Is this property of a functor stronger than a monad?

Question

While thinking about how to generalize monads, I came up with the following property of a functor F:

inject :: (a -> F b) -> F(a -> b) 

-- which should be a natural transformation in both a and b.

In absence of a better name, I call the functor F bindable if there exists a natural transformation inject shown above.

The main question is, whether this property is already known and has a name, and how is it related to other well-known properties of functors (such as, being applicative, monadic, pointed, traversable, etc.)

The motivation for the name "bindable" comes from the following consideration: Suppose M is a monad and F is a "bindable" functor. Then one has the following natural morphism:

fbind :: M a -> (a -> F(M b)) -> F(M b)

This is similar to the monadic "bind",

bind :: M a -> (a -> M b) -> M b

except the result is decorated with the functor F.

The idea behind fbind was that a generalized monadic operation can produce not just a single result M b but a "functor-ful" F of such results. I want to express the situation when a monadic operation yields several "strands of computation" rather than just one; each "strand of computation" being again a monadic computation.

Note that every functor F has the morphism

eject :: F(a -> b) -> a -> F b

which is converse to "inject". But not every functor F has "inject".

Examples of functors that have "inject": F t = (t,t,t) or F t = c -> (t,t) where c is a constant type. Functors F t = c (constant functor) or F t = (c,t) are not "bindable" (i.e. do not have "inject"). The continuation functor F t = (t -> r) -> r also does not seem to have inject.

The existence of "inject" can be formulated in a different way. Consider the "reader" functor R t = c -> t where c is a constant type. (This functor is applicative and monadic, but that's beside the point.) The "inject" property then means R (F t) -> F (R t), in other words, that R commutes with F. Note that this is not the same as the requirement that F be traversable; that would have been F (R t) -> R (F t), which is always satisfied for any functor F with respect to R.

So far, I was able to show that "inject" implies "fbind" for any monad M.

In addition, I showed that every functor F that has "inject" will also have these additional properties:

• it is pointed

point :: t -> F t

• if F is "bindable" and applicative then F is also a monad

• if F and G are "bindable" then so is the pair functor F * G (but not F + G)

• if F is "bindable" and A is any profunctor then the (pro)functor G t = A t -> F t is bindable

• the identity functor is bindable.

Open questions:

• is the property of being "bindable" equivalent to some other well-known properties, or is it a new property of a functor that is not usually considered?

• are there any other properties of the functor "F" that follow from the existence of "inject"?

• do we need any laws for "inject", would that be useful? For instance, we could require that R (F t) be isomorphic to F (R t) in one or both directions.

Answer 1

To improve terminology a little bit, I propose to call these functors "rigid" instead of "bindable". The motivation for saying "rigid" will be explained below.

Definition

A functor f is called rigid if it has the inject method as shown. Note that every functor has the eject method.

class (Functor f) => Rigid f where
  inject :: (a -> f b) -> f(a -> b)

  eject :: f(a -> b) -> a -> f b
  eject fab x = fmap (\ab -> ab x) fab

The law of "nondegeneracy" must hold:

eject . inject = id

Properties

A rigid functor is always pointed:

instance (Rigid f) => Pointed f where
  point :: t -> f t
  point x = fmap (const x) (inject id)

If a rigid functor is applicative then it is automatically monadic:

instance (Rigid f, Applicative f) => Monad f where
  bind :: f a -> (a -> f b) -> f b
  bind fa afb = (inject afb) <*> fa

The property of being rigid is not comparable (neither weaker nor stronger) than the property of being monadic: If a functor is rigid, it does not seem to follow that it is automatically monadic (although I don't know specific counterexamples for this case). If a functor is monadic, it does not follow that it is rigid (there are counterexamples).

Basic counterexamples of monadic functors that are not rigid are Maybe and List. These are functors that have more than one constructor: such functors cannot be rigid.

The problem with implementing inject for Maybe is that inject must transform a function of type a -> Maybe b into Maybe(a -> b) while Maybe has two constructors. A function of type a -> Maybe b could return different constructors for different values of a. However, we are supposed to construct a value of type Maybe(a -> b). If for some a the given function produces Nothing, we don't have a b so we can't produce a total function a->b. Thus we cannot return Just(a->b); we are forced to return Nothing as long as the given function produces Nothing even for one value of a. But we cannot check that a given function of type a -> Maybe b produces Just(...) for all values of a. Therefore we are forced to return Nothing in all cases. This will not satisfy the law of nondegeneracy.

