I'm trying to prove the following for inverse semigroups

$\bf Def:$ an inverse semigroup $S$is a semigruop such that for each $x\in S$ the exists a unique $y\in S$ such that $xyx=x$ and $yxy=y$.

An involution $*$ on $S$ can be defined as follows for $x\in S$ define $x^* = y.$

I'm trying to show that this is an involution and then If $S$ is a topological inverse semigroup this involution is continuous. I searched around and found this sketch of the proof:

1) Let $E(S)=\{x\in S : x^2 =x \}$ then for every $x\in E(S)$ we have $x^* =x$, no problem here it is easy to check.

2) $E(S)$ is a commutative subsemigroup of $S$ and for $s\in S$ and $e\in E(S)$ we have $ses^*\in E(S)$, again no problem here I checked it.

3) from 1 and 2 one can ched directly that for $s,t\in S$ we have $(s^*)^*=s$ and $(st)^*=t^*s^*$ and I'm stuck here.

and I have no Idea how to check if the involution is continuous

any ideas.

Thank you

  • $\begingroup$ Are you saying you have not checked $(s^*)^* = s$? $\endgroup$ Mar 27, 2014 at 13:44
  • $\begingroup$ I checked that the part about $(st)^*$ is where I have a problem $\endgroup$
    – user100478
    Mar 27, 2014 at 17:31

2 Answers 2


The key thing is that $ss^* \in E(S)$.

Proof of that:

$(ss^*)^2 = (ss^*) (ss^*) = s(s^* s s^*) = s(s^*)$

How does it help?



Since $ss^*$, $s^*s = s^*(s^*)^*$, $tt^*$, and $t^*t$ are in $E(S)$, so they all commute. That means $(t^* s^*) (st ) (t^* s^*) = t^* (s^*s)(tt^*) s^* = t^* (t^*t)(s^*s) s^* = (t^* t t^*)(s^* s s^*) = t^* s^*$. Similarly $(st)(t^* s^*)(st) = s(tt^*)(s^*s)t = s(s^*s)(tt^*) t = (ss^*s)(tt^*t) = st$, so $(st)^* = t^*s^*$ by uniquness.

  • $\begingroup$ is there a way to show that this involotion is continuous if $S$ have a $T_2$ topology? @Jack Schmidt $\endgroup$
    – user100478
    Mar 27, 2014 at 18:09
  • $\begingroup$ Sorry, no idea on the topological question. If it is true for groups (and I guess it has to be, as they are inverse semigroups), then try replacing the group $G$ with the inverse monoid $\{ (xx^*) y (xx^*) : y \in S \}$ when you are looking at some element $x$. $\endgroup$ Mar 27, 2014 at 18:17
  • $\begingroup$ Its true for topological groups by assumption ( a topological group must have the map $g\to g^{-1}$ is continuous) so if we take the involution to be the inverse map we are all set. but its different for semi-groups. Thank you for answering the first part though. $\endgroup$
    – user100478
    Mar 27, 2014 at 18:32
  • $\begingroup$ By "it is true", I meant this statement "if a set is simultaneously a group and a topological space, and if it has a continuous multiplication map, then its inverse map must also be continuous." $\endgroup$ Mar 27, 2014 at 18:33

If $S$ is a compact topological semigroup which is algebraically an inverse semigroup, then the map $S \rightarrow S, x \mapsto x^*$ is continuous, i.e. $S$ is a compact topological inverse semigroup. For a proof, see e.g. Proposition 1.6.7 in Hajji (2011).

I am wondering whether this still holds if one replaces "compact" by "locally compact" or "completely metrizable".


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