In the following link there is an answer to a question I am working on but I'm nut sure I understand it fully.
Exercise 1.9.3 of Howie's “Fundamentals of Semigroup Theory”.
The second question:
Let $X$ be a countably infinite set and let S be the set of one-to-one maps $\alpha:X\rightarrow X$ with the property that $X\setminus X\alpha$ is infinite.
(b) Show that for all $\alpha\in S$ there exists a bijection between $X\setminus X\alpha$ and $X\alpha \setminus X\alpha^2$.
Which was answered with:
(b) In general, $X\alpha\setminus X\alpha^2\subseteq (X\setminus X\alpha)\alpha.$
The reverse inclusion holds because α is injective. α restricts to a bijection $X \setminus X\alpha\rightarrow X\alpha \setminus X\alpha^2$.
I'm unsure as to why the first line in general is true?