So, we can implement inject if f t is a container of "fixed shape" (having only one constructor). Hence the name "rigid".

Another explanation as to why rigidity is more restrictive than monadicity is to consider the naturally defined expression

(inject id) :: f(f a -> a) 

where id :: f a -> f a. This shows that we can have an f-algebra f a -> a for any type a, as long as it is wrapped inside f. It is not true that any monad has an algebra; for example, the various "future" monads as well as the IO monad describe computations of type f a that do not allow us to extract values of type a - we shouldn't be able to have a method of type f a -> a even if wrapped inside an f-container. This shows that the "future" monads and the IO monad are not rigid.

A property that is strictly stronger than rigidity is distributivity from one of E. Kmett's packages. A functor f is distributive if we can interchange the order as in p (f t) -> f (p t) for any functor p. Rigidity is the same as being able to interchange the order only with respect to the "reader" functor r t = a -> t. So, all distributive functors are rigid.

All distributive functors are necessarily representable, which means they are equivalent to the "reader" functor c -> t with some fixed type c. However, not all rigid functors are representable. An example is the functor g defined by

type g t = (t -> r) -> t

The functor g are not equivalent to c -> t with a fixed type c.

Constructions and examples

Further examples of rigid functors that are not representable (i.e. not "distributive") are functors of the form a t -> f t where a is any contrafunctor and f is a rigid functor. Also, the Cartesian product and the composition of two rigid functors is again rigid. In this way, we can produce many examples of rigid functors within the exponential-polynomial class of functors.

My answer to What is the general case of QuickCheck's promote function? also lists the constructions of rigid functors:

1. f = Identity
2. if f and g are both rigid then the functor product h t = (f t, g t) is also rigid
3. if f and g are both rigid then the composition h t = f (g t) is also rigid
4. if f is rigid and g is any contravariant functor then the functor h t = g t -> f t is rigid

One other property of rigid functors is that the type r () is equivalent to (), i.e. there is only one distinct value of the type r (). This value is point (), where point is defined above for any rigid functor r. (I have a proof but I will not write it here, because I could not find an easy one-line proof.) A consequence is that a rigid functor must have only one constructor. This immediately shows that Maybe, Either, List etc. cannot be rigid.

Connection with monads

If f is a monad that has a monad transformer of the "composed-outside" kind, t m a = f (m a), then f is a rigid functor.

The "rigid monads" are possibly a subset of rigid functors because construction 4 only yields a rigid monad if f is also a rigid monad rather than an arbitrary rigid functor (but the contravariant functor g can still be arbitrary). However, I do not have any examples of a rigid functor that is not also a monad.

The simplest example of a rigid monad is type r a = (a -> p) -> a, the "search monad". (Here p is a fixed type.)

To prove that a monad f with the "composed-outside" transformer t m a = f (m a) also has an inject method, we consider the transformer t m a with the foreign monad m chosen as the reader monad, m a = r -> a. Then the function inject with the correct type signature is defined as

 inject = join @t . return @r . (fmap @m (fmap @f return @m))

with appropriate choices of type parameters.

The non-degeneracy law follows from the monadic naturality of t: the monadic morphism m -> Identity (substituting a value of type r into the reader) is lifted to the monadic morphism t m a -> t Id a. I omit the details of this proof.

Use cases

Finally, I found two use cases for rigid functors.

The first use case was the original motivation for considering rigid functors: we would like to return several monadic results at once. If m is a monad and we want to have fbind as shown in the question, we need f to be rigid. Then we can implement fbind as

fbind :: m a -> (a -> f (m b)) -> f (m b)
fbind ma afmb = fmap (bind ma) (inject afmb)

We can use fbind to have monadic operations that return more than one monadic result (or, more generally, a rigid functor-ful of monadic results), for any monad m.

The second use case grows out of the following consideration. Suppose we have a program p :: a that internally uses a function f :: b -> c. Now, we notice that the function f is very slow, and we would like to refactor the program by replacing f with a monadic "future" or "task", or generally with a Kleisli arrow f' :: b -> m c for some monad m. We, of course, expect that the program p will become monadic as well: p' :: m a. Our task is to refactor p into p'.

