Monadicity
From Algebras to Monads and back to Algebras
In this post I'd like to introduce you to how mathematicians think about monads, and how they are useful. We will study a general case alongside with a simple example of the category of monoids. I hope you will get from this post how a monad describes category of monoids. We will need some simple category theory tools like: adjoint functors, falgebras, free algebras (or just free monoids) and monads. This blog post is based on a library I recently wrote, and it will use snippets of code from it.
LetHask
be
the category of Haskell types and Monoid
be the category of
monoids in Hask
. There are two functors that connect this two
categories:
$$U:\mathrm{Monoid}\rightarrow\mathrm{Hask}$$
which sends a monoid $m$ to its underlying type forgetting the $Monad$
constraint; and the free monoid functor:
$$F:\mathrm{Hask}\rightarrow\mathrm{Monoid}$$
which takes a type $a$ and sends it to the list monoid: $[a]$. Let's try
to analyse the relationship between the two functors. If you read
my previous post on
free algebras
it won't be a surprise to you that there is a close relationship
between these two constructions.
 
 Type family which for each free algebra @m@ returns a type level
 lambda from types to constraints. It is describe the class of
 algebras for which this free algebra is free.
type family AlgebraType (f :: k) (a :: l) :: Constraint
 
 Type family which limits Hask to its full subcategory which
 satisfies a given constraints. Some free algebras, like free
 groups, or free abelian semigroups have additional constraints on
 on generators, like @Eq@ or @Ord@.
type family AlgebraType0 (f :: k) (a :: l) :: Constraint
 
 A proof that constraint @c@ holds for type @a@.
newtype Proof (c :: Constraint) (a :: l) = Proof (Dict c)
 
 A lawful instance has to guarantee that @'unFoldFree'@ is an inverse
 of @'foldMapFree'@.

 This in turn guaranties that @m@ is a left adjoint functor from Hask
 to algebras of type @'AlgebraType m'@. The right adjoint is the
 forgetful functor. The composition of left adjoin and the right one
 is always a monad, this is why we will be able to build monad instance
 for @m@.
class FreeAlgebra (m :: Type > Type) where
  Injective map that embeds generators @a@ into @m@.
returnFree :: a > m a
  The freeness property.
foldMapFree
:: forall d a
. ( AlgebraType m d
, AlgebraType0 m a
)
=> (a > d)  ^ map generators of @m@ into @d@
> (m a > d)  ^ returns a homomorphism from @m a@ to @d@
 
 Proof that @AlgebraType0 m a => m a@ is an algebra of type
 @AlgebraType m@. This proves that @m@ is a mapping from the full
 subcategory of @Hask@ of types satisfyling @AlgebraType0 m a@
 constraint to the full subcategory satisfying @AlgebraType m a@,
 @fmapFree@ below proves that it's a functor.
proof :: forall a. AlgebraType0 m a
=> Proof (AlgebraType m (m a)) (m a)
 
 Proof that the forgetful functor from types @a@ satisfying
 @AgelbraType m a@ to @AlgebraType0 m a@ is well defiend.
forget :: forall a. AlgebraType m a
=> Proof (AlgebraType0 m a) (m a)
 
 All types which satisfy @'FreeAlgebra'@ constraint are foldable.

 prop> foldFree . returnFree == id

foldFree
:: forall m a .
( FreeAlgebra m
, AlgebraType m a
)
=> m a
> a
foldFree ma = case forget @m @a of
Proof Dict > foldMapFree id ma
type instance AlgebraType0 [] a = ()
type instance AlgebraType [] m = Monoid m
instance FreeAlgebra [] where
returnFree a = [a]
foldMapFree = foldMap
proof = Proof Dict
forget = Proof Dict
The type class FreeAlgebra
has many more instances, e.g. for semigroups, groups, idempotent semigroups
(semilattices), abelian
semigroups / monoids / groups, and there is even a version for functors,
applicative functors and monads. For simplicity, in this post we will not
touch the higher kinded types, but all of this works perfectly well for
such data types too.
You may ask what this type instance mean for monoids? We have two maps
returnFree @[] :: a > [a]
and a bijective map
foldMapFree @[] :: forall m a. Monoid m => (a > m) > ([a] > m)
given by Prelude.foldMap
with inverse
unFoldMapFree
:: FreeAlgebra m
=> (m a > d)
> (a > d)
unFoldMapFree f = f . returnFree
Note that foldMap
(also
foldMapFree
)
has some nice properties:

