[Haskell-cafe] Categories (cont.)

[Jay Sulzberger]

On Thu, 20 Dec 2012, Gershom Bazerman <gershomb at gmail.com> wrote:

> 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.