The refactoring proceeds in two steps: First, we refactor the program p so that the function f is explicitly an argument of p. Assume that this has been done, so that now we have p = q f where

q :: (b -> c) -> a

Second, we replace f by f'. We now assume that q and f' are given. We would like to construct the new program q' of the type

q' :: (b -> m c) -> m a

so that p' = q' f'. The question is whether we can define a general combinator that will refactor q into q',

refactor :: ((b -> c) -> a) -> (b -> m c) -> m a

It turns out that refactor can be constructed only if m is a rigid functor. In trying to implement refactor, we find essentially the same problem as when we tried to implement inject for Maybe: we are given a function f' :: b -> m c that could return different monadic effects m c for different b, but we are required to construct m a, which must represent the same monadic effect for all b. This cannot work, for instance, if m is a monad with more than one constructor.

If m is rigid (and we do not need to require that m be a monad), we can implement refactor:

refactor bca bmc = fmap bca (inject bmc)

If m is not rigid, we cannot refactor arbitrary programs. So far we have seen that the continuation monad is rigid, but the "future"-like monads and the IO monad are not rigid. This again shows that rigidity is, in a sense, a stronger property than monadicity.

Answer 2

I have been doing some experiments lately to better understand Distributive. Happily enough, my results appear closely related to your rigid functors, in a way that clarifies them both.

To begin with, here is one possible presentation of rigid functors. I have taken the liberty to bikeshed your names a bit, for reasons I'll soon get to:

flap :: Functor f => f (a -> b) -> a -> f b
flap u a = ($ a) <$> u 

class Functor g => Rigid g where
    fflip :: (a -> g b) -> g (a -> b)
    fflip f = (. f) <$> extractors

    extractors :: g (g a -> a)
    extractors = fflip id

-- "Left inverse"/non-degeneracy law: flap . fflip = id

instance Rigid ((->) r) where
    fflip = flip

Some remarks on my phrasing:

• I have changed the names of inject and eject to fflip and flap, mainly because, to my eyes, flap looks more like injecting, due to things like this:

  sweep :: Functor f => f a -> b -> f (a, b)
  sweep u b = flap ((,) <$> u) b
  

• I took the flap name from protolude. It is a play on flip, which is fitting because it is one of two symmetrical ways of generalising it. (We can either pull the function outside of an arbitrary Functor, as in flap, or pull a Rigid functor outside of a function, as in fflip.)

• I first realised the significance of extractors while playing with Distributive, but it hadn't occured to me that it might make sense as part of a different class. extractors and fflip are interdefinable, making it possible to write, for example, this rather neat instance for the search/selection monad:

  newtype Sel r a = Sel { runSel :: (a -> r) -> a }
      deriving (Functor, Applicative, Monad) via SelectT r Identity
  
  instance Rigid (Sel r) where
      -- Sel r (Sel r a -> a) ~ ((Sel r a -> a) -> r) -> Sel r a -> a
      extractors = Sel $ \k m -> m `runSel` \a -> k (const a)
  

Every distributive functor is rigid:

fflipDistrib :: Distributive g => (a -> g b) -> g (a -> b)
fflipDistrib = distribute @_ @((->) _)
-- From this point on, I will pretend Rigid is a superclass of Distributive.
-- There would be some tough questions about interface ergonomics if we were
-- writing this into a library. We don't have to worry about that right now,
-- though.

From the other direction, we can write a function which imitates the signature of distribute using Rigid:

infuse :: (Rigid g, Functor f) => f (g a) -> g (f a)
infuse u = (<$> u) <$> extractors

infuse, however, is not distribute. As you note, there are rigid functors that are not distributive, such as Sel. Therefore, we have to conclude that infuse, in the general case, does not follow the distributive laws.

(An aside: that infuse is not a lawful distribute in the case of Sel can be established by a cardinality argument. If infuse followed the distributive laws, we would have infuse . infuse = id for any two rigid functors. However, something like infuse @((->) Bool) @(Sel r) leads to a result type with fewer inhabitants than the argument type; therefore, there is no way it can have a left inverse.)

A place in the constellation

At this point, it would be relevant to sharpen our picture of exactly what distinguishes Distributive from Rigid. Given that your rigid law is flap . fflip = id, intuition suggests the other half of an isomorphism, fflip . flap = id, might hold in the case of Distributive. Checking that hypothesis requires a detour through Distributive.