For each
f :: Monoid m => a > m
the mapfoldMap f :: Monoid m => [a] > m
is a homomorphism of monoids, i.e.
This is simply becausefoldMap f (as <> bs) = foldMap f as <> foldMap f bs
foldMap
maps the elements ofas :: [a]
to a list of type[m]
and then folds them using monoid multiplication. Being a monoid homomorphism in this case just means that we can take two listsas :: [a]
andbs :: [a]
andfoldMap
over both of them and then multiply the results inm
and we will get the same result as computingfoldMap f (as ++ bs :: [a])
. The result will be the same because monoids (and list concatenation) satisfy the associativity law.When you will define an instance of
FreeAlgebra
you should check that this law is obeyed. Otherwise all the theory that we are uncovering here will fail.You may think that it will fail to hold for nonassociative monoids, since we used associativity here; but we used in two places at the same time: associativity of list concatenation and monoids, and this is crucial. For non associative monoids, the functor which will appear instead of a list will not obey the associativity law and it will not be required to prove that
foldMapFree
preserves non associative multiplication. 
foldMap :: Monoid m => (a > m) > ([a] > m)
is natural inm
which in turn means that
For any mapfoldMap (f . g) = f . foldMap g
g :: Monoid m => a > m
and any monoid homomorphisms:f :: (Monoid m, Monoid n) => m > n
. 
foldMap :: Monoid m => (a > m) > ([a] > m)
is natural ind
which means that
For any mapfoldMap (g . h) = foldMap g . map h
h :: b > a
andg :: Monoid m => a > m
.
The last two properties mean that
foldMapFree
(and foldMap
)
is a natural transformation of bifunctors
from
Hom m
to AlgHom m
which we define below.
data Hom m a b where
Hom :: (AlgebraType0 m a, AlgebraType0 m b)
=> (a > b)
> Hom m a b
bimapHom :: forall m a' a b b'.
( AlgebraType0 m a'
, AlgebraType0 m b'
)
=> (a' > a)
> (b > b')
> Hom m a b
> Hom m a' b'
bimapHom f g (Hom ab) = Hom (g . ab . f)
data AlgHom m a b where
AlgHom :: ( AlgebraType m a
, AlgebraType m b
)
=> (a > b)
> AlgHom m a b
unAlgHom :: AlgHom m a b > a > b
unAlgHom (AlgHom f) = f
bimapAlgHom :: forall m a' a b b'.
( AlgebraType m a'
, AlgebraType m b'
)
=> (a' > a)
> (b > b')
> AlgHom m a b
> AlgHom m a' b'
bimapAlgHom f g (AlgHom ab) = AlgHom (g . ab . f)
Note that you could define (full sub) categories of Hask
using these bifunctors.
foldMapFree
being a bijection), into single statement:
foldMapFree
is a natural isomorphism between Hom m
and AlgHom m
:
 
ψ :: forall m a d .
( FreeAlgebra m
, AlgebraType0 m a
)
=> AlgHom m (m a) d
> Hom m a d
ψ (AlgHom f) = case forget @m @d of
Proof Dict > Hom $ unFoldMapFree f
with inverse
 
φ :: forall m a d .
( FreeAlgebra m
, AlgebraType m d
, AlgebraType0 m a
)
=> Hom m a d
> AlgHom m (m a) d
φ (Hom f) = case proof @m @a of
Proof Dict > case forget @m @(m a) of
Proof Dict > AlgHom $ foldMapFree f
Such natural bijections arising for a pair of functors F
and
U
are called adjoint functors. In case of monoids the
functor F
is simply the list functor: []
and it is the left adjoin (it appears in the left argument of
AlgHom
) to the forgetful functor from
the category of monoids
to Hask
(it's on the right hand since of
Hom m a d
though in an invisible way,
only because we are dropping a constraint).
Unit and counit of an adjunction
Every adjunction can be characterised by two maps called unit and counit. The unit is defined as
unit :: forall m a .
( FreeAlgebra m
, AlgebraType0 m a
)
=> Hom m a (m a)
unit = case proof @m @a of
Proof Dict > case forget @m @(m a) of
Proof Dict > ψ (AlgHom id)
and
counit :: forall m d .
( FreeAlgebra m
, AlgebraType m d
)
=> AlgHom m (m d) d
counit = case forget @m @d of
Proof Dict > φ (Hom id)
Let us decipher these morphisms. Let's start with counit
counit = φ (Hom id)
= AlgHom (foldMapFree id)
= AlgHom foldFree
So this looks familiar! In case of monoids this is just the good old
friend fold
. The unit
is even simpler:
unit = ψ (AlgHom id)
= Hom (unFoldMapFree id)
= Hom returnFree
In case of monoids the unit is just (:[]) :: a > [a]
.
Monad associated with an adjunction
Every adjunction gives rise to a monad. The left adjoint functor followed by the right adjoint functor is an endofunctor and a monad. In our case the left adjoin functor is the list functor[]
from the
Hask
category to the category of monoids in Hask
,
while the right adjoin is the functor that forgets the Monoid
constraint. So the composition is simply the list endofunctor on
Hask
: (:[]) :: a > [a]
I am sure you know that list is a monad, and it's the same monad as one
given by the (unAlgHom counit) @[] :: [[a]] = [a]
. Note that
the map unAlgHo counit = foldFree
and it can be recasted to
a monad's join
:
 
