TITLE: Quotient-free monoidal categories
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: 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.