# Definition of a monoid object

Wikpedia defines a monoid object in a monoidal category ($\mathbb{C}$, ⊗, I) as an object $M$ with two morphisms :

• μ: M ⊗ M → M called multiplication,
• η: I → M called unit,

such that the following diagrams commute

I have some troubles understanding this definition :

1) What is $1$ in both diagrams ? Shouldn't we write something like $\mu \otimes id_M$ instead ?

2) In the second diagram, how should one read/understand the arrows $$I \otimes M \overset{\eta \otimes 1}{\rightarrow} M \otimes M \quad\quad \text{and} \quad\quad M \otimes M \overset{1 \otimes \eta}{\leftarrow} M \otimes I$$ ?

3) According to Awodey's "Category Theory" (page 65, 66) : A group consists of objects and morphisms $\quad\quad G \times G \overset{m}{\rightarrow} G$, $\quad\quad G \overset{i}{\rightarrow} G$, $\quad\quad 1 \overset{u}{\rightarrow} G$. $\quad\quad$ A particular diagram should commute which is similar to the first one I provided but with $\times$ instead of $\otimes$ and $m \times 1$ instead of $\mu \otimes 1$ etc

But the definition doesn't seem correct to me. Isn't $1$ a terminal object and $u$ a morphism ? If so, what is $u \times 1$ ?

• The $1$ here is the identity morphism on $M$, so indeed, $1=\text{Id}_M$ as you say. As for the second diagram, $\eta:I\rightarrow M$ should be thought of as the unit of your monoid. – Mathematician 42 Oct 26 '17 at 8:25
• Let $A$ be an unital $\mathbb{C}$-algebra with unit $1_A$, then $\eta:\mathbb{C}\rightarrow A:\lambda\rightarrow \lambda\cdot 1_A$. Clearly $\eta(1)=1_A$. So the map $\eta$ sees all multiples of the unit of $A$. Moreover $A$ is a monoid object in the monoidal category of $\mathbb{C}$-vector spaces. You could think of monoids as generalizing algebra objects to different monoidal categories. – Mathematician 42 Oct 26 '17 at 8:28
• A group object is usually defined by starting with a category with products, and a final object (denoted $1$). Then you have the unit map $u: 1 \to G$, a multiplication map $m: G \times G \to G$, and so on. This does not use or require a monoidal structure on the category. – Joppy Oct 26 '17 at 8:39
• @Joppy Thanks for your answer. What is troubling me is that Awodey writes $u \times 1$ in his first diagram. $u$ is a morphism and $1$ a terminal object so it doesn't seems correct to me. – Boris E. Oct 26 '17 at 8:47
• Which diagram has $u\times 1$? Is it in your question? – Joppy Oct 26 '17 at 8:53

As Mathematician 42 states, $1$ in the image is this article's notation for the identity morphism.
In Awodey's book, $1$ is being used also to indicate the terminal object. There isn't really much chance for confusion due to this notational overloading, but it's not something I would do. (I, personally, use $1$ for the terminal object, but $id$ for identity arrows.) The cartesian structure of $(\times,1)$ is a special case of monoidal structure.
You can gain some intuition for these by considering the cartesian case. The map $\eta\otimes 1_M : I\otimes M \to M\otimes M$ in $\mathbf{Set}$ say with the cartesian monoidal structure (so an actual monoid), would be a function $\langle\langle\rangle,m\rangle\mapsto\langle e,m\rangle$ where $\langle\rangle\in 1$ and $e$ is the unit element of the monoid.
The cartesian case may be a bit too special, so, as the notation suggests, you can also consider vector spaces over a field $k$ and the monoidal structure $(\otimes, k)$ where $\otimes$ is the usual tensor product. Then $\eta$ essentially picks out a vector of $M$ and $\eta\otimes 1_M$ sends the vector $m$ to $\eta\otimes m$. All the structure together makes $M$ an (associative, unital) algebra.
Another perspective I like is to connect monoidal categories to multicategories. A multicategory is simply a category-like thing except arrows may have multiple inputs (and thus there's a multi-composition). Every monoidal category gives rise to a multicategory by representing a multi-arrow $(A_1,\dots,A_n)\to B$ via $A_1\otimes\cdots\otimes A_n \to B$. From this perspective, a monoid object looks like an object $M$ equipped with two multi-arrows $\eta : () \to M$ and $\mu: (M,M)\to M$. With multi-composition, the laws look a bit more normal: $\mu \circ (\eta,f) = f = \mu \circ (f,\eta)$, $\mu \circ (\mu, id) = \mu \circ (id,\mu)$. You can go even further and use type theoretic concepts to produce a notation that looks just like traditional notation: $\mu(\eta,x)=x=\mu(x,\eta)$, $\mu(\mu(x,y),z) = \mu(x,\mu(y,z))$. One way of understanding this is that the theory of monoids can be interpreted in any monoidal category (that's exactly what a monoid object is!) This is true because the axioms defining a monoid use each variable exactly once (per side of the equation). (Any algebraic theory with this property can be interpreted into any monoidal category. The axioms of a group, however, do not satisfy this property because an axiom like $x\cdot x^{-1} = e$ duplicates the variable on the left and drops it on the right, so it is no accident Awodey uses the cartesian monoidal structure for the axioms of a group.)