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$Y\subseteq X$ and $\exists f:X\to Y$ injection, then $X$ and $Y$ are in bijective correspondence

if $ Y=X$, we're done, since Identity map servers our purpose

assume $Y\neq X \rightarrow X\setminus Y \neq 0$

define $A= f^{n}(X\setminus Y)$

define $g:X\to Y$

$x \mapsto f(x) $ if $x\in A$ and

$x \mapsto x $ if $x\notin A$

It is well defined.

if $g(x)=g(y)$

then if both $x,y \in A\rightarrow f(x)=f(y) \rightarrow x=y $

else if both $x,y \notin A \rightarrow x=y$

if $x\in A $ and $y\notin A$

then $g(x)=g(y)\rightarrow f(x)=y$

as $x\in A \rightarrow \exists m\in N\cup\{0\} $ and $\exists z\in X\setminus Y$ such that $x=f^m (z)\rightarrow y=f(x)=f^{m+1}(z) \in A$ contradiction

so $g$ is injective

How can I show that this map is surjective?

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I assume you mean $A = \bigcup_{n=0}^\infty f^n(X\setminus Y).$ To see $g$ is surjective, let $y\in Y.$ If $y\notin A,$ then $y=g(y).$ If $y\in A,$ then since $y\notin X\setminus Y,$ $y=f(x)$ for some $x\in A,$ so $y=g(x).$

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As $Y\subset X$, you can consider the injective embedding $Y \rightarrow X$. As there is alzó An injective function $X \rightarrow Y$, it follows by Cantor-Bernstein theorem that $|Y|=|X|$, i.e., there is a bijection between these two sets.

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    $\begingroup$ I am using this to prove Cantor Bernstein theorem $\endgroup$ – Abhay Sep 17 '19 at 21:58

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