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This is more of a request for pointers to relevant literature than a question per se. I am, erm, looking at a paper which uses a kind of iterated pushout construction to obtain a commutative monoid with certain desired properties. The particular "gluing" construction the authors want to do is handled by quite direct means, and while that's probably fine for this particular problem, it would be nice to put it in the proper context.

My copy of Howie's "Introduction to Semigroup Theory" discusses the general notion of "semigroup amalgams" but doesn't seem to say anything about the pushouts in the smaller category of commutative monoids. The closest I can find via MathSciNet is a 1968 paper of Howie, which seems to be working in the category of commutative semigroups, but I can't get hold of a copy at the moment.

Anyway: I'm hoping that someone reading might know of a sensible place to look for a summary of some general results. (The issue is whether the constituent pieces "embed" into the pushout, not the mere existence of the pushout.)

EDIT: On further reflection, while pushouts are undoubtedly relevant, for the purposes of the paper I'm looking at, faithfulness of the embedding is more important than the universal property of pushouts. So I'm changing the title of the question to reflect this. I also think that while the suggestions below have been useful in a broad sense they don't quite address the case I need - which could just be an illustration that said case falls between the two extremes most commonly looked at by semigroup theorists.

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5 Answers 5

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Arthur Ogus wrote a book on logarithmic geometry, apparently soon to be published, and there is a preprint version on his webpage. The first chapter is about commutative monoids, and in particular, it has a bit about pushouts (starting on page 12, and you can tell your computer to search for other instances of the word).

General pushouts can be quite pathological, but you can say interesting things if the monoids satisfy some properties such as integrality. For example, if all of the monoids are integral and one of the monoids is a group, then the pushout is integral, and its group completion is the pushout of the group completions. This reduces Reid's example to a calculation with groups.

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Thanks Scott. Having had a quick look at the Ogus bookdraft, I gather that "integral" = "cancellative". As it happens, the examples in question have nilpotents (so can't be cancellative) but do satisfy $(rx= sx \neq 0) \implies (r=s)$ –  Yemon Choi Nov 29 '09 at 1:06

I'm not sure exactly what you're looking for, but here is a somewhat weird example of a pushout in commutative monoids I recently came across:

$\begin{matrix} \mathbb{N}&\to&\mathbb{Z}\\\\ \downarrow&&\downarrow\\\\ 0&\to&0 \end{matrix}$

This can't happen in abelian groups (more precisely a pushout square with 0 in the lower left whose top arrow is a monomorphism is also a pullback square).

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Well, this kind of thing shows why pushouts of (commutative) monoids or semigroups can be a fiddly business. Really what I wanted is to be able to say that "in this paper, the gluing construction that's being done directly is a special case of harder/nastier results that can be found in X's paper/book". It would IMHO just be weird if it hadn't been written up somewhere... –  Yemon Choi Nov 25 '09 at 7:02

I don't know a reference, but in case you never find one, here's a possible strategy for figuring it out yourself.

Let $F$ be the functor from commutative monoids to abelian groups that adjoins formal inverses to every element. Then $F$ is left adjoint to the forgetful functor $U$, so preserves colimits and in particular pushouts. I believe it's the case that for a commutative monoid $M$, the canonical map $M \to UF(M)$ is injective iff $M$ is cancellable, i.e. $m + n = m' + n$ implies $m = m'$ (for $m, m', n \in M$). This might allow you to transfer known results about pushouts of abelian groups over to the category of commutative monoids.

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Have you looked at the book Monoids, Acts and Categories by Mati Kilp, Ulrich Knauer and Alexander V. Mikhalev? Pushouts are discussed in chapter II.

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From what I can see, these appear to be pushouts in the category of S-acts where S is the core of your amalgam. It's not clear to me that this will recover the notion of semigroup amalgam –  Yemon Choi Dec 1 '09 at 20:56

Pushouts are expressible in terms of coproducts and coequalizers (just as all other colimits).

Since commutative monoids are the same as $\mathbb{N}$-semimodules, the coproduct $A+B$ is the direct sum with pairs $(a,b)$ as elements along with the injections $i : A \to A+B$ sending $a \mapsto (a,0_B)$ and $j : B \to A+B$ sending $b \mapsto (0_A,b)$.

The pushout of $A \stackrel{f}{\leftarrow} X \stackrel{g}{\rightarrow} B$ is then the coequalizer of the parallel maps $X \stackrel{f}{\rightarrow} A \stackrel{i}{\rightarrow} A+B$ and $Y \stackrel{g}{\rightarrow} B \stackrel{j}{\rightarrow} A+B$. Following the recipe for Set given in MacLane, CWM, Sec. III.3, the coequalizer is $(A+B)/E$ where $E \subseteq (A+B) \times (A+B)$ is the least monoid congruence relation containing pairs $\langle i(f(x)), j(g(y)) \rangle$ for all $x \in X$.

Except for replacing "equivalence relation" by "monoid congruence relation", there is nothing mysterious here.

[Note added: If one were working with non-commutative monoids, the coproduct would be the free product instead of the direct sum, but everything else would be the same.]

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Thanks. Having read the basics of MacLane, I know that one can set up all the machinery, but for instance, why is it obvious that the maps from A and B into the pushout are monomorphisms? THAT was what is important in the paper I'm looking at. –  Yemon Choi Aug 25 '13 at 23:23
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As I said in the question: "The issue is whether the constituent pieces 'embed' into the pushout, not the mere existence of the pushout." The paper was constructing a counter-example using iterated amalgams/pushouts, and one needed to make sure that the construction didn't accidentally collapse. Of course in the group case this is all standard, but semigroup amalgams always looked trickier to me, cf. the chapter in Howie's book. –  Yemon Choi Aug 25 '13 at 23:26
    
Indeed, sorry that I missed your parenthetical part of the question. I am not sure of an easy reference but I will investigate. –  Uday Reddy Aug 26 '13 at 8:34
    
Thanks: if you do find something, I am still interested in this question! –  Yemon Choi Aug 26 '13 at 8:36

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