Exercise $1.9.3$ of Howie's “Fundamentals of Semigroup Theory” follow up 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?

The first line follows from general properties of maps between sets. Let $\alpha:E \to F$ be a map (not necessarily injective) and let $A, B$ be subsets of $E$. Then
$A = (A \setminus B) \cup B$, whence $A\alpha = (A \setminus B)\alpha \cup B\alpha$
and finally $A\alpha \setminus B\alpha \subseteq (A \setminus B)\alpha$.
The inclusion $X\alpha\setminus X\alpha^2\subseteq (X\setminus X\alpha)\alpha$   now follows by taking $A = X$ and $B = X\alpha$.
