a natural transformation which embeds generators of `m` into `d`
-> (m f a -> d a)
-- ^ a morphism from `m f` to `d`
```

In this blog post whenever we will refer to `liftFree`

and `foldNatFree`

we will actually refer to this instance:

```
instance FreeAlgebra1 Free where
liftFree fa = Free (Return <$> fa)
foldNatFree _nat (Return a) = return a
foldNatFree nat (Free ff) =
join $ nat $
(foldNatFree nat) <$> ff -- induction step
```

You probably recognise `Free.liftF`

and `Free.foldFree`

from the free package.

Instances of this class have the property that to construct a natural transformation from `FreeAlgebra1 m => m f`

to a monad `Monad d => d`

is enought to come up with a natural transformation of functors `forall x. f a -> d a`

. If you’d like to explore more why this class is the right one to speak about freeness, checkout one of my previous posts. Note that the instance for `Free`

follows the structure, and there is no room how to implement the methods: that’s a very common feeling in category theory, which means one is on the right path.

This particular instance satisfies rather interesting laws:

or as an equation:

`fmap g . liftFree == liftFree . fmap g`

This is more or less streightforward, just take a look at the definition of `liftFree`

.

or as an equation:

`fmap g . foldNatFree nat == foldNatFree nat . fmap g`

This one is slightly more involved, and lets prove it:

```
(foldNatFree nat . fmap g) (Return a)
= foldNatFree nat (Return (g a))
= return (g a)
= fmap g (return a)
= fmap g (foldNatFree nat (Return a))
= (fmap g . foldNatFree nat) (Return a)
```

and

```
(foldNatFree nat . fmap g) (Free ff)
= foldNatFree nat (fmap g (Free ff))
= foldNatFree nat (Free (fmap (fmap g) ff))
= join $ nat $ foldNatFree nat <$> (fmap (fmap g) ff)
= join $ nat $ foldNatFree nat <$> fmap g <$> ff
= join $ nat $ (foldNatFree nat . fmap g) <$> ff
-- by induction hypotesis
= join $ nat $ (fmap g . foldNatFree nat) <$> ff
= join $ nat $ fmap g <$> (foldNatFree nat <$> ff)
= join $ nat $ fmap (fmap g) (foldNatFree nat <$> ff)
-- since nat is a natural transformation
= join $ fmap (fmap g) $ nat (foldNatFree nat <$> ff)
-- join is a natural transformation
-- join :: Monad m => m (m a) -> m a
= fmap g $ join $ $ nat (foldNatFree nat <$<> ff)
= fmap g $ foldNatFree nat (Free ff)
= (fmpg g . foldNatFree nat) (Free ff)
```

Since in our case `Free`

is a functor from the category of (endo-)functors into the category of monads. `foldNatFree nat`

has to ba monad morphism for any `nat`

. This means that the following diagrams commute (or equations hold):

or as an equation:

`return = return . foldNatFree nat`

This should be clear from definition of `foldNatFree`

. Moreover, the following diagram commutes:

or as an equation:

```
join . (foldNatFree nat . fmap (foldNatFree nat))
== foldNatFree nat . join
```

Let us prove this it:

```
join . (foldNatFree nat . fmap (foldNatFree nat)) (Return (Return a))
== join (foldNatFree nat (Return (foldNatFree nat (Retrun a))))
== join (foldNatFree nat (Return (return a))
== join (return (return a))
== return a
== foldNatFree nat (Return a)
== foldNatFree nat (join (Return (Return a)))
```

And the other one, which we prove starting from the right hand side:

```
foldNatFree nat (join (Free ff)))
-- by definition of join
= foldNatFree nat (Free (fmap join ff))
-- by definition of foldNatFree
= join $ nat $ fmap (foldNatFree nat) (fmap join ff)
= join $ nat $ fmap (foldNatFree nat . join) ff
-- by induction hypotesis
= join $ nat $ fmap (join . foldNatFree nat . fmap (foldNatFree nat)) ff
= join $ nat $ fmap join $ fmap (foldNatFree nat) $
fmap (fmap foldNatFree nat)) ff
-- since nat is a natural transformation
= join $ fmap join $ nat $ fmap (foldNatFree nat) $
fmap (fmap (foldNatFree nat)) ff
-- by associativity of join
= join $ join $ nat $ fmap (foldNatFree nat) $
fmap (fmap (foldNatFree nat)) ff
-- by definition of foldNatFree (the outer one)
= join $ foldNatFree nat $ Free $ fmap (fmap (foldNatFree nat)) ff
-- by definition of functor instance for Free
= join $ foldNatFree nat $ fmap (foldNatFree nat) $ Free ff
= (join . foldNatFree nat . fmap (foldNatFree nat)) $ Free ff
```

Note that by the natural transformation law for `foldNatFree`

we have:

```
foldNatFree nat . fmap (foldNatFree nat)
== fmap (foldNatFree nat) . foldNatFree nat
```

For any monad morphism `fn :: (Monad m, Monad n) => m a -> n a`

, we will show that:

`fn . (f <=< g) == (fn . f) <=< (fn . g)`

in particular this is true for `foldNatFree nat`

.

```
(fn . f <=< fn . g)
== \a -> fn . f =<< (fn . g) a
== \a -> fn . f =<< fn (g a)
-- by definition of join (or =<< in terms of a join)
== \a -> join (fn . f <$> fn (g a))
-- by functor associativity law
== \a -> join (fn <$> (f <$> fn (g a)))
-- since fn is a morphism of monads it is a natural transformation
-- and thus it commutes with fmap/<$>
== \a -> join (fn <$> (fn (f <$> g a)))
-- since fn is a morphism of monads: join (fmap fn . fn) == fn . join
== \a -> fn (join (f <$> g a))
-- by definition of join
== \a -> fn (f =<< g a)
== \a -> fn (f <=< g) a
== fn (f <=< g)
```

The proof could be much shorter if we use monad morphism law in terms of binds. The equivalne form of `join (fmap fn . fn) == fn . join`

expressed with bind is `fn (f =<< ma) = (fn . f) =<< fn ma`

. A reason to use `join`

is that the law take the same form as for monoid homomorphisms, so it is very easy to remember them.

## Kleisli categories for free monads

```
liftKleisli
:: Monad m
=> Kleisli m a b
-> Kleisli (Free m) a b
liftKleisli (Kleisli f) = Kleisli (liftFree . f)
```

```
foldKleisli
:: ( Functor f
, Monad m
)
=> (forall x. f x -> m x)
-> Kleisli (Free f) a b
-> Kleisli m a b
foldKleisli nat (Kleisli f) = Kleisli $ foldNatFree nat . f
```

This means that `Kleisli (Free f)`

is a free category for the class of graphs of type

`Functor f => Kleisli f`

(`Keisli f`

is a category only when `f`

is a monad). Both morphisms: `liftKleisli`

is marely a morphisms of graphs, while `foldKleisli`

is a functor, which means it preserves `id`

and the composition `(.) :: Category c => c y z -> c x y -> c x y`

.

```
foldKleisli nat id
== foldKleisli nat (Kleisli Return)
== Kleisli (foldNatFree nat . Return)
== Kleisli return
== id
```

```
foldKleisli nat (Kleisli f . Kleisli g)
== foldKleisli nat (Kleisli f <=< g)
== Kleisli (foldNatFree nat (f <=< g))
-- foldNatFree nat is a morphism of monads thus
== Kleisli (foldNatFree nat f <=< foldNatFree nat g)
== Kleisli (foldNatFree nat f) . Kleisli (foldNatFree nat g)
== foldKleisli nat f . foldKleisli nat g
```