# Monoid as a single object category

I'm struggling with comprehending what monoids are in terms of category theory.

In examples they view integer numbers as a monoid. I think I get the set theoretic definition. We have a set and a associative binary operator (addition) and the neutral element (zero).

Then they are saying something like that - view the whole set as a single object and binary operator as bunch of morphisms for every element of the set.

Like add0 is an identity morphism. Which would really give us the same object i.e. the same set of all integer numbers. I think I understand this.

But let's view the morphism add1. After applying it to the our single object (the set of all integers) we would have a set {1,2,3…} not the {0,1,2,3…}. Aren't domain and codomain different in that case?

That's what's bothering me. Can someone clarify that to me?

Here is the text that gives me problems.

The text

No, that's not the way to view it. In order to view a monoid as a category, you have a single object $\mathsf{Andreas}$, and each element of the monoid is one morphism $\mathsf{Andreas}\to\mathsf{Andreas}$ in the category. The monoid operation is the composition in the category.

So for the integers, you don't have a morphism "add 1", but a morphism that is simply called $1$. And composition in the category works such that $1$ composed with $1$ is the morphism called $2$.

This is an example of a category where the morphisms are not functions.

• @user1685095: The domain of each morphism is $\bullet$ (that is, the object of the category), and so is the codomain. The very point here is that a morphism is not necessarily a function. Commented Jun 20, 2015 at 15:13
• @user1685095: $\bullet$ is the name of the object in the category. Its precise identity is not important. Commented Jun 20, 2015 at 15:26
• @user1685095 add$1$ can be looked at as a function having $\mathbb N$ as domain and as codomain. It is prescribed by $n\mapsto n+1$. There is a category with objectset $\{\mathbb N\}$ and morphismset $\{\text{add}n\mid n\in\mathbb N\}$. So there is only one object and it can equally well be denoted as $\bullet$. Monoids can be identified as categories that have exactly one object. Composition of morphisms corresponds with addition: $\text{add}k\circ\text{add}m=\text{add}(k+m)$. Commented Jun 20, 2015 at 15:47
• Since the unique but unspecified object seems to cause difficulties for the OP, I hereby volunteer to make things specific by being that object. That is, the category has one object, namely me, and its morphisms are the integers. Composition of morphisms is addition of integers. Commented Jun 20, 2015 at 16:00
• @codeshot: No, I mean each element of the monoid. An object in category theory does not have elements (or at least not elements that are visible at the category level). Commented Nov 4, 2017 at 15:03

You already know that a monoid $M$ is a set with a unit $e$ and a binary operation. More precisely, if $a,b,c\in M$ then $$a\circ b\in M$$ $$(a\circ b)\circ c=a\circ (b\circ c)$$ $$e\circ a=a\circ e=a$$

Now, take any category $C$ with one object, $c$. Since $C$ is a category, we need to say what the arrows are. That is, what the morphisms $c\rightarrow c$ are. There must be a unit $1_{c}:c\rightarrow c$, the arrows must be composeable and the composition must be associative. More precisely, if $f,g$ and $h$ are arrows, then $$f\circ g\in Morph(C)$$ $$(f\circ g)\circ h= f\circ (g\circ h)$$ $$1_{c}\circ f=f\circ 1_{c}=f$$Notice we are $\textit not$ talking about sets here. Just objects and arrows, in the abstract.

But now if we just observe that the operations on $M$ are $\textit exactly$ the same as the operations on $Morph(C)$, we may regard the category $C$ as the monoid $M$. This correspondence is reversible: given category $C=\left \{ c \right \}$ we obtain a monoid $M$ whose elements are the arrows of $C$.

Thus the two descriptions are equivalent.

All this works because the binary operations are the same for both structures.

• What's special about the case when $Ob(C)=\{c\}$? Why can't we interpret a monoid as a category with more than one object? The same axioms will hold for $Morph(C)$. Commented Jan 27, 2019 at 16:15
• You want to identify Morph$C$ with $M$? But in an arbitrary category, not all morphisms are composable, right? If you have $f,g:A\to B$ then $f\circ g$ doesn't even exist. What's special about the above construction is that there is an $\textit {exact}$ i.e. bijective correspondence between the operations. Commented Jan 27, 2019 at 16:37
• Then we can consider a category with more than one object and the set of all morphisms from an object $A$ to $A$. This set with the operation of composition should be a monoid. Can we then say that conversely, a monoid can be interpreted as the set of arrows from a fixed object to itself in an arbitrary category? This is also a bijective correspondence between the arrows from that object to itself and the monoid operation. Commented Jan 27, 2019 at 17:00
• "...a monoid can be interpreted as the set of arrows from a fixed object to itself in an arbitrary category"...yes, and this is the essence of my answer. If you have a category $C$ with small hom-sets, then for each $c\in C$, then each Hom$[c,c]$ will give a monoid, distinct if the cardinalities are different. Commented Jan 27, 2019 at 17:06
• This should be the accepted answer, the accepted one uses the integers as a confusing example. Commented Feb 10, 2023 at 23:25

In addition to Henning Makholm's crisp and clear answer, you might find the opening six pages of my Notes on Category Theory helpful. They too give the example of a monoid as a category, but also give some other examples of categories where the arrows are not functions in any ordinary sense. Another important illustration is the case of a posets treated as a category.

In fact these examples suggest why we might well prefer to talk of 'arrows' rather than 'morphisms' (because the very term 'morphism' comes with baggage, and almost inevitably makes us think of a function -- but to repeat, arrows need not be functions).

• A footnote to another solution should, well, be a footnote on that solution (or rather a comment or two.) Commented Jun 20, 2015 at 16:03
• Nice notes Peter. And nice blog too. I will contact you there. Commented Jun 24, 2015 at 2:03