[Haskell-cafe] A quick question about distribution of finite maps.
David Feuer
david.feuer at gmail.com
Sat Jan 9 21:37:38 UTC 2021
Where do you get
() -> Maybe (X^Y) ~
OneOrBoth () Y -> Maybe X
On Sat, Jan 9, 2021, 4:26 PM MigMit <migmit at gmail.com> wrote:
> 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]:
> >>
> https://hackage.haskell.org/package/distributive-0.6.2.1/docs/Data-Distributive.html#t:Distributive
> >
> > 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
> > every monad m, where
> > * 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
> >
> >
> >
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