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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?


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?

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Take a non-trivial commutative group M and let N be M with an (absorbing) element adjoined. Then let i_1 be the inclusion of G and i_2 the trivial map. Then the L is trivial. But the groupified sequence is 0->G->0 so not exact –  Benjamin Steinberg May 14 at 21:32
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Groupification is the left adjoint to the inclusion functor, so it's cocontinuous, but there's no reason to expect that it has nice behavior with respect to limits like equalizers. –  Qiaochu Yuan May 15 at 1:14

2 Answers 2

up vote 6 down vote accepted

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.

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You can even get an example where the map $N\to\widehat{N}$ has trivial kernel, so there are no absorbing elements in a rather strong way. Let $M=\mathbb{N}$ and let $N$ be $M\oplus M$ together with elements $ax+b$ for all $a,b\in\mathbb{N}$ with $a>0$. The sum of $ax+b$ and $(m,n)\in M$ is defined as $ax+(b+m+n)$. This $N$ can also be described as what you get by freely adjoining an element $x$ to $M\oplus M$ such that $x+(0,1)=x+(1,0)$. –  Eric Wofsey May 14 at 22:19
    
This last example is in some sense the universal example: it is initial among all pairs of commutative monoids $M$ and $N$ equipped with a pair of maps $M\rightrightarrows N$, an element $m\in M$, and an element of $N$ which witnesses that the two maps equalize $m$ after group-completion. –  Eric Wofsey May 14 at 22:40
    
Sorry, but I don't see why your $i_1 - i_2$ is not injective. Could you elaborate? –  Matthias Ludewig May 14 at 22:46
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In the group-completion $\widehat{N}$, $(0,1)$ and $(1,0)$ are identified because $x+(0,1)=x+(1,0)=x+1$. So $i_1$ and $i_2$ are actually the same map after group-completion. –  Eric Wofsey May 14 at 22:48
    
However, the map $N \longrightarrow \widehat{N}$ is not injective, by the same argument... –  Matthias Ludewig May 15 at 8:20

This is not true. This might be a simple example. Let $M=(\mathbb{N}, +)$ be the monoid of natural numbers (under addition) and let $N=(\mathbb{N}, \times)$ be the monoid of natural numbers under multiplication. Put $i(n)=2^n$ and $j(n)=3^n$. Then $L=0$. However, the groupified sequence will be $0\to\mathbb{Z}\to 0$.

Of course, I assume that $0$ is a natural number.

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