There is an alternative presentation of Distributive (and Rigid) in which distribute (or fflip) is factored through the function functor. More specifically, any functorial value of type g a can be converted into a CPS suspension that takes a forall x. g x -> x extractor:

-- The existential wrapper is needed to prevent undue specialisation by GHC.
-- With pen and paper, we can leave it implicit.
-- Note this isn't necessarily the best implementation available; see also
-- https://stackoverflow.com/q/56826733/2751851
data Ev g a where
    Ev :: ((g x -> x) -> a) -> Ev g a

-- Existential aside, this is ultimately just a function type.
deriving instance Functor (Ev g)

-- Morally, evert = flip id
evert :: g a -> Ev g a
evert u = Ev $ \e -> e u

If g is Rigid, we can go in the other direction and recover the functorial value from the suspension:

-- Morally, revert = flip fmap extractors
revert :: Rigid g => Ev g a -> g a
revert (Ev s) = s <$> extractors

Ev g itself is Distributive, regardless of what g is -- after all, it is just a function:

-- We need unsafeCoerce (yikes!) because GHC can't be persuaded that we aren't
-- doing anything untoward with the existential.
-- Note that flip = fflip @((->) _)
instance Rigid (Ev g) where
    fflip = Ev . flip . fmap (\(Ev s) -> unsafeCoerce s)

-- Analogously, flap = distribute @((->) _)
instance Distributive (Ev g) where
    distribute = Ev . flap . fmap (\(Ev s) -> unsafeCoerce s) 

Further, fflip and distribute for arbitrary Rigid/Distributive functors can be routed through evert and revert:

-- fflip @(Ev g) ~ flip = distribute @((->) _) @((->) _)
fflipEv :: Rigid g => (a -> g b) -> g (a -> b)
fflipEv = revert . fflip . fmap evert

-- distribute @(Ev g) ~ flap = distribute @((->) _) _
distributeEv :: (Rigid g, Functor f) => f (g a) -> g (f a) 
distributeEv = revert . distribute . fmap evert

revert, in fact, would be enough for implementing Distributive. In such terms, the distributive laws amount to requiring evert and revert being inverses:

revert . evert = id  -- "home" roundtrip, right inverse law
evert . revert = id  -- "away" roundtrip, left inverse law

The two roundtrips correspond, respectively, to the two non-free distributive laws:

fmap runIdentity . distribute = runIdentity                               -- identity
fmap getCompose . distribute = distribute . fmap distribute . getCompose  -- composition

(The distribute . distribute = id requirement stated in the Data.Distributive docs ultimately amounts to those two laws, plus naturality.)

Earlier on, I speculated about an isomorphism involving fflip:

flap . fflip = id  -- "home" roundtrip, left inverse Rigid law  
fflip . flap = id  -- "away" roundtrip, would-be right inverse law

It can be verified directly that the rigid law, flap . fflip = id, is equivalent to the other "home" roundtrip, revert . evert = id. The other direction is trickier. The purported isomorphisms can be chained like this:

                        g (a -> b)        
    {fflip => <= flap}              {evert => <= revert}
a -> g b                                                   Ev g (a -> b)
    {fmap evert => <= fmap revert} {distribute => <= distribute}
                             a -> Ev g b

Let's assume the rigid law holds. We want to prove that fflip . flap = id if and only if evert . revert = id, so we must handle both directions:

• Firstly, let's assume evert . revert = id. The counterclockwise way of going around the square from a -> g b to g (a -> b) amounts to fflip (see the definition of fflipEv above). As the conterclockwise way is made out of three isomorphisms, it follows that fflip has an inverse. Since flap is its left inverse (by the rigid law), it must also be its inverse. Therefore fflip . flap = id.

• Secondly, let's assume fflip . flap = id. Again, the counterclockwise way from a -> g b to g (a -> b) is fflip, but now we know that it has an inverse, namely flap. It follows that each of the functions composed to make up the counterclockwise way must have an inverse. In particular, revert must have an inverse. Since evert is its right inverse (by the rigid law), it must also be its inverse. Therefore, evert . revert = id.

