Independence of axioms for biordered sets
Biordered sets were introduced by \textit{K. S. S. Nambooripad} as an abstraction of the partial semigroup of idempotents of a semigroup [Mem. Am. Math. Soc. 224 (1979; Zbl 0457.20051]. Let X, Y be sets, \(\rho \subseteq X\times Y\) and put \(\rho(y)=\{x\in X:\quad x\rho y\}.\) A partial algebra is defined to be a set E equipped with a partial binary operation. The set \(\{(e,f):\quad ef\quad exists\quad in\quad E\}\) is called the domain of this partial operation and is denoted by \(D_ E\). On E define \(\omega^ r=\{(e,f):\quad fe=e\}, \omega^{\ell}=\{e,f):\quad ef=e\}.\) The author gives two new sets of axioms for a biordered set (examples show independence of each set of axioms). Let E be a partial algebra. Then E is a biordered set if and only if the following axioms and their duals hold. (BS1) \(\omega^ r\), \(\omega^{\ell}\) are quasiorders on E and \(D_ E=(\omega^ r\cup \omega^{\ell})\cup(\omega^ r\cup \omega^{\ell})^{-1}.\) (BS2) \(e\omega^ rf\omega^ rg\) implies \(eg\omega^ re\) and \((eg)f=ef.\) (BS3) If \(e\omega^{\ell}f\), \(e,f\in \omega^ r(g)\) then (fg)(eg) exists and \((fg)(eg)=(fe)g.\) (BS4) If \(e,f\in \omega^ r(g)\) and \(eg\omega^{\ell}fg\) then there exists \(e_ 1\in \omega^{\ell}(f)\cap \omega^ r(g)\) such that \(e_ 1g=eg.\) Let M(e,f) denote the quasiordered set \((\omega^{\ell}(e)\cap \omega^ r(f),\prec)\) where \(\prec\) is defined by \(g\prec h\) if and only if \(eg\omega^ reh\) and \(gf\omega^{\ell}hf\). Then \(S(e,f)=\{h\in M(e,f):\quad g\prec h\quad for\quad all\quad g\in M(e,f)\}\) is called the sandwich set of e and f. The second set of axioms given for a biordered set is selfdual. (A1) \(=\) (BS1) (A2) \((e,f)\in D_ E\) implies \((e,ef),(ef,f)\in D_ E\) (A3) If ef, fg, e(fg) and (ef)g exist then \(e(fg)=(ef)g.\) (A4) \(If\quad ef\quad exists\quad and\quad if\quad both\quad e\quad and\quad f\quad belong\quad to\quad either\quad \omega^ r(g)\quad or\quad \omega^{\ell}(g)\quad then\quad(geg)(gfg)\quad exists\quad and\quad(geg)(gfg)=g(ef)g.\) (A5) If both e and f belong to \(\omega^ r(g)\) or \(\omega^{\ell}(g)\) then \(gS(e,f)g=S(geg,gfg)\).
scientific article; zbMATH DE number 3845843