If $f:A\to B\subseteq A$ is injective, then there exists a bijection $h:A\to B$

Question

Please check whether my proof has any error! Thank you so much!

Theorem:

If $f:A\to B\subseteq A$ is injective, then there exists a bijection $h:A\to B$.

Proof:

Let $Y=A \setminus B$, and $X=\bigcup_{i\in \mathbb{N}}f^i(Y)$ where $ f^i(Y)=f(f(...f(Y)...)), i$ times.

$f(X)=f(\bigcup_{i \in \mathbb{N}}f^i(Y))=\bigcup_{i \in \mathbb{N}}f(f^i(Y))=\bigcup_{i \geq1}f^i(Y)$.

$X=\bigcup_{i \in \mathbb{N}}f^i(Y) = f^0(Y) \cup (\bigcup_{i \geq1}f^i(Y))=Y \cup (\bigcup_{i \geq1}f^i(Y))=Y \cup f(X)$.

To sum sup, $X=Y \cup f(X)$.

$A \setminus X=(Y\cup B) \setminus (Y\cup f(X))=B \setminus f(X)$.

We define $h:A\to B$ such that $h(x)=f(x)$ for all $x \in X$, and $h(x)=x$ for all $x \in A \setminus X$, then $h$ is bijective.

Answer 1

Since $f:A \to B$ is injective, this implies that $|A| \leq |B|$. Now, since $B \subset A$, this implies that $|B| \leq |A|$. Combining these two inequalities, we have that $|A| =|B|$. Then, since $f$ is an injection, it is also a surjection and hence a bijection.