I have a question about a passage from this paper. First, some definitions

A Semigroup $S$ is said to be an inverse semigroup provided that for every $x \in X$, there exists a unique element $x^{-1}$ such that $x^{-1}xx^{-1} = x^{-1}$ and $xx^{-1}x=x$.

The semilattice $E$ of idempotents in an inverse semigroup $S$ is given by $E = \{x^{-1}x \mid x \in S\} = \{xx^{-1} \mid x \in S\} = \{e \in S \mid e = e^2 \}$.

On page 18 of the above linked paper, the author writes:

In the case of partial bijections, the semilattice of idempotents is given by all domains and images. Multiplication in this semilattice is intersection of sets, and $\le$ is $\subseteq$ for sets, i.e., containment.

Given a set $X$, the collection of all partial bijections on $X$ is denoted as $I(X)$. The semilattice of idempotents $I(X)$ is a subset of $I(X)$, so it itself should consist of partial bijections. But how can the semilattice of idempotents consist of sets? Shouldn't it consist of all partial bijections $f$ such that $f^2 = f$? Are we identifying $f$ with its domain and image? I don't think this makes sense because there are distinct functions with the same domain and image.

  • $\begingroup$ My Master's dissertation might interest you. You can find a link to it in my profile here. $\endgroup$
    – Shaun
    Jan 25, 2021 at 21:18

1 Answer 1


The idempotents in this semigroup are those partial functions $f$ such that the range is a subset of the domain and $f$ maps every element of the range to itself.

CORRECTION: I misread "partial bijections" as "partial functions". So the requirements on $f$ in my answer together with the "bijection" requirement imply that $f$ is simply the identity function on some subset of $X$. Apparently the author of the quoted text identifies "the identity function on a subset $Y$ of $X$" with simply $Y$ and thus gets that the idempotents amount to subsets of $X$.

  • $\begingroup$ Ah, okay. That is useful. But how do you explain the passage I quoted? Because on that same page he looks at the "left inverse hull" $I_{l}(P)$ attached to a left cancellative semigroup $P$, and defines $\mathcal{J}_{P}$ as the semilattice of idempotents of $I_{l}(P)$; and then he says it's "easy" to show that $$\mathcal{J}_{P} = \{p_n ... q_1^{-1} p_1 (P) \mid q_i,p_i \in P \} \cup \{ q_n^{-1} p_n ... q_1^{-1}p_1 (P) \mid q_i,p_i \in P\}.$$ But this means that $\mathcal{J}_{P}$ consists of subsets of $P$, not partial functions on it. How does one make sense of this? $\endgroup$
    – user193319
    Jan 25, 2021 at 23:42
  • $\begingroup$ @user193319 See the correction that I just added to my answer. $\endgroup$ Jan 26, 2021 at 1:10
  • $\begingroup$ Wait, can't we actually prove that the range and domain are equal? If $f$ is an idempotent, then $f^2=f$ which means that $f^2$ and $f$ have the same domain. Since $dom(f^2) = dom(f) \cap f^{-1}(dom(f)) = f^{-1}(dom(f))$, we have $dom(f) = f^{-1}(dom(f))$. Applying $f$ to both sides and using injectivity of $f$, we obtain $$f(dom(f)) = f(f^{-1}(dom(f))) = dom(f).$$ By definition, $f(dom(f))$ is the image/range of $f$, so $im(f) = dom(f)$. So, idempotents really can be uniquely identified with subsets of $X$...right? $\endgroup$
    – user193319
    Jan 27, 2021 at 15:59
  • $\begingroup$ @user193319 I think that's a complicated version of what I said in my "Correction". More simply: I had already said in the original answer, that $f$ maps each element of the range to itself. If, in addition, $f$ is one-to-one, then it can't map anything outside the range into the range, so the domain has to equal the range. $\endgroup$ Jan 27, 2021 at 16:19

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