# Inverse Semigroups, Partial Bijections, and Semilattice of Idempotents

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.

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

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 $$X$$.
• 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? Jan 25, 2021 at 23:42
• 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? Jan 27, 2021 at 15:59
• @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. Jan 27, 2021 at 16:19