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What is the right definition of an exact sequence of monoid homomorphisms?

I can't seem to find a consistent in my searches; indeed Balmer (Remark 2.6, calls it a "slippery notion".

Among other things, I'll know that it's a good definition if I can take an exact sequence of groups and forget inverses and obtain an exact sequence of monoids.

One possible answer is just to take the definition for groups: a sequence $$A \xrightarrow{f} B \xrightarrow{g} C $$ of monoid homomorphisms is exact if $\mathrm{ker}\ g = \mathrm{img}\ f$, i.e., $$ \{ y \in B : g(y) = 1 \} = \{ f(x) : x \in A \}. $$ But, according to Bergman (3.10.6--7,, this definition of kernel does not determine the image. Instead, we look at the \emph{kernel congruence} $$ K_g = \{(y,z) \in B \times B : g(y)=g(z)\}. $$ The set $K_f$ defines a \emph{congruence} on $A$, an equivalence relation compatible with the operation on $A$, i.e., if $(y,z),(y',z') \in K_g$ then $(yy',zz') \in K_g$.

So then how should we define the image $I_f$ so that our sequence is exact if and only if $I_f=K_g$. One proposal is given by Arturo Magidin ( he defines $$ I_f = (f(A) \times f(A)) \cup \Delta(B) $$ where $\Delta(B)=\{(b,b) : b \in B\}$.

This notion, though probably useful in some ways, has a grave defect. Let $$ A \xrightarrow{f} B \xrightarrow{g} C, $$ be an exact sequence of groups, so $$ \mathrm{ker} g = \{y \in B : g(y) = 1\} = \mathrm{img} f = \{f(x) : x \in A\}. $$ I would like to verify that the sequence is exact as a sequence of monoids, so $I_f=K_g$. We have $(y,z) \in K_g$ if and only if $g(yz^{-1}) = 1$ since g is a group homomorphism, if and only if $yz^{-1} \in K = \mathrm{ker} g$ if and only if $y = kz$ for some $k \in K$. Thus $$ K_g = \{(kz,z) \in B \times B : k \in K, z \in B \}. $$

On the other hand, suppose g is not injective and that f is not surjective (not an atypical situation). Then there exists $z \in B$ such that $z \not\in f(A)$ and there exists $k \in K$ such that $k \neq 1$. Then $(kz,z) \in K_g$ by definition, but $(kz,z) \not\in I_f$: it does not belong to $f(A) \times f(A)$ since $z \not \in f(A)$ and does not belong to $\Delta(B)$ since $kz \neq z$. So $K_g \not\subseteq I_f$.

I can see why one must add the diagonal to $f(A) \times f(A)$, since one wants a reflexive relation on B. It follows that $I_f$ is an equivalence relation. I would think that one should go farther and take the image to be the congruence closure of $I_f$, i.e., the submonoid $\overline{I}_f$ of $B \times B$ generated by $f(A) \times f(A)$ and $\Delta(B)$. Then it is clear that $(kz,z) = (k,1)(z,z)$ in $\overline{I}_f$, and so $K_g \subseteq \overline{I}_f$. Conversely, it is clear that $\overline{I}_f$ is a subset of $K_g$, so then I'm happy again.

But this definition has a different problem: an exact sequence $$ A \xrightarrow{f} B \to 1$$ of monoids is exact if $f$ is surjective but not conversely: the inclusion map $$ \mathbb{N} \xrightarrow{f} \mathbb{Z} $$ of the natural numbers in the integers has $\overline{I}_f = \mathbb{Z} \times \mathbb{Z}$ but is not surjective. Arturo Magidin pointed out to me in an e-mail that this map $f$ is an epimorphism in the category of monoids; maybe I'll have to accept that.

Can anyone shed some light on this matter? I'd already be pretty happy with a reference.

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The category of monoids is an exact category in the sense of Barr, see So it has a natural notion of exact sequence. – Benjamin Steinberg Dec 9 '11 at 23:02
To add to Benjamin's and James's responses, any category that is algebraic (or monadic) over $Set$ is Barr-exact. – Todd Trimble Dec 9 '11 at 23:50
I am always wary of the word 'right' in the first line of the question. "Horses for courses", but given that there was a very nice preprint by Charlie Wells: which might be of use. The original dated to 1980. – Tim Porter Dec 10 '11 at 7:00
up vote 14 down vote accepted

Hi John. I'd say there is no generalization of short exact sequence to the category of monoids, although I suppose it really depends on what you want to do with it. What you probably want is an internal equivalence relation. So you could say a diagram $A\rightrightarrows B \to C$ of monoid maps (where the two compositions $A\rightrightarrows C$ agree) is short exact if the map $B\to C$ is surjective and the induced map $A\to B\times_C B$ is an isomorphism. This is equivalent to requiring the induced map $A\to B\times B$ to be injective, its image to be an equivalence relation, and the induced map $B/A\to C$ is to be an isomorphism.

My general feeling is that this is the right concept in most categories of sets with algebraic structure (e.g. the category of sets itself, semi-rings). It's only in categories where the objects have some group structure that you can re-express it using kernels.

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I see Ben said the same thing in a comment to the original post. – JBorger Dec 9 '11 at 23:17
This doesn't exactly answer my question, since it only gives me short exact sequences, but I'll take it. Thanks James (and Ben)! – John Voight Dec 10 '11 at 17:17
John, the monoid analogue of a long exact sequence is probably a simplicial monoid that is contractible (as a simplicial set), or something like that. – JBorger Dec 11 '11 at 5:55

For the special case of commutative monoids, or more generally semimodules over a semiring, my related preprints on arXiv might be helpful (see below). For arbitrary monoids, the general definition is similar, however it is difficult to apply since the notion of the cokernel of a morphism of monoids is really "slippery":

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