I'm trying to prove the following characterization for inverse semigroups:
Let $S$ be a regular semigroup. For each $a \in S$ define $A(a) = \{x \in S : a = axa \}.$ The semigroup $S$ is inverse if and only if, for each a $\in S$, $A(a)$ contains a least element with respect to the natural partial order. (For $a,b \in S$, we have $a\leq b$ if there exists an idempotent $e$ such that $a=eb $).
It's easy to see that if $S$ is inverse, then for each $ x \in A(a)$ we have that $a^{-1}\leq x$ since
$$a^{-1}=a^{-1}aa^{-1}=a^{-1}(ax)(aa^{-1})=a^{-1}(aa^{-1})(ax)=(a^{-1}a)(a^{-1}a)x=(a^{-1}a)x,$$ and therefor $a^{-1} $ is the least element of $A(a).$
Reciprocally, it's also easy to see that if $A(a)$ has least element $m$ then $mam$ must equal $m$, otherwise we would have that $mam=k, k \neq m$ and therefor $k\leq m$ which is absurd since $m$ is the least of $A(a)$. So we conclude that the least element of $A(a)$ is one of the inverses of $a$. Note that if we assume that $n$ is an inverse of $a$ and $m=en$ and $m=nf$ then $$n=nan=naman=na(en)an=naen=nama(en)=na(nf)a(en)=nfaen=mam=m $$
Since $m$ is the least element of $A(a)$ we have that $m=en$ for some idempotent $e$. So we only need to prove that there is a $f$ such that $m=nf.$
For some reason I'm having a hard time finishing this. It should be easy though. Any help would be greatly appreciated.