Why this function defines a bijection?

Question

Show that if $B\subset A$ and there is an injective function $f:A\to B$, then $\operatorname{card}(A)=\operatorname{card}(B)$.

This exercise suggest a way to solve the problem: define $A_1=A, B_1=B$ and $A_n=f(A_{n-1}),B_n=f(B_{n-1})$. Since $A \subset B$ we have $A_{n+1}\subset B_n\subset A_{n-1}$ for $n\geq2$.

Then define $$h(x) = \left\lbrace \begin{array}{l} f(x) &\text{ if } x \in A_n-B_n \text{ for some n}\\ x &\text{ otherwise } \\ \end{array} \right.$$

And the above function is supposed to define a bijection, but I cannot see why the function is injective, I mean, why isn't this possible:

I imagine the problem about injectivity as in the picture I draw, $x\in A_1 - B_1$ then $h(x)=f(x)=y$, but since $y\notin A_n-B_n$ for any $n$ and $y\in A$ must be $h(y)=y$ as well.

Answer 1

In your picture, if $f(x)=y\in B_2$, then $x$ has to be in $f^{-1}(B_2) = B$, so your $x\notin A_1-B_1$.

You can probably generalise this a bit to get a proof for injectivity.

Answer 2

I am trying to suggest an alternative solution.I will use Cantor–Bernstein–Schroeder theorem.

$B\subset A\Rightarrow g:B\to A$ by $g(x)=x$ $\forall x \in B$ is an injection. Also given that $f:A\to B$ is an injection.

Thus by Cantor–Bernstein–Schroeder theorem we conclude that there is a bijection between $A$ and $B$.

hence $card(A)=card(B)$

NOTE: According to OP and user18921 , the question actually leads to prove Cantor–Bernstein–Schroeder theorem.