CMon is not coregular

database/data/004_property-assignments/CMon.sql

@@ -76,4 +76,10 @@ VALUES
 	'regular subobject classifier',
 	FALSE,
 	'Assume that $\Omega$ is a regular subobject classifier. Since the trivial monoid is a zero object, every regular submonoid $U \subseteq M$ of any commutative monoid $M$ would have the form $\{m \in M : h(m) = 1 \}$ for some homomorphism $M \to \Omega$. Now take any commutative monoid $M$ with zero that has two different homomorphisms with zero $f,g : M \rightrightarrows N$ (for example, let $M = N = \{0\} \cup \{x^n : n \geq 0\}$ be the free monoid with zero on one generator, $f(x) = 0$,and $g(x) = x$). Take their equalizer $U \subseteq M$, and choose a homomorphism $h : M \to \Omega$ with $U = \{m \in M : h(m) = 1\}$. Since $0 \in U$, we have $h(0)=1$. But then for all $m \in M$ we have $h(m) = h(m) h(0) = h(m 0) = h(0) = 1$, i.e. $U = M$, which yields the contradiction $f = g$.'
+),
+(
+	'CMon',
+	'coregular',
+	FALSE,
+	'We can show this analogously to the case of commutative rings <a href="https://math.stackexchange.com/a/3746890" target="_blank">MSE/3746890</a>. Consider the commutative monoid $\mathbb{N}^2$ and its submonoid $U\coloneqq\{(m,n)\mid m\ge n\}$ with the inclusion $i\colon U\hookrightarrow\mathbb{N}^2$. Then, the pushout of $i$ along itself is $\langle x,y,z : x+y=x+z \rangle$, and the equalizer of the cokernel pair of $i$ is $D\coloneqq\{(m,n)\mid m=0 \implies n=0 \}$. If the category $\mathbf{CMon}$ were coregular, the canonical inclusion $j\colon U\hookrightarrow D$ would have to be an epimorphism. However, it is not: let $I\coloneqq\{0,1\}$ be the two-element commutative monoid with $1+1=1$, and let $u,v\colon D\to I$ be the morphisms defined by $u^{-1}(0)=\{(0,0)\}$ and $v^{-1}(0)=\{(0,0),(1,2)\}$; then we have $u\circ j = v\circ j$.'
 );