The results above allow us to precisely situate where Rigid stands in relation to Distributive. A rigid functor is a would-be distributive, except that it only follows the identity law of distributive, and not the composition one. Making fflip an isomorphism, with flap as its inverse, amounts to upgrading Rigid to Distributive.

Miscellaneous remarks

• Looking at fflip and flap from a monadic point of view, we might say that rigid monads are equipped with an injective conversion from Kleisli arrows to static arrows. With distributive monads, the conversion is upgraded to an isomorphism, which is a generalisation of how Applicative and Monad are equivalent for Reader.

• extractors condenses much of what Distributive is about. For any distributive functor g, there is a g (g a -> a) value in which each position is filled with a matching g a -> a extractor function. It seems accurate to say that when we move from Distributive to Rigid we lose this guarantee that position and extractor will match, and, with it, the ability to reconstruct an adequate functorial shape out of nothing. In this context, it is worth having a second look at the extractors implementation for Sel early in this answer. Any a -> r function corresponds to a Sel r a -> a extractor, which means there generally will be a myriad of extractors we can't enumerate, so we have to satisfy ourselves with non-isomorphic fflip and infuse (in hindsight, the const that shows up in the implementation of extractors already gives the game away). This feels a bit like the lack of a Traversable instance for functions. (In that case, though, there is a way to cheat if the domain type of the function is enumerable, Data.Universe style. I'm not sure if there actually is such a workaround, however impractical, for Sel.)

• I obtained the results about the revert isomorphism for Distributive largely by mirroring how the shape-and-contents decomposition of Traversable, the dual class, works. (A very readable paper that explores the shape-and-contents theme is Understanding Idiomatic Traversals Backwards and Forwards, by Bird et. al.). While covering that in more detail would probably be better left for a separate post, there is at least one question worth posing here: does a notion analogous to Rigid make sense for Traversable? I believe it does, albeit my feeling is that it sounds less useful than Rigid might be. One example of a "co-rigid" pseudo-traversable would be a data structure equipped with a traversal that duplicates effects, but then discards the corresponding duplicate elements upon rebuilding the structure under the applicative layer, so that the identity law is followed -- but not the composition one.

• Speaking of revert, the Ev construction is in itself quite meaningful: it is an encoding of the free distributive functor. In particular, evert and revert are comparable to liftF and retract for free monads, as well as to similar functions for other free constructions. (In such a context, revert being a full inverse to evert hints at how strong Distributive is. It is more usual for the retraction to discard information in some cases, as it happens in the general case of Rigid.)

• Last, but not least, there is another way still of making sense of Ev: it means the polymorphic extractor type represents the distributive functor, in the Representable sense, with evert corresponding to index, and revert, to tabulate. Unfortunately, the quantification makes it very awkward to express that in Haskell with the actual Representable interface. (It is symptomatic that I had to reach for unsafeCoerce to give Ev its natural Rigid and Distributive instances.) If it serves as solace, it wouldn't be a terribly practical representation anyway: if I already have a polymorphic extractor in hands, I don't actually need index for extracting values.

Answer 3

We are all familiar with the Traversable typeclass, which can be boiled down to the following:

class Functor t => Traversable t
  where
  sequenceA :: Applicative f => t (f a) -> f (t a)

This makes use of the concept of an Applicative functor. There is a laws-only strengthening of the categorical concept underlying Applicative that goes like this:

-- Laxities of a lax monoidal endofunctor on Hask under (,)
zip :: Applicative f => (f a, f b) -> f (a, b)
zip = uncurry $ liftA2 (,)

husk :: Applicative f => () -> f ()
husk = pure

-- Oplaxities of an oplax monoidal endofunctor on ... (this is trivial for all endofunctors on Hask)
unzip :: Functor f => f (a, b) -> (f a, f b)
unzip fab = (fst <$> fab, snd <$> fab)

unhusk :: f () -> ()
unhusk = const ()

-- The class
class Applicative f => StrongApplicative f

-- The laws
-- zip . unzip = id
-- unzip . zip = id
-- husk . unhusk = id
-- unhusk . husk = id -- this one is trivial

The linked question and its answers have more details, but the gist is that StrongApplicatives model some notion of "fixed size" for functors. This typeclass has an interesting connection to Representable functors. For reference, Representable is:

class Functor f => Representable x f | f -> x
  where
  rep :: f a -> (x -> a)
  unrep :: (x -> a) -> f a

instance Representable a ((->) a)
  where
  rep = id
  unrep = id

An argument by @Daniel Wagner shows that StrongApplicative is a generalization of Representable, in that every Representable is StrongApplicative. Whether there are any StrongApplicatives that are not Representable is not yet clear.

