If $f^{-1}(f(E)) = E$ for all $E \subseteq X$ then $f(A-B) = f(A) - f(B)$ for all $A,B\subseteq X$

Question

Let $f:X\rightarrow Y$.

Prove: If $f^{-1}(f(E)) = E$ for all $E \subseteq X$, then $f(A-B) = f(A) - f(B)$ for all $A,B\subseteq X$.

Note: $A-B = \{x: x\in A, \text{ but } x\notin B \}$.

Proof: Suppose $f^{-1}(f(E)) = E$ for all $E \subseteq X$. Let $y \in f(A-B)$. Then $\exists x \in A-B$ such that $y = f(x)$. Since $x \in A-B$, $x\in A$, but $x\notin B$. Then $f(x)\in f(A)$, but $f(x) \notin f(B)$. Thus, $y = f(x) \in f(A) - f(B)$.

On the other hand, let $y \in f(A)-f(B)$. Then $y \in f(A)$, but $y \notin f(B)$. Then $\exists x \in A$ such that $y = f(x)$. Since $y \notin f(B)$, $x \notin B$. Thus, $x \in A -B$, so $y = f(x) \in f(A-B)$.

My proof that $f(A-B) = f(A) - f(B)$ didn't use the antecedent at all. So I'm wondering if I need to use it directly.

Answer 1

Let $y\in f(A-B)$, then there exists some $x\in A-B$ such that $f(x)=y$. Note $x\in A$, therefore $f(x)=y\in f(A)$. But $f(x)$ may be not in $f(B)$.(why?)

Suppose it were true that $f(x)\in f(B)$.
Then there exists some $x_0\in B$ such that $f(x_0)=f(x)$. Then consider the pre-image set of $f(x)$,

$$f^{-1}(f\{x_0\})=f^{-1}(f\{x\})\Rightarrow x_0=x\text{ (By assumption)}$$ Then contradiction arises.