 @'FreeAlgebra'@ constraint implies @Monad@ constraint.
joinFree :: forall m a .
( FreeAlgebra m
, AlgebraType0 m a
)
=> m (m a)
> m a
joinFree mma = case proof @m @a of
Proof Dict > foldFree mma
From a join you can get the bind operator in a standard way:
 
 
 The monadic @'bind'@ operator. @'returnFree'@ is the corresponding
 @'return'@ for this monad.
bindFree :: forall m a b .
( FreeAlgebra m
, AlgebraType0 m a
, AlgebraType0 m b
)
=> m a
> (a > m b)
> m b
bindFree ma f = case proof @m @b of
Proof Dict > foldMapFree f ma
Together with
returnFree :: FreeAlgebra m => a > m a
we have a a very general formula for constructing monads; actually any
monad arise in this way: from a pair of adjoint functors; the
counit
morphism defines
join
and the unit
defines
return
.
Let us put this monad in a wrapper type, we are using GADTs
to
capture the
FreeAlgebra m
constraint.
 
 The composition of @Free@ followed by @U@
data FreeMAlg (m :: * > *) (a :: *) where
FreeMAlg :: ( FreeAlgebra m, AlgebraType m a ) => m a > FreeMAlg m a
runFreeMAlg :: FreeMAlg m a > m a
runFreeMAlg (FreeMAlg ma) = ma
 
 @FreeMAlg@ is a functor in the category @Hom m@.
fmapF :: forall m a b .
Hom m a b
> FreeMAlg m a
> FreeMAlg m b
fmapF (Hom fn) (FreeMAlg ma) = FreeMAlg $ fmapFree fn ma
 
 unit of the @FreeMAlg@ monad (i.e. @return@ in Haskell)
returnF :: forall m a .
( FreeAlgebra m
, AlgebraType0 m a
, AlgebraType0 m (FreeMAlg m a)
)
=> Hom m a (FreeMAlg m a)
returnF = case unit :: Hom m a (m a) of Hom f > Hom (FreeMAlg . f)
 
 join of the @FreeMAlg@ monad
joinF :: forall m a .
( FreeAlgebra m
, AlgebraType0 m a
, AlgebraType0 m (FreeMAlg m a)
, AlgebraType0 m (FreeMAlg m (FreeMAlg m a))
)
=> Hom m (FreeMAlg m (FreeMAlg m a)) (FreeMAlg m a)
joinF = case proof @m @a of
Proof Dict > case forget @m @(m a) of
Proof Dict >
Hom $ \(FreeMAlg mma)
> FreeMAlg $ joinFree $ fmapFree runFreeMAlg mma
 
