Say I've got a Distributive
instance written for some complex custom type, Foo
.
Is it possible to write Foo
's Representable
instance using only the properties available from its Distributive
instance? And, if not, then why is Distributive
a superclass of Representable
?
The superclass relationship between Distributive
and Representable
...
class Distributive f => Representable f where
... means that if f
is a Representable
, then it also must be a Distributive
, and not the other way around. When used for subclassing, =>
should be read as "is a prerequisite of", rather than "implies". (This is, in fact, opposite to how it is when =>
is used for constraints in type signatures. Purescript uses <=
for subclassing for this very reason.)
For most other pairs of superclass and subclass, the story would end here. Distributive
and Representable
, however, have a special relationship, in that Distributive
functors are actually representable, as stated by the documentation of both Distributive
...
Categorically every
Distributive
functor is actually a right adjoint, and so it must beRepresentable
endofunctor and preserve all limits. This is a fancy way of saying it isomorphic to(->) x
for somex
.
... and Representable
:
A
Functor
f
isRepresentable
iftabulate
andindex
witness an isomorphism to(->) x
.Every
Distributive
Functor
is actuallyRepresentable
.Every
Representable
Functor
from Hask to Hask is a right adjoint.
The hierarchy is set up the way it is, with Distributive
as the superclass, because Distributive
is meant to have a simpler interface which is expressible in Haskell 98, unlike Representable
(which uses a type family) and Adjunction
(which is a multi-parameter type class). From a more conceptual point of view, while the distributive laws imply that every Distributive
is representable, they are not enough for figuring out what the representation is. Getting our hands on the representation requires specifying it, either directly (as in Representable
) or indirectly (as in Adjunction
).
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