[Haskell-cafe] Comments from OCaml Hacker Brian Hurt

[Miguel Mitrofanov]

On 16 Jan 2009, at 01:10, Dan Weston wrote:

> Maybe you can explain that again?

Sure.

Consider the following setting: a category C and a bifunctor T : C x C  
-> C, which is associative and have a (left and right) unit I. This is  
what is called "monoidal category".

A "monoid" is an object X in C with two morphisms: I -> X and T(X, X) - 
 > X, satisfying two relatively simple conditions (I don't want to  
draw commutative diagrams).

If your category is a category of sets, and T is a cartesian product,  
then you have ordinary monoids (I is a one-element set, first morphism  
is a unit of a monoid, and second morphism is monoid multiplication).

If, however, you category is a category of endofunctors of some  
category D (that is, functors D -> D), and T is composition, then our  
"monoids" become monads on D: I is an identity functor, first morphism  
is "return", and second one is "join".

>
>
> I see how the subset of Kleisli arrows (a -> m a) forms a monoid (a,  
> return . id, >>=), but what to do with (a -> m b)? (>>=) is not  
> closed under this larger set.
>
> Dan
>
> Miguel Mitrofanov wrote:
>>> Notice that "monoid" sounds almost *exactly* like "monad". And  
>>> yet,  what you use them for is wildly unrelated.
>> Well, monads are monoids. I remember explaining you that...
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