On Thu, 20 Dec 2012, Gershom Bazerman
On 12/20/12 10:05 PM, Jay Sulzberger wrote:
What does the code for going backwards looks like? That is, suppose we have an instance of Category with only one object. What is the Haskell code for the function which takes the category instance and produces a monoid thing, like your integers with 1 and usual integer multiplication? Could we use a "constraint" at the level of types, or at some other level, to write the code? Here by "constraint" I mean something like a declaration that is a piece of Haskell source code, and not something the human author of the code uses to write the code. instance C.Category k => Monoid (k a a) where mempty = C.id mappend = (C..)
The above gives witness to the fact that, if I'm using the language correctly, if we choose any object (our "a") in any given category, this induces a monoid with the identity morphism as unit and composition of endomorphisms as append.
The standard libraries in fact provide this instance for the function arrow category (under a newtype wrapper):
newtype Endo a = Endo { appEndo :: a -> a }
instance Monoid (Endo a) where mempty = Endo id Endo f `mappend` Endo g = Endo (f . g)
--Gershom
Thanks, Gershom! I think I see. The Haskell code picks out the "isotropy/holonomy" monoid at the object a of any Haskell Category instance. actual old fashioned types remark: To get the holonomy semigroup^Wmonoid, interpolate a functor. I am glad that Haskell today smoothly handles this. ad paper on polymorphisms: I hope to post a rant against the misleading distinction between "parametric polymorphism" and "ad hoc polymorphism". Lisp will be used as a bludgeon in the only argument in the rant. The Four Things Which Must Be Distinguished will perform the opening number. oo--JS.