[Haskell-cafe] From monads to monoids in a small category

Alberto G. Corona agocorona at gmail.com
Wed Sep 5 18:47:17 CEST 2012


Alexander,


In my post (excuses for my dyslexia)  I try to demonstrate that  the
codomain (m a), from the point of view of C. Theory,  can be seen as
'a' plus, optionally, some additional element,  so  a monadic morphism
(a -> m a) is part of a endofunctor in (m a -> m a)

When considering the concept of arrow from category theory,  (not the
concept of function), a point in the domain can "send" more than one
arrow to the codomain, while a function do not.  I think that this is
the most interesting part of the interpretation, if I´m right.

About this, I found this article revealing:

http://cdsmith.wordpress.com/2012/04/18/why-do-monads-matter/

Therefore, codomains of (a -> m a) which are containers with multiple
a elements can be considered as multi-arrow morphisms from 'a' to 'a'
with the optional addition of some special elements that denote
special conditions.


For example

(a -> [a])

May be considered as the general signature of the morphisms from the
set 'a' to the set  ('a' + [])

So a monadic transformation  (a -> [a])  can be considered as a point
transformation of a endofunctor within the set (a + []).


Alberto


2012/9/5 Alexander Solla <alex.solla at gmail.com>:
>
>
> On Tue, Sep 4, 2012 at 4:21 PM, Alexander Solla <alex.solla at gmail.com>
> wrote:
>>
>>
>>
>> On Tue, Sep 4, 2012 at 3:39 AM, Alberto G. Corona <agocorona at gmail.com>
>> wrote:
>>>
>>> "Monads are monoids in the category of endofunctors"
>>>
>>> This Monoid instance for the endofunctors of the set of all  elements
>>> of (m a)   typematch in Haskell with FlexibleInstances:
>>>
>>> instance Monad m => Monoid  (a -> m a) where
>>>    mappend = (>=>)   -- kleisly operator
>>>    mempty  = return
>>
>>
>> The objects of a Kliesli category for a monad m aren't endofunctors.  You
>> want something like:
>>
>> instance Monad m => Monoid (m a -> m (m a)) where ...
>>
>> /These/ are endofunctors, in virtue of join transforming an m (m a) into
>> an (m a).
>
>
> Actually, even these aren't endofunctors, for a similar reason that :  you
> "really" want something like
>
> instance Monad m => Monoid (m a -> m a) where
>     mempty = id
>     mappend = undefined -- exercise left to the reader
>
> (i.e., you want to do plumbing through the Eilenberg-Moore category for a
> monad, instead of the Kliesli category for a monad -- my last message
> exposes the kind of plumping you want, but not the right types.)



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