 bind of the @'FreeMAlg'@ monad
bindF :: forall m a b .
( FreeAlgebra m
, AlgebraType0 m b
)
=> FreeMAlg m a
> Hom m a (FreeMAlg m b)
> FreeMAlg m b
bindF (FreeMAlg ma) (Hom f) = case proof @m @a of
Proof Dict > case forget @m @(m a) of
Proof Dict > FreeMAlg $ ma `bindFree` (runFreeMAlg . f)
Algebras for a monad
Ifm
is a monad then an m
algebra is a map:
m a > a
. m
algebras form a category where
arrows are commutative squares.
class MAlg m a where
alg :: m a > a
instance MAlg [] [a] where
alg = concat
instance MAlg NonEmpty (NonEmpty a) where
alg = NE.fromList . concatMap NE.toList
We can define MAlg
instance for FreeMAlg
:
instance ( FreeAlgebra m
, AlgebraType0 m a
, AlgebraType0 m (FreeMAlg m a)
) => MAlg m (FreeMAlg m a) where
alg ma = case proof @m @a of
Proof Dict > FreeMAlg $ joinFree $ fmapFree runFreeMAlg ma
Note that this is an undecidable instance, since the
constraints are not smaller than the instance head.
It turns out that FreeMAlg m
is
a left adjoin functor to the forgetful functor $\mathrm{U_{MAlg}}$ from the
category of m
algebras to $\mathrm{Hask}$ which sends an
m
algebra f :: m a > a
to a
. This
is simply because we can write
FreeAlgebra
instance for this type. Well, not quite, only because Haskell computes
types families in a lazy way, but we can define all the morphisms we need
(we leave checking all the necessary properties though to the curious
reader):
returnFreeMAlg
:: FreeAlgebra m
=> a
> FreeMAlg m a
returnFreeMAlg = FreeMAlg . returnFree
foldMapFreeMAlg
:: ( AlgebraType0 m a
, AlgebraType0 m d
, MAlg m d
)
=> (a > d)
> (FreeMAlg m a > d)
foldMapFreeMAlg fn (FreeMAlg ma) = alg $ fmapFree fn ma
foldFreeMAlg
:: ( AlgebraType0 m a
, MAlg m a
)
=> FreeMAlg m a > a
foldFreeMAlg = foldMapFreeMAlg id
We're kind of at the starting point of this post again. Since we have an
instance of
FreeAlgebra
we can define an adjunction in the same way, but for the class
MAlg
.
So we end up with the following picture:
k
and it takes an object in Alg
(which is the category of monoids in our example case). k
can
be defined as follows:
k :: ( FreeAlgebra m
, AlgebraType m a
)
=> Proxy a
> (m a > a)
k _ = foldFree
At the end we have a commutative diagram
Monadicity
A right adjoin functor $\mathrm{U}$ is called monadic iff the
comparison functor is an equivalence of categories. This means that
the two categories Alg
and
MAlg
have exactly the same
categorical properties. It turns out that all algebraic theories which
have free algebras (equivalently equational theories) are monadic.
This includes the categories of semigroups, monoids, groups, etc are
all monadic, but also the categories of functors, applicative functors
and monads on Hask
are monadic too.
From technical point of view, maybe the most important fact is that the
free monad encodes algebraic operations that the initial category was
using. So in out example, we should be able to define
<>
and
mempty
only using the list monad. And it's quite trivial:
 
 @'mempty'@ deduced from @FreeMAlg []@ @MAlg@ instance.
k_inv_monoid_mempty :: MAlg [] a => a
k_inv_monoid_mempty = foldFreeAlg (FreeMAlg [])
 
 @'mappend'@ deduced from @FreeMAlg []@ @MAlg@ instance.
k_inv_monoid_mappend :: MAlg [] a => a > a > a
k_inv_monoid_mappend a b = foldFreeAlg (FreeMAlg [a, b])
 
 @'<>'@ deduced from @FreeMAlg NonEmpty@ @MAAlg@ instance.
k_inv_semigroup_append :: MAlg NonEmpty a => a > a > a
k_inv_semigroup_append a b = foldFreeAlg (FreeMAlg (a : [b]))
k_inv_pointed :: MAlg Maybe a => a
k_inv_pointed = foldFreeAlg (FreeMAlg Nothing)
 
 @'invert'@ deduced from @FreeMAlg FreeGroup@ @MAlg@ instance.
k_inv_group_invert :: (MAlg FreeGroup a, Eq a) => a > a
k_inv_group_invert a
= foldFreeAlg (FreeMAlg (FreeGroup.fromList [Left a]))
 
 @'act'@ deduced from @FreeMAlg FreeMSet@ @MAlg@ instance.
k_inv_act :: (MAlg (FreeMSet m) a, Monoid m) => m > a > a
k_inv_act m a = foldFreeAlg $ FreeMAlg $ FreeMSet (m, a)
A reference for free
group: newtype FreeGroup a = FreeGroup [Either
a a]
.
Laws
There is one more interesting thing here. Not only the operations are
deduced from FreeMAlg m
, but also the laws. For example,
we use FreeMAlg []
for monoids and it encodes both
associativity of the multiplication and unit law for
mempty
and this is simply cerried by the free
monoid [a]
. The same will hold for any other
free algebra and thus by the monad FreeAlgebra m =>
m
. This is just because the free algebra carries all
the laws of the category of given algebra types (monoids
/ semigroups / lattices, ...) and no other equality is built
in.
Final remarks
A working example on which this post is based is published on github. This monadicity works well for higher kinded algebras, e.g. functors, applicative functors or monads. The changes that one needs to make are really minimal: useFreeAlgebra1
type class of FreeAlgebra
. Please check the code.