Exact sequence of monoids - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:41:07Z http://mathoverflow.net/feeds/question/83080 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83080/exact-sequence-of-monoids Exact sequence of monoids John Voight 2011-12-09T22:36:35Z 2012-10-23T23:06:02Z <p>What is the right definition of an exact sequence of monoid homomorphisms? </p> <p>I can't seem to find a consistent in my searches; indeed Balmer (Remark 2.6, <a href="http://www.math.ucla.edu/~balmer/research/Pubfile/Prod.pdf" rel="nofollow">http://www.math.ucla.edu/~balmer/research/Pubfile/Prod.pdf</a>) calls it a "slippery notion". </p> <p>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. </p> <p>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., <code>$$\{ y \in B : g(y) = 1 \} = \{ f(x) : x \in A \}.$$</code> But, according to Bergman (3.10.6--7, <a href="http://math.berkeley.edu/~gbergman/245/Ch.3.pdf" rel="nofollow">http://math.berkeley.edu/~gbergman/245/Ch.3.pdf</a>), this definition of kernel does not determine the image. Instead, we look at the \emph{kernel congruence} <code>$$K_g = \{(y,z) \in B \times B : g(y)=g(z)\}.$$</code> 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 $(yy',zz') \in K_g$. </p> <p>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 (http://math.stackexchange.com/questions/18387/is-there-an-analogue-of-short-exact-sequences-for-semigroups): he defines $$I_f = (f(A) \times f(A)) \cup \Delta(B)$$ where <code>$\Delta(B)=\{(b,b) : b \in B\}$</code>. </p> <p>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 <code>$$\mathrm{ker} g = \{y \in B : g(y) = 1\} = \mathrm{img} f = \{f(x) : x \in A\}.$$</code> 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 <code>$$K_g = \{(kz,z) \in B \times B : k \in K, z \in B \}.$$</code></p> <p>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 $K_g \not\subseteq I_f$. </p> <p>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.</p> <p>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.</p> <p>Can anyone shed some light on this matter? I'd already be pretty happy with a reference.</p> http://mathoverflow.net/questions/83080/exact-sequence-of-monoids/83087#83087 Answer by James Borger for Exact sequence of monoids James Borger 2011-12-09T23:16:21Z 2011-12-09T23:16:21Z <p>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 <code>$A\rightrightarrows B \to C$</code> of monoid maps (where the two compositions <code>$A\rightrightarrows C$</code> agree) is short exact if the map <code>$B\to C$</code> is surjective and the induced map <code>$A\to B\times_C B$</code> is an isomorphism. This is equivalent to requiring the induced map <code>$A\to B\times B$</code> to be injective, its image to be an equivalence relation, and the induced map <code>$B/A\to C$</code> is to be an isomorphism.</p> <p>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.</p> http://mathoverflow.net/questions/83080/exact-sequence-of-monoids/110489#110489 Answer by Jawad Abuhlail for Exact sequence of monoids Jawad Abuhlail 2012-10-23T23:06:02Z 2012-10-23T23:06:02Z <p>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":</p> <p><a href="http://arxiv.org/abs/1111.0330" rel="nofollow">http://arxiv.org/abs/1111.0330</a></p> <p><a href="http://arxiv.org/abs/1210.4566" rel="nofollow">http://arxiv.org/abs/1210.4566</a></p>