# Why are the two natural transformations in the definition of monad called the unity and multiplication?

Categories for the Working Mathematician says

Definition. A monad $$T= \langle T, \eta, \mu\rangle$$ in a category $$X$$ consists of a functor $$T: X \to X$$ and two natural transformations

$$\eta : I_X \Rightarrow T, \mu : T^2 \Rightarrow T$$

which make the following diagrams commute

We shall thus call $$\eta$$ the unit and $$\mu$$ the multiplication of the monad $$T$$; the first commutative diagram of (2) is then the associative law for the monad, while the second and third diagrams express the left and right unit laws, respectively.

Why is $$\eta$$ called the unity, and $$\mu$$ multiplication?

How is the first diagram the associative law? It says $$μ∘Tμ=μ∘μT$$, which equates $$Tμ$$ and $$μT$$ up to $$μ$$, so seems to me the commutative law instead.

Why are the second and third diagrams the second the left and right unit laws? They say that $$μ∘ηT=μ∘Tη=1_T$$, They seem to me that $$ηT$$ and $$Tη$$ are the same up to $$\mu$$ and they are both inverse of $$μ$$.

Thanks.

Given a category $$\mathcal C$$, the category $$[\mathcal C,\mathcal C]$$ of endofunctor is naturally a (non symmetric) strict monoidal category: the tensor product is the composition, the unit object if the identity functor and the associators and unitors are actual identities.
A monad on $$\mathcal C$$ is exactly a monoid in this monoidal category $$[\mathcal C,\mathcal C]$$.
• Some other questions. (2) How is the first diagram the associative law? It says $μ∘Tμ=μ∘μT$, which equates $Tμ$ and $μT$ up to $μ$, so seems to me the commutative law instead. (3) Why are the second and third diagrams the second the left and right unit laws? They say that $μ∘ηT=μ∘Tη=1_T$, They seem to me that $ηT$ and $Tη$ are both inverse of $\mu$. – Tim Jul 24 at 12:01
• A monoid object $C$ in a monoidal category $(\mathcal C, \otimes, I)$ has unit $I\to C$ and composition $C\otimes C\to C$, satisfying the same diagrams. Consider first what it means in $(Set,\times,\{*\})$, and then in $([\mathcal C,\mathcal C],\,\circ,\,id_{\mathcal C})$. – Berci Jul 24 at 15:07