What are the domains of the multiplication and unit morphisms of a monoid object?

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

In category theory, a monoid (or monoid object) $$(M, \mu, \eta)$$ in a monoidal category $$(\mathcal{C}, \otimes, I)$$ is an object $$M$$ together with two morphisms

• $$\mu: M \otimes M \to M$$ called multiplication,
• $$\eta: I \to M$$ called unit, [...]

My confusion is about morphism notation. Why is the binary operation $$\otimes$$ a part of the morphism notation? My understanding of morphism is that it's a kind of function that can map one type to another (domain to codomain) like $$M \to M$$... Why is the operation $$\otimes$$ a part of the domain in the definition?

The second confusion is about $$I$$. Why $$I$$ is a domain... there is no $$I$$ object in a Monoid at all. It's just the neutral element of object $$M$$.

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

Is $$M \otimes M$$ somehow related to cartesian product, so the domain of the morphism is defined as $$M \times M$$ ?

• You should focus on understanding what a monoidal category is first, that seems to be your main problem. May 18 '19 at 15:56

An ordinary monoid is a set $$M$$ equipped with a map $$\mu\colon M\times M\to M$$ and an element $$e\in M$$ such that certain axioms are satisfied.

Now the goal of the definition you quoted (monoid object in a monoidal category) is to generalize the notion of ordinary monoid to the most general categorical context possible.

The first step is to observe that in a general category it doesn't make any sense to talk about an "element" of an object. So let's replace the element $$e\in M$$ with a map $$e\colon 1\to M$$, where $$1 = \{*\}$$ is a set with a single element. The point is that (in the category of sets) specifying a map $$1\to M$$ is exactly the same as specifying an element of $$M$$: given an element $$e\in M$$, we can define a map $$1\to M$$ by $$*\mapsto e$$, and given a map $$e\colon 1\to M$$, we can define an element of $$M$$ by $$e(*)$$.

Ok, now we want to generalize away from the category of sets and talk about monoids living in other categories. So let $$\mathcal{C}$$ be a category. A monoid in $$\mathcal{C}$$ should be an object $$M$$ in $$\mathcal{C}$$ together with maps $$\mu$$ and $$e$$ satisfying certain axioms. To make sense of the maps $$\mu$$ and $$e$$, we need:

• A specified way of putting two copies of $$M$$ together to make the domain of $$\mu$$, analogous to $$M\times M$$ in the case of sets. Let's denote this by $$M\otimes M$$.
• A special object to be the domain of $$e$$, analogous to $$1$$ in the case of sets. Let's denote this by $$I$$.

The operation $$\otimes$$ on objects and the special object $$I$$ are the basic data of a monoidal category.

If $$\mathcal{C}$$ has finite products, you're welcome to define $$\otimes$$ to be the product $$\times$$ and $$I$$ to be the terminal object $$1$$. This is called the Cartesian monoidal structure. But the point is that we don't need $$M\otimes M$$ to be the product and $$I$$ to be the terminal object in order to make sense of the definition of monoid.

We do need to assume a little bit more about $$\otimes$$ and $$I$$ in order to make sense of the monoid axioms. In particular, we need $$\otimes$$ to be associative (up to isomorphism), $$M\otimes(M\otimes M) \cong (M\otimes M)\otimes M,$$ and we need $$I$$ to act as an identity for $$\otimes$$ up to isomorphism, i.e. $$M\cong I\otimes M\cong M\otimes I.$$ Making all of this appropriately functorial and natural leads directly to the full definition of monoidal category.