Monoid as a category

Question

I saw this note in a book and I don't fully understand it:

A monoid is, essentially, the same thing as a category with a single object.

I came up with an example that you can think of a monoid of real numbers with multiplication and $1$ ($(\mathbb{R}, \cdot, 1)$) as a category with single object - real numbers and functions from real to real, where all elements are multiplied by real numbers ($(\{\mathbb{R}\}, \{A \Rightarrow \{a \cdot r, a \in A\}, r \in \mathbb{R}\})$).

However it seems a bit counterintuitive for me. Let's consider a monoid $(C, \times, c_0)$, where $C$ is a set of its elements, $\times$ is a composable function $\times: (c_1, c_2) \rightarrow c_3$, where $c_1, c_2, c_3 \in C$ and c_0 is a unit element: $c \times c_0 = c_0 \times c = c$ for $c \in C$. We can construct a category with objects $C$ and morphisms being functions $f_{c'}: c \rightarrow c \times c'$, where $c' \in C$. Then the function $f_0: c \rightarrow c \times c_0$ is an isomorphism. Morphisms are composable: $f_{c_1} \circ f_{c_2} = f_{c_1 \times c_2}$.

For example having a monoid with natural numbers with zero, addition and zero ($(\mathbb{N} \cup \{0\}, +, 0)$) we can construct a category with natural numbers with zero and morphisms being functions $f_{n'}: n \rightarrow n + n'$ where $n' \in \mathbb{N} \cup \{0\}$ and $f_{0}$ as isomorphism: $(\mathbb{N} \cup \{0\}, \{n \rightarrow n + n', n' \in \mathbb{N} \cup \{0\}\})$.

Is my inderstanding correct or am I missing something?

Answer 1

The association is a bit different.

• Start with a monoid $(M,\cdot,1)$. Associate the category $C$ with one object, i will call it $*$, and with morphisms between objects of $C$ - well we need to set only $\operatorname{Hom}_C(*,*)$ to be (a copy of) the set $M$. Define the compositon of morphisms to correspond to the operation $\cdot$. So $m\circ m':=m\cdot m'$. Use the associativity to have the right property for the composition of functions. Define the identity morphism from $*$ to $*$ to be the $1$ of $M$.

• Conversely, start with a category with one object, extract from it the Hom-set $M$ of morphisms from this one object to itself. The composition of homomorphisms becomes the monoid operation on $M$, the unit is the identity morphism...