Don't understand notation of morphisms in Monoid definition

12
votes

I'm trying to understand what Monoid is from a category theory perspective, but I'm a bit confused with the notation used to describe it. Here is Wikipedia:

In category theory, a monoid (or monoid object) (M, μ, η) in a monoidal category (C, ⊗, I) is an object M together with two morphisms

μ: M ⊗ M → M called multiplication,

η: I → M called unit

My confusion is about the morphism notation. Why is the binary operation ⊗ a part of the morphism notation? My understanding of a morphism is that it's a kind of function that can map from one type to another (domain to codomain), like M → M. Why is the operation ⊗ a part of the domain in the definition? The second confusion is about I. Why is I a domain? There is no I object in a Monoid at all. It's just a neutral element of the object M.

I understand that Monoid is a category with one object, an identity morphism, and a binary operation defined on this object, but the notation makes me think that I don't understand something.

Is M ⊗ M somehow related to the cartesian product, so that the domain of the morphism is defined as M x M?

Edit: I got a really helpful answer for my question on the Mathematics Stack Exchange.

1
haskellcategory-theorymonoids
Do you understand what monoid is from a set theory perspective? Definitions just are -- there's little answer that can be given to "why is the definition that way" other than "because we observe there are a bunch of things we want to talk about uniformly that are that way". So I can't really imagine answering that question sensibly. But I could imagine plausibly answering "How does this definition correspond with the various parts of the set theory definition?". – Daniel Wagner
Okay. Since your (now deleted) comment says you grok the set theory definition, when I get back from lunch if there's no answer yet I'll write up a description of how the parts of the two definitions correspond; and the connection to the category with one object that's induced by the definition. – Daniel Wagner
@daniel-wanger Yes, I understand what monoid is from set theory (at least the main idea). I even have some "non-stable" understanding of what it is in category theory. The main problem for me now (at least I think it's the main one)) is that I don't understand how to read notations. Is there kind of rule that say that on the left side (domain) there can be object only?. Because definition like this I -> M looks very strange for me. Like it's morphism from object I (which is not object) to M. Or M ⊗ M is kind of domain. But I would really appreciated for any answer. – Bogdan Vakulenko
Warning! There are different but related notions at work! A category with one object corresponds to the traditional notion of a monoid (e.g., the ways you can get from where you are to where you are by hopping on the spot). The notion of monoidal category is something else: they can have much more interesting collections of objects in which (X) and I induce monoid-like structure (the way (,) on Haskell types is associative and absorbs () up to isomorphism). – pigworker
The words "monoidal category" are essential, and it's where these operators you are asking about come from. You seem to be ignoring them. – luqui

1 Answers

21
votes

Is M ⊗ M some how related to cartesian product, so domain of the morphism is defined as M x M ?

Exactly. More specifically, we get those monoids that are expressed by the Monoid class from base by picking Hask (the category with all Haskell types as objects and all Haskell functions as morphisms) as C, (,) (the pair type constructor) as ⊗, and () (the unit type) as I. The signatures of μ and η, translated to Haskell, then become:

μ :: (M, M) -> M
η :: () -> M

By currying μ, and making use of how () -> M functions are in one-to-one correspondence to M values (all of them look like \() -> m for some m), we get the familiar Monoid methods:

mappend :: M -> M -> M
mempty :: M

Note that the categorical definition is far more general than just Monoid. For instance, we might keep working in Hask while replacing (,) and () with their duals, Either and Void, thus getting:

μ :: Either A A -> A
η :: Void -> A

Every Haskell type is a monoid in this particular manner (μ is either id id, and η is absurd).

Another example is taking C to be the category of Haskell Functors (with natural transformations between them -- which I will write as type f ~> g = forall a. f a -> g a -- as morphisms), Compose as ⊗, and Identity as I:

-- Note the arrows here are ~>, and not ->
μ :: Compose M M ~> M
η :: Identity ~> M

These two are commonly written as:

-- "Inlining" the definitions of Compose, Identity, and ~>
join :: M (M a) -> M a
return :: a -> M a

In other words, a Monad is a monoid in the category of Functors (which is the Hask-specific version of "a monad is a monoid in the category of endofucntors"). It is worth mentioning that, as in the other example, this is not the only way to get monoids out of that category (see the final paragraph of this answer for pointers -- the rest of it, in fact, might be relevant reading, as it discusses the notion of monoidal category).