MigMit migmit at gmail.com
Sat Jan 9 21:26:28 UTC 2021

```Actually, it's pretty easy to construct a type `P x y`, so that Maybe (P x y) ~ (Maybe x, Maybe y). It would be

data OneOrBoth x y = Left' x | Right' y | Both x y

The isomorphism is, I think, obvious

iso1 :: Maybe (OneOrBoth x y) -> (Maybe x, Maybe y)
iso1 Nothing = (Nothing, Nothing)
iso1 (Just (Left' x)) = (Just x, Nothing)
iso1 (Just (Right' y)) = (Nothing, Just y)
iso1 (Just (Both x y)) = (Just x, Just y)

iso2 :: (Maybe x, Maybe y) -> Maybe (OneOrBoth x y)
iso2 = -- left as an excersize for the reader

And indeed, "OneOrBoth" would be a cartesian product functor in the category of finite types (and maps).

But it won't be cartesian closed. If it were, then for any finite X and Y we should have

Maybe (X^Y) ~
() -> Maybe (X^Y) ~
OneOrBoth () Y -> Maybe X ~
(() -> Maybe X, Y -> Maybe X, ((), Y) -> Maybe X) ~
(Maybe X, Y -> Maybe X, Y -> Maybe X)

so

X^Y ~ (X, Y -> Maybe X, Y -> Maybe X)

But then

Z -> Maybe (X^Y) ~
Z -> (Maybe X, Y -> Maybe X, Y -> Maybe X) ~
(Z -> Maybe X, (Z, Y) -> Maybe X, (Z, Y) -> Maybe X) ~

and

OneOrBoth Z Y -> Maybe X ~
(Z -> Maybe X, Y -> Maybe X, (Z, Y) -> Maybe X)

We see that those aren't the same, they have a different number of elements, so, no luck.

> On 9 Jan 2021, at 22:01, Olaf Klinke <olf at aatal-apotheke.de> wrote:
>
>> Hello!
>>
>> Finite maps from `"containers" Data.Map` look like they may form a
>> Cartesian closed category. So it makes sense to ask if the rule _α ⇸
>> (β ⇸ γ) ≡ ⟨α; β⟩ ⇸ γ ≡ ⟨β; α⟩ ⇸ γ ≡ β ⇸ (α ⇸ γ)_ that holds in such
>> categories does hold for finite maps. Note that, a map being a
>> functor, this also may be seen as _f (g α) ≡ g (f α)_, which would
>> work if maps were `Distributive` [1].
>>
>> It looks to me as though the following definition might work:
>>
>>    distribute = unionsWith union . mapWithKey (fmap . singleton)
>>
>> — And it works on simple examples. _(I checked the law `distribute ∘
>> distribute ≡ id` — it appears to be the only law required.)_
>>
>> Is this definition correct? Is it already known and defined
>> elsewhere?
>>
>> [1]:
>
> Hi Ignat,
>
> TL;DR: No and no.
>
> The documentation says that every distributive functor is of the form
> (->) x for some x, and (Map a) is not like this.
>
> If Maps were a category, what is the identity morphism?
>
> Let's put the Ord constraint on the keys aside, Tom Smeding has already
> commented on that. Next, a Map is always finite, hence let's pretend
> that we are working inside the category of finite types and functions.
> Then the problems of missing identity and missing Ord go away. Once
> that all types are finite, we can assume an enumerator. That is, each
> type x has an operation
> enumerate :: [x]
> which we will use to construct the inverse of
> flip Map.lookup :: Map a b -> a -> Maybe b
> thereby showing that a Map is nothing but a memoized version of a
> Kleisli map (a -> Maybe b). Convince yourself that Map concatenation
> has the same semantics as Kleisli composition (<=<). Given a Kleisli
> map k between finite types, we build a Map as follows.
> \k -> Map.fromList (enumerate >>= (\a -> maybe [] (pure.(,) a) (k a)))
>
> With that knowledge, we can answer your question by deciding: Is the
> Kleisli category of the Maybe monad on finite types Cartesian closed?
> Short answer: It is not even Cartesian.
> There is an adjunction between the categories (->) and (Kleisli m) for
> * The left adjoint functor takes
>   types x to x,
>   functions f to return.f
> * The right adjoint functor takes
>   types x to m x,
>   Kleisli maps f to (=<<) f
> Right adjoint functors preserve all existing limits, which includes
> products. Therefore, if (Kleisli m) has binary products, then m must
> preserve them. So if P x y was the product of x and y in Kleisli m,
> then m (P x y) would be isomorphic to (m x,m y). This seems not to hold
> for m = Maybe: I can not imagine a type constructor P where
> Maybe (P x y) ~ (Maybe x,Maybe y).
> In particular, P can not be (,). The only sensible Kleisli projections
> from (x,y) would be fst' = return.fst and snd' = return.snd. Now think
> of two Kleisli maps f :: Bool -> Maybe x, g :: Bool -> Maybe y. Assume
> that f True = Just x for some x and g True = Nothing. In order to
> satisfy g True = (snd' <=< (f&&&g))True, the unique pair arrow (f&&&g)
> would need to map True to Nothing, but then f True = (fst' <=< (f&&&g))
> True can not hold. We conclude that (Kleisli Maybe) does not even have
> categorical products, so asking for Cartesian closure does not make
> sense.
>
> You might ask for a weaker property: For every type x, ((,) x) is a
> functor on (Kleisli Maybe). Indeed, the following works because ((,) x)
> is a polynomial functor.
> fmapKleisli :: Functor m => (a -> m b) -> (x,a) -> m (x,b)
> fmapKleisli f (x,a) = fmap ((,) x) (f a)
> Thus you may ask whether this functor has a right adjoint in the
> Kleisli category of Maybe. This would be a type constructor g with a
> natural isomorphism
>
> (x,a) -> Maybe b   ~   a -> Maybe (g b).
>
> The first thing that comes to mind is to try
> g b = x -> Maybe b and indeed djinn can provide two functions going
> back and forth that have the right type, but they do not establish an
> isomorphism. I doubt there is such a right adjoint g, but can not prove
> it at the moment. The idea is that a function (x,a) -> Maybe b may
> decide for Nothing depending on both x and a, and therefore the image
> function under the isomorphism must map every a to Just (g b) and delay
> the Nothing-decision to the g b. But for the reverse isomorphism you
> can have functions that do not always return Just (g b) and there is no
> preimage for these.
>
> Regards,
> Olaf
>
>
>
> _______________________________________________