By a commutative monoid, let us mean a monoid $A$ in which $a,b \in A$ implies $ab=ba$. Its not at all obvious how to generalize this to the case of an arbitrary category; we cannot just assume that all morphisms $f$ and $g$ satisfy $fg=gf$, since the domain/codomain conditions won't be met. We can however assume that for all objects $X$ and all endomorphisms $f,g \in \mathrm{End}(X),$ $fg=gf.$ Lets call this the naive definition of "commutative category."

Does it have any interesting consequences?

  • $\begingroup$ Related: en.wikipedia.org/wiki/Abelian_category $\endgroup$ Commented Feb 16, 2014 at 1:00
  • 7
    $\begingroup$ @IsaacSolomon, $\mathbb{R}$-$\mathbf{Mod}$ isn't a commutative category as defined in the question, since not all real matrices commute. So I don't think Abelian categories are really relevant here; its an altogether different concept. $\endgroup$ Commented Feb 16, 2014 at 1:04
  • 3
    $\begingroup$ Sometimes I wish to give -1 for certain comments, but this isn't built into the software. $\endgroup$ Commented Feb 16, 2014 at 2:07
  • 2
    $\begingroup$ @MartinBrandenburg Although you can't "punish" bad comments with a down-vote, you can help prevent them from misleading people, by adding a corrective comment of your own (unless someone else has already done so, in which case you can up-vote the correction). $\endgroup$ Commented Feb 16, 2014 at 2:16
  • 1
    $\begingroup$ This doesn't seem like a useful thing to do. Do you have any particular motivation in mind? $\endgroup$ Commented Feb 16, 2014 at 6:00

2 Answers 2


I don't think that this is a useful definition. You don't consider the morphisms which are no endomorphisms.

Commutative monoids are precisely the monoid objects in the category of monoids (Eckmann-Hilton argument). If $M$ is a commutative monoid, the multiplication map is a homomorphism $M \times M \to M$.

Since categories are many-objects-monoids, we could expect that (small) "commutative categories" are the monoid objects in the category of (small) categories, i.e. the strict monoidal categories. This has the correct decategorification: A monoidal category with one object (even not assumed to be symmetric) is the same as a commutative monoid.

I have to admit that this isn't a satisfactory answer, because we would rather expect that "commutative" is a property, not an extra structure.

Edit: There is a good reason why there is no natural definition of a commutative category: If $O$ is a set, the category of $O$-graphs has as objects pairs $(M,M \rightrightarrows O)$, consisting of a set $M$ and a pair of parallel morphisms $s,t : M \to O$. This category is monoidal with unit $(O,O = O)$ and tensor product $(M,s,t) \otimes (M',s',t') = (M \times_{t,s'} M', s \circ \mathrm{pr}_1,t \circ \mathrm{pr}_2)$. Then a category with object set $U$ is the same as a monoid object in this monoidal category (this is easy to see, for a reference see II.7 in Mac Lane's CWM). In order to define commutative monoids, we need a symmetric monoidal category. However, the category of $O$-graphs is not symmetric monoidal. Only the full subcategory consisting of those $O$-graphs with $s=t$ is symmetric monoidal. The commutative monoid objects therein correspond to those categories which are disjoint unions of commutative monoids.

  • 4
    $\begingroup$ Commutativity only looks like a property at the $1$-categorical level. At the higher categorical level it is very clearly extra structure because you need to specify what the "commutators" are. $\endgroup$ Commented Feb 16, 2014 at 5:58
  • 1
    $\begingroup$ (E.g. the commutativity of the second homotopy group $\pi_2(X)$ reflects extra structure on the double loop space $\Omega^2 X$, namely the structure of being an $E_2$-algebra.) $\endgroup$ Commented Feb 16, 2014 at 8:47
  • $\begingroup$ Sure, I know that, but nevertheless it would be interesting to find a natural property for categories which comes down to commutativity in the case of monoids. $\endgroup$ Commented Feb 16, 2014 at 15:23
  • 1
    $\begingroup$ Here is one. If $C$ is a category, on every endomorphism monoid $\text{End}(c)$ impose the equivalence relation $fg \sim gf$ where $g : c \to d, f : d \to c$ and $d$ is any other object. The resulting map to equivalence classes can be regarded as a universal trace on $C$ (a kind of Hochschild homology), and the condition that the universal trace is injective reduces to commutativity in the one-object case. $\endgroup$ Commented Feb 16, 2014 at 18:45
  • 2
    $\begingroup$ Why not making this an answer so that it gets more readers? $\endgroup$ Commented Feb 17, 2014 at 18:24

I have arrived here from Google in the search of a reference on such categories, so would be very happy if someone turns up some "interesting consequences".

The property that such a "commutative category" might have that I want to take advantage of is that you can actually compute them from a finite presentation. In fact I require a slightly stronger condition, that the endomorphism monoids are abelian groups. When I say 'presentation' I have in mind a simplicial set, and I am interested in the fundamental category. Now categories are too hard to work with for the application I have in mind and I don't want to invert any more edges than I have to. Imposing the commutativity and invertibility relations on endomorphisms gives a category which is much richer than the first homology, but still possible to actually work with.

I agree with the other contributors here that the condition may be a little artificial from a category theory perspective, but it still may have some practical use.

EDIT: I have since found a little on this topic under the subject of graph congruences. A graph congruence is essentially an equivalence relation on the set of paths in a directed graph, which respects endpoints and composition of paths. i.e. the set of equivalence classes of paths forms a category.

I'm not really familiar enough with the terms used in the papers I've found (languages, varieties etc.) to say anything about them, but see:

Arkadev Chattopadhyay and Denis Therien Locally Commutative Categories

and its references.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .