Let $(\mathsf{Rel},\otimes,1)$ denote the monoidal category of sets and relations, where $1$ is the one-element set. I once conjectured (with a little help from Jamie Vicary) that $\mathsf{Rel}$ is "quotient-free" in the sense that if a strong monoidal functor $F\colon\mathsf{Rel}\to S$ identifies any parallel pair of morphisms, then $F$ identifies every parallel pair of morphisms, and hence it factors through the terminal monoidal category (since $\mathsf{Rel}$ has a zero-object). [I'd be happy to hear suggestions for a better name than "quotient-free monoidal category", or for a better way of thinking of such things.]

Definition: We say a monoidal category $M$ is quotient-free if for any monoidal category $S$ and strong monoidal functor $F\colon M\to S$, if $F(f_0)=F(f_1)$ for distinct morphisms $f_0\neq f_1\colon A\to B$ then $F$ factors through a terminal monoidal category.

Explaining the conjecture to Tobias Fritz, he quickly proved it (by contradiction) as follows.

Proposition: The monoidal category $(\mathsf{Rel},\otimes,1)$ is quotient-free.

Proof (Fritz): Suppose that $A$ and $B$ are sets and that $R_0,R_1\colon A\to B$ are relations such that $R_0\neq R_1$. Then there exists $a\in A$ and $b\in B$ such that $(a,b)\notin R_0$ and $(a,b)\in R_1$ (without loss of generality).

Let $e_a\colon 1\to A$ and $e_b\colon 1\to B$ correspond to the relations characterizing the subsets $\{a\}\subseteq A$ and $\{b\}\subseteq B$, respectively, and let $e'_b\colon B\to 1$ be the transpose of $e_b$. Then we have two different relations $$1\xrightarrow{e_a}A\xrightarrow{R_0\ ,\ R_1}B\xrightarrow{e'_b}1.$$ These ($e'_bR_0e_a$ and $e'_bR_1e_a$) are the only two relations $1\to 1$, equaling the "null" relation $\emptyset_{1,1}$ and the identity $\mathrm{id}_1$, respectively.

Assuming now that $F(R_0)=F(R_1)$, we have $F(\mathrm{id}_1)=F(\emptyset_{1,1})$. It follows that $F$ identifies any given relation $X\colon C\to D$ with the null relation $\emptyset_{C,D}\colon C\to D$, because $$ FX\cong F(X)\otimes F(\mathrm{id}_1)= F(X)\otimes F(\emptyset_{1,1})\cong F(X\otimes\emptyset_{1,1})= F(\emptyset_{C,D}). $$ Thus for any set $A$, we obtain an isomorphism $F(A)\cong F(\emptyset)$, where $\emptyset$ is the zero-object of $\mathsf{Rel}$. $\square$

Question: What are other examples of quotient-free monoidal categories?

Question: Might we consider quotient-free monoidal categories as acting like fields, which are also somehow quotient-free? That is, maps to quotient-free monoidal categories would be analogous to points? Any thoughts on this would be useful.

  • 3
    $\begingroup$ Sorry if this is off-topic, but it amuses me that fields are quotient-free because they are fields of quotients. $\endgroup$ – Todd Trimble Jan 2 '17 at 12:57
  • 1
    $\begingroup$ I feel like you should call these categories "entangled" or something. Or maybe "delicate." $\endgroup$ – Jonathan Beardsley Jan 3 '17 at 0:42
  • 1
    $\begingroup$ Here's a question: what conditions must one put on a subcategory so that quotienting by it yields a quotient-free category? I.e. what are the analogues of maximal ideals? $\endgroup$ – Jonathan Beardsley Jan 3 '17 at 0:44
  • 1
    $\begingroup$ Possibly the analogy with fields is a bit more than just an analogy: it may have to do with noncommutative algebraic geometry, which is (in one approach) about abelian monoidal categories, arxiv.org/abs/math/0611806. While $\mathsf{Rel}$ is not abelian, maybe this could at least provide a useful perspective. Perhaps somebody who knows more about this can weigh in. $\endgroup$ – Tobias Fritz Jan 3 '17 at 9:28
  • 2
    $\begingroup$ I don't think one really needs abelian. For instance in this paper: arxiv.org/pdf/1105.3104v4.pdf one replaces commutative rings with presentable symmetric monoidal categories. Thus one might able to do something meaningful just working with presentable monoidal categories. But I agree that this connects to noncommutative algebraic geometry. $\endgroup$ – Jonathan Beardsley Jan 3 '17 at 21:11

Here are some additional thoughts and examples.

1) The argument presented in the OP applies whenever one has a monoidal category $(\mathsf{C},\otimes,1)$ in which the unit object $1$ is a separator and coseparator, and the monoid of scalars $\mathsf{C}(1,1)$ is itself quotient-free. So under these assumptions, also $\mathsf{C}$ is quotient-free as a monoidal category.

2) A similar argument shows that for any field $K\neq\mathbb{Z}/2$, the category $\mathsf{Vect}_K$ has exactly one nontrivial monoidal quotient, namely a category that I will denote by $\mathsf{Vect}_K/K^\times$. This quotient is given by identifying any two parallel morphisms that are scalar multiples of each other. (For $K=\mathbb{Z}/2$, this quotient coincides with the original category, and already $\mathsf{Vect}_{\mathbb{Z}/2}$ itself is quotient-free by the same token.)

To prove this claim, note again that the monoidal unit $K\in\mathsf{Vect}_K$ is a separator and coseparator, so that any nontrivial quotient of $\mathsf{Vect}_K$ must also identify two different scalars. But the scalars are the multiplicative monoid $(K,\cdot)$, which consists of the group of units $(K^\times,\cdot)$ together with an additional absorbing element $0\in K$. Its only nontrivial quotient is the one where all units are identified, with quotient $(\mathbb{Z}/2,\cdot)$, which is accidentally also the monoid of scalars of $\mathsf{Rel}$. By strong monoidality, our original quotient of $\mathsf{Vect}_K$ must therefore also be a quotient of $\mathsf{Vect}_K/K^\times$.

Finally, it remains to be shown that $\mathsf{Vect}_K/K^\times$ is quotient-free. So suppose that we have a quotient that identifies two parallel morphisms $[f],[g]:V\to W$, where I'm working with representatives $f,g\in\mathsf{Vect}_K(V,W)$. By suitably pre- and postcomposing with a subspace inclusion and quotient space projection, we may assume that $f=\mathrm{id}_V$, while $g$ is not a scalar multiple of the identity. In particular, we can choose a basis in which $g$ is not diagonal, resulting in an element $x:K\to V$ and a functional $t:V\to K$ such that $tx = tfx =0$, but $tgx\neq 0$. Hence the quotient must identify the two remaining scalars $0,1\in\mathsf{Vect}_K/K^\times(K,K)$, and we are again done.

3) The same statement and proof apply to the category $\mathsf{FVect}_K$ of finite-dimensional vector spaces. But this category is also the free additive $K$-linear category, which makes it quotient-free in a different sense, namely with respect to $K$-linear functors into categories enriched over $\mathsf{Vect}_K$.


I would call these monoidal categories simple. But one should exclude the terminal monoidal category, which are too simple to be simple.

Here is a baby example: A commutative monoid, considered as a strict one-object monoidal category (for this to make sense, we need commutativity!), is simple iff it has exactly two regular quotient monoids, the trivial monoid and the monoid itself. Equivalenty, it has exactly two congruence relations.

The only simple commutative groups are $C_p$ for primes $p$. The only finite simple (commutative) monoid, which is not a group, is the multiplicative monoid $(\{0,1\},\cdot,1)$; see Proposition 2.1 in the paper The kernel of monoid morphisms by Rhodes and Tilson. Unfortunately, I could not find more about simple commutative monoids.

  • 1
    $\begingroup$ Those are indeed the only simple commutative monoids. To see this, note that every commutative monoid has a join-semilattice quotient given by writing $a\geq b$ whenever there are $n\in\mathbb{N}$ and $c$ such that $a^n = bc$, and identifying two elements whenever they are ordered both ways. Thus in order to be simple, this semilattice quotient has to be trivial (group case), or the original monoid must already have been a semilattice. But since any semilattice can be quotiented by a principal upper set, the two-element semilattice that you mention is the only simple semilattice. $\endgroup$ – Tobias Fritz Jan 4 '17 at 13:28
  • $\begingroup$ BTW, that semilattice quotient of my previous comment is well-known, but a quick search hasn't turned up a good reference. It has come up repeatedly e.g. at this MO question: mathoverflow.net/questions/131110/… $\endgroup$ – Tobias Fritz Jan 4 '17 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.