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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?

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  • $\begingroup$ I have tried to type your question properly. Is this edit correct? $\endgroup$ – MRT Oct 30 '18 at 18:28
  • $\begingroup$ @MRT Sorry to say that your edit made the question incomprehensible. You confused the symbols $\setminus$ and $/$. $\endgroup$ – J.-E. Pin Nov 9 '18 at 15:18
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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 $B = X\alpha$.

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