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.