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.
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 $X$.