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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?

Andronicus
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    A categorical monoid is a category with a single object meant to model an algebraic monoid. In your example with the natural numbers, If you add $1$ to every element of $\mathbb N\cup{0}$ you get $\mathbb N\setminus{0}$ which is a different object so you no longer have a categorical monoid. – John Douma Jan 04 '23 at 11:49
  • @JohnDouma what about the example with real numbers and multiplication? The set of real numbers is dense, so every time it's the same set. Also in my example I'm not considering a function from set to set, but rather from a natural number to natural number and all natural numbers are objects of this category. – Andronicus Jan 04 '23 at 12:36
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    If all natural numbers are objects in the category then it is not a categorical monoid. As for your example with the real numbers and even with the naturals if we consider $\mathbb N\cup{0}$ as one object and ignore that the morphisms are non-surjective functions, we end up with the same definition as in the other examples with '*' as the single object. In these cases the sets, $\mathbb R$ for example, are the single objects. – John Douma Jan 04 '23 at 12:46
  • @JohnDouma I think I understand everything apart from "". My intuition is that morphism is a function from to *. But I don't understand what this star is. Is it the set of all objects of the monoid? – Andronicus Jan 04 '23 at 12:52
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    No. It is the only object in the category. Remember, morphisms do not have to be functions. Consider a category whose elements consist of the elements of a partially ordered set. We say a morphism exists between $A$ and $B$ if $A\le B$. That is a perfectly valid category. This may be the cause of your confusion. Morphisms are not required to be functions although they are in many useful categories. – John Douma Jan 04 '23 at 15:07
  • @JohnDouma now I understand it! Thank you! Can I ask you one more thing? I've given an example of a category constructed using a monoid in this answer and got downvoted, do you know what is the problem with my understanding: https://math.stackexchange.com/a/4611301/528171 ? – Andronicus Jan 04 '23 at 15:12
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    I am not sure why you got downvoted. They may not have understood the construction because you started with an algebraic monoid. I know I have been downvoted for posts that I thought were a lot better than posts for which I received upvotes. I will sometimes remove a post that gets downvotes but if I believe my answer is a good one I will leave it. – John Douma Jan 04 '23 at 15:19
  • @JohnDouma thank you very much for your help! – Andronicus Jan 04 '23 at 15:57

1 Answers1

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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...

dan_fulea
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  • I'm sorry, but I feel like I'm rereading all the answers regarding single-object categories. What about the one I described - a category where the objects are the objects of the monoid? I don't understand what is this *, could you give a more concrete example? – Andronicus Jan 04 '23 at 11:17
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    So you want to make a connection between the following two structures: (A) Monoids. (Forget now that they build a category and have morphisms, and so on.) (B) Categories with the set of objects being a one element set. Your question was about your understanding of this / any correspondence. Yes, you give an example. That one with $\Bbb R$ with multiplication is not fitting into the scheme. You start with a monoid, yes, which is the category for it? Few lines later, you start with a monoid $C$, and the associated category has as set of objects $C$. Does it fit (B)? Does it have one object? – dan_fulea Jan 04 '23 at 13:53