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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