I know there is a splitting lemma for groups, but is there a similar lemma for semigroups or monoids? And do you have the proof of that?
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$\begingroup$ Well, I'm not sure what exactness ought to mean in the context of monoids, since the notion of "kernel" is a bit odd - for example, consider the monoid $N^+$ of natural numbers (including 0) together with $\infty$, and the monoid $T=(0, \infty)$. Then there is a homomorphism from $N^+$ to $T$ taking 0 to 0 and everything else to $\infty$ - clearly this map is not injective, but the "kernel" (preimage of 0) is $\{0\}$. I don't know, but I suspect that the "right" notion of exactness for semigroups/monoids is one which directly implies a splitting lemma. $\endgroup$– Noah SchweberJun 3, 2014 at 23:05
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1$\begingroup$ (cont'd) See, however, mathoverflow.net/questions/83080/exact-sequence-of-monoids, where JBorger argues that there is no right notion of exactness for monoids. So maybe the answer is "mu" :). $\endgroup$– Noah SchweberJun 3, 2014 at 23:05
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You might look at Tilson's approach http://www.sciencedirect.com/science/article/pii/0022404987901083 where the kernel of a monoid morphism is a category and wreath product decompositions are obtained.
answered Jun 4, 2014 at 0:26