In the paper "Schreier split epimorphisms between monoids" by Bourn, Nelson, Martins-Ferreira, Montoli and Sobral, Semigroup Forum June 2014, the authors prove a characterization of groups among monoids. I am hoping for a simpler proof of one direction.

Definitions. A splitting is a monoid homomorphism $f:A\to B$ with a specified section $s:B\to A$. A splitting is left homogeneous if for every $b\in B$, multiplication on the left by $s(b)$ is bijective $\mathrm{Ker}f\to f^{-1}(b)$. Analogously for right homogeneous. A splitting is homogeneous if it's both left and right homogeneous.

Remark. For the characterization, I think it suffices to assume the multiplication maps are surjective, not necessarily bijective.

Theorem. For a monoid $B$, TFAE

  1. $B$ is a group;
  2. Every splitting $A\overset{s}{\leftrightarrows}B$ over $B$ is homogeneous.

That 1$\implies$2 is straightforward and appears in proposition 3.4. The converse implication 2$\implies$1 is part of corollary 5.7 and involves an ordeal with internal relations.

Is there a simple(r) proof that 2$\implies$1?

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    $\begingroup$ Their proof actually shows that it suffices to consider $A=B\times B$, with $f$ (either first or second) projection and $s$ the diagonal. Indeed if this is homogeneous then for any $(b',b)\in B\times B$ there is an $x\in B$ such that $(b',b)=(x,1)(b,b)$, i. e. that $b'=xb$. $\endgroup$ May 3, 2017 at 10:22
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    $\begingroup$ Is the section a monoid homomorphism or just a set map? $\endgroup$ May 3, 2017 at 12:35
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    $\begingroup$ @მამუკაჯიბლაძე : Maybe post as an answer? (It's enough to let $b'=1$, by the way.) $\endgroup$ May 3, 2017 at 15:33
  • $\begingroup$ @BenjaminSteinberg the section is a monoid homomorphism. $\endgroup$
    – Arrow
    May 3, 2017 at 15:43

1 Answer 1


Frieder Ladisch noticed, that even less of 2. is needed than I thought. By his suggestion I make my comment into an answer.

Given 2. (in fact even less than one-sided homogeneity of one particular splitting suffices), let $A=B\times B$, let $f:B\times B\to B$ be the projection $(b',b)\mapsto b$, and let $s$ be the diagonal $s(b)=(b,b)$. Given any $b\in B$, multiplication on the left by $s(b)$ on $\operatorname{Ker}(f)\to f^{-1}(b)$ is the map $B\times\{1\}\to B\times\{b\}$ with $(x,1)\mapsto(b,b)(x,1)=(bx,b)$. From left homogeneity of $s$ we only need that $(1,b)\in f^{-1}(b)$ is in the image of this map, i. e. there is an $(x,1)\in B\times\{1\}$ with $(bx,b)=(1,b)$; that is, there is an $x\in B$ with $bx=1$.

Thus every element has a right inverse, so $B$ is a group.


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