Now, we know that Traversable is formulated in terms of Applicative, and runs in one direction. Since StrongApplicative promotes the Applicative laxities to isomorphisms, perhaps we want to use this extra equiment to invert the distributive law that Traversable supplies:

class Functor f => Something f
  where
  unsequence :: StrongApplicative f => f (t a) -> t (f a)

It just so happens that (->) a is a StrongApplicative, and in fact a representative specimen (if you'll pardon the pun) of the genus of Representable StrongApplicative functors. Hence we can write your inject/promote operation as:

promote :: Something f => (a -> f b) -> f (a -> b)
promote = unsequence

We mentioned before that StrongApplicative is a superclass of the family of Representative functors. From examining the type of unsequence, it is obvious that the stronger a constraint we place on the polymorphic applicative, the easier it will be to implement unsequence (and hence the more instances of the resulting class).

So in a sense there is a hierarchy of "detraversable" functors that flows in the opposite direction to a hierarchy of applicative effects with respect to which you might wish to detraverse them. The hierarchy of "inner" functors would go like this:

Functor f => Applicative f => StrongApplicative f => Representable x f

And the corresponding hierarchy of detraversable/distributive functors might go like this:

Distributive t <= ADistributive t <= SADistributive t <= RDistributive t

With definitions:

class RDistributive t
  where
  rdistribute :: Representable x f => f (t a) -> t (f a)

  default rdistribute :: (SADistributive t, StrongApplicative f) => f (t a) -> t (f a)
  rdistribute = sadistribute

class RDistributive t => SADistributive t
  where
  sadistribute :: StrongApplicative f => f (t a) -> t (f a)

  default sadistribute :: (ADistributive t, Applicative f) => f (t a) -> t (f a)
  sadistribute = adistribute

class SADistributive t => ADistributive t
  where
  adistribute :: Applicative f => f (t a) -> t (f a)

  default adistribute :: (Distributive t, Functor f) => f (t a) -> t (f a)
  adistribute = distribute

class ADistributive t => Distributive t
  where
  distribute :: Functor f => f (t a) -> t (f a)

Our definition of promote can be generalized to depend on RDistributive (since (->) a itself is indeed a representable functor):

promote :: RDistributive f => (a -> f b) -> f (a -> b)
promote = rdistribute

In a strange turn of events, once you get down to the bottom of this hierarchy (i.e. to Distributive), your promise of detraversability has become so strong relative to your demands that the only functors for which you can implement it are themselves Representable. An example of such a distributive, representable (and hence rigid) functor is that of pairs:

data Pair a = Pair { pfst :: a, psnd :: a }
  deriving Functor

instance RDistributive Pair
instance SADistributive Pair
instance ADistributive Pair
instance Distributive Pair
  where
  distribute x = Pair (pfst <$> x) (psnd <$> x)

Of course if you make a strong demand of the polymorphic "inner functor", for example Representable x f in RDistributive, instances like this become possible:

newtype Weird r a = Weird { runWeird :: (a -> r) -> a }
  deriving Functor

instance RDistributive (Weird r)
  where
  rdistribute = fmap unrep . promoteWeird . rep
    where
    promoteWeird :: (x -> Weird r a) -> Weird r (x -> a)
    promoteWeird f = fmap (. f) $ Weird $ \k m -> m `runWeird` \a -> k (const a)

TODO: Check where (if anywhere) in the hierarchy all the other examples of rigid functors fall.

As I said I haven't thought about it super carefully, so maybe the folks here that have devoted some thought to the rigid functor concept can immediately poke holes in it. Alternately, maybe it makes things fall into place that I can't yet see.

It's probably worthwhile thinking about some laws for these untraversing typeclasses. An obvious one that suggests itself is sequence . unsequence = id and unsequence . sequence = id wherever the functor supports both Traversable and Untraverse.

It's also worth mentioning that the interaction of "distributive law"s of functors with monads and comonads is quite well studied, so that might have some relevance to the monad related discussion in your posts.