[Haskell-cafe] A quick question about distribution of finite maps.

David Feuer david.feuer at gmail.com
Sat Jan 9 23:04:26 UTC 2021 

`These` is not my own invention. It's in the `these` package.

On Sat, Jan 9, 2021, 5:41 PM Olaf Klinke <olf at aatal-apotheke.de> wrote:

> On Sat, 2021-01-09 at 22:26 +0100, MigMit 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).
>
> I stand corrected. Below is the full code, in case anyone wants to play
> with it.
>
> import Control.Arrow ((&&&))
>
> -- due to David Feuer
> data OneOrBoth x y = Left' x | Right' y | Both x y
>
> iso2 :: (Maybe x,Maybe y) -> Maybe (OneOrBoth x y)
> iso2 p = case p of
>     (Nothing,Nothing) -> Nothing
>     (Just x,Nothing) -> (Just (Left' x))
>     (Nothing,Just y) -> (Just (Right' y))
>     (Just x, Just y) -> (Just (Both x y))
>
> -- (OneOrBoth x) is a functor on Kleisli Maybe
> fmapKleisli :: (a -> Maybe b) ->
>     OneOrBoth x a -> Maybe (OneOrBoth x b)
> fmapKleisli k (Left' x) = Just (Left' x)
> fmapKleisli k (Right' a) = fmap Right' (k a)
> fmapKleisli k (Both x a) = fmap (Both x) (k a)
>
> -- is OneOrBoth the cartesian product? Seems so:
>
> pairKleisli :: (a -> Maybe x) ->
>     (a -> Maybe y) ->
>     a -> Maybe (OneOrBoth x y)
> pairKleisli f g = iso2 . (f &&& g)
>
> fstMaybe :: OneOrBoth x y -> Maybe x
> fstMaybe (Left' x) = Just x
> fstMaybe (Both x _) = Just x
> fstMaybe (Right' _) = Nothing
>
> sndMaybe :: OneOrBoth x y -> Maybe y
> sndMaybe (Left' _) = Nothing
> sndMaybe (Right' y) = Just y
> sndMaybe (Both _ y) = Just y
>
> >
> > 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.
>
> Doesn't this chain of isomorphisms require () to be the terminal
> object, or did you take () as synonym for the terminal object in the
> category? For example, there are several functions of the type
> Bool -> Maybe ().
> So the terminal object in Kleisli Maybe would be Void, because Maybe
> Void ~ (). We'd need to make fields in OneOrBoth strict to have an
> isomorphism OneOrBoth Void a ~ a, just as the true categorical product
> in (->) is the strict pair.
>
> Olaf
>
> >
> > > 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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>
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