For each commutative monoid $M$, there exists a "groupification" $\widehat{M}$, i.e. an abelian group that satisfies an obvious universal property.


I tried to prove the following: If in the diagram of Monoids
$$ L \stackrel{j}{\rightarrow} M \stackrel{i\_1, i\_2}{\rightrightarrows} N$$
the morphism of monoids $j$ is an equaliser of $i\_1$ and $i\_2$, then the sequence of abelian groups
$$ \widehat{L} \stackrel{\widehat{j}}{\longrightarrow} \widehat{M} \stackrel{\widehat{i}\_1 - \widehat{i}\_2}{\longrightarrow} \widehat{N}$$
is exact in the middle.


I thought the proof would be quite easy, but I didn't manage to do it, so I thought that probably the statement is just wrong.


Do you have a proof (or a reference) or a counterexample respectively?




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Edit: Thank you for the answers so far. All the examples include absorbing elements. Is this fixable by requiring that none of the monoids has absorbing elements, or in some other way?



"Exactness" of groupify functor
Let $A=\mathbb{N}\cup\{\infty\}$, considered as a monoid under addition. Let $M=\mathbb{N}$, $N=A\oplus A$, $i\_1(n)=(n,0)$ and $i\_2(n)=(0,n)$. Then the equalizer of $i\_1$ and $i\_2$ is $0\to\mathbb{N}$. But $\widehat{N}=0$, so $0\to\widehat{M}\to\widehat{N}$ is not exact.


If you want an example without absorbing elements, you can replace $A$ by $\mathbb{N}\cup\{\infty,2\infty,3\infty,\dots\}$ with the obvious monoid structure.


