Thinking about monoids as single-object categories, I got to wondering about what distinguishes the arrows of a single-object category, especially given that the object might have no internal structure (e.g., it might not be a set with elements).

For instance, the trivial monoid consisting of just the identity element under some binary operation would be a category $\mathbf{M_0}$ with object $M$ and the identity arrow $\mathit{id}_M : M \rightarrow M$ as its sole arrow. But since the natural numbers under addition form a commutative monoid $(\mathbb{N}, +)$ it should be possible to construe that monoid as a single-object category $\mathbf{M}_\mathbb{N}$ with object $M$. Suppose that the single-object is the same in both categories, so that we get from $\mathbf{M_0}$ to $\mathbf{M}_\mathbb{N}$ simply by adding a countable infinity of arrows $1, 2, \dotsc : M \rightarrow M$, identifying $0$ with the identity arrow, requiring composition to be commutative, and ensuring additional things like that $1 \circ 1 = 2$ so that composition behaves like addition.

But assuming that $M$ has no "elements" or any other such internal structure, what differentiates the countable infinity of arrows that make up $\mathbf{M}_\mathbb{N}$? I've read, like in the comments on the linked question, that arrows can be thought of as ordered triples so that $1$ would be $(1, M, M)$ and $2$ would be $(2, M, M)$. Does the "name" (e.g., "1") of an arrow play a distinctive role in distinguishing it from other arrows with the same source and target (a single object, in this case)?

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    $\begingroup$ The answer to your last question is "yes". Usually, if we're being careful, we restrict "category" to mean "locally small category" meaning the class of arrows between any two objects is a set. This implicitly means we've been provided a notion of equality of arrows between any given pair of objects. More compactly, $M$ doesn't need to be a set with elements for $\text{Hom}(M,M)$ to be a set with elements. $\endgroup$ Apr 24, 2017 at 0:36

3 Answers 3


It's composition that plays a role, not the 'names'.

Without thinking your monoid as a one-object category, we can pose exactly the same question:

We have a set $\Bbb N$ of countably infinitely many elements, and an addition (and a distinguished additive identity element). What does differentiate its elements?

We have the $0$ that acts as the additive identity.
Besides that we have $1$ which is not a result of any $a+b$ for nonzero $a,b$.
Then we have $2=1+1$, and so on.

These do describe these elements, and you are free to replace 'addition' to 'composition' everywhere in the above.

  • $\begingroup$ So is it best to think of the categorical rendition of the commutative monoid of natural numbers under addition as having only two non-composite arrows, 0 (the identity arrow) and 1 (the "successor arrow" of 0), and then to think of all the other arrows as iterated compositions of 1? $2 = 1 \circ 1$, $3 = 1 \circ 1 \circ 1$, etc.? Then the "names", as I called them, are just convenient abbreviations for iterated compositions of 1 with itself and could be dropped entirely? $\endgroup$
    – Dennis
    Apr 24, 2017 at 0:47
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    $\begingroup$ Yes, exactly. It's the one element generated free commutative monoid. $\endgroup$
    – Berci
    Apr 26, 2017 at 19:56
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    $\begingroup$ This is sort of true but mostly misleading. What "distinguishes" the arrows in a monoid is their names (or rather, the arrows in a monoid literally are nothing but their "names"). For instance, in $\mathbb{Z}$, the elements $1$ and $-1$ are distinct, even though they cannot be distinguished using only the composition operation. $\endgroup$ Dec 16, 2017 at 19:04

When defining a category, you can take any set to be the set of morphisms between two objects (as long as you then define composition appropriately). So if you have your favorite monoid $(A,\cdot_A)$, you can simply define a category which has one object $M$ and whose set of morphisms from $M$ to $M$ is the set $A$, with composition defined by the operation $\cdot_A$. So, if you like, the arrows are distinguished by their "names", but that's not really the right way to think about it. Rather, the arrows are nothing but their names: the arrows literally are just elements of the set $A$ (which could be any set at all, as long as you have an appropriate binary operation on it).


For variety, it's worth noting that any category can be given an extensional-flavored notion of equality:

If $f$ and $g$ are two arrows $X \to Y$, then the following are equivalent:

  • $f=g$
  • $fx = gx$ for every arrow $x$ with codomain $X$
  • $yf = yg$ for every arrow $y$ with domain $Y$

This is one of many ways in which the notion "arrow with codomain $X$" acts as a good substitute for the notion of "element of $X$". When thinking this way, we call the arrow a "generalized element".

So, despite the theorem being a rather trivial one, it's still a rather useful point of view.

As an example, the generalized elements of the object $M$ in $\mathbf{M}_{\mathbb{N}}$ are precisely the natural numbers. (because $\hom(M,M)$ is, by definition, the natural numbers)

Also, many categories allow a useful (and less trivial) variation:

A set of objects $\mathcal{G}$ is a generating set for the category if and only if, for every pair $f,g$ of arrows $X \to Y$, the following are equivalent:

  • $f = g$
  • $fx = gx$ for every arrow $x:G \to X$, with $G \in \mathcal{G}$

There are a lot of categories where the usual notion of elements coincides with the notion of a generalized element with domain in a generating set. For example, in the category of groups, take $\mathcal{G} = \{ \mathbb{Z} \}$. The notions of "an element of the group $G$" and "an arrow $\mathbb{Z} \to G$" are basically the same.

Another example is $\mathbf{Cat}$. Take $\mathcal{G}$ to consist of the two categories $\bullet$ and $\bullet \to \bullet$. The two kinds of generalized elements are precisely the objects and the arrows.


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