If $f:X\rightarrow Y$ is bijective, $A\subset X $ and $B \subset Y$, prove that $f(C_{X}A)=C_{Y}f(A)$ I want to prove the following proposition:
"If $f:X\rightarrow Y$ is bijective, $A\subset X $ and $B \subset Y$, prove that $f(C_{X}A)=C_{Y}f(A)$".
Obs>:  the symbol $C_{X}$, $C_{Y}$ represents the complement set.
Proof
$y \in f(C_{X}A) \rightarrow$ there exists $x \in C_{X}A$ such that $y=f(x)$. $x \in C_{X}A \rightarrow x \notin A \rightarrow y=f(x) \notin f(A) \rightarrow y \in C_{Y} f(A)$. Therefore $f(C_{X}A) \subset C_{Y}f(A)$.
On the other hand, $x \in C_{Y}f(A) \rightarrow x \notin f(A) \rightarrow f^{-1}(x) \notin A \rightarrow f^{-1}(x) \in C_{X}A \rightarrow x \in f(C_{X}A)$, hence $C_{Y}f(A)\subset f(C_{X}A)$.
As you see, I did not use the fact that $f$ is bijective, so my proof is probably incorrect. Please, let me know where is the mistake. Thanks!
 A: $$x \notin A \rightarrow y=f(x) \notin f(A) \rightarrow y \in C_{Y} f(A)$$
This relies on the fact that $f$ is injective.  If $f$ were not injective, then it would be possible that $x\not\in A$ but $f(x)\in f(A)$ since $f(x)=f(w)$ and $w\in A$.
$$x \in C_{Y}f(A) \rightarrow x \notin f(A) \rightarrow f^{-1}(x) \notin A \rightarrow f^{-1}(x) \in C_{X}A \rightarrow x \in f(C_{X}A)$$
This relies on the fact that $f$ is surjective since if $f$ were not surjective then it could happen that $x\in C_X A$ but $f^{-1}(\{x\})=\varnothing$.
I'd include in the proof an explicit mention or explanation of the ways in which injectivity and surjectivity are used.
A: You always have $f(C_X A)\supset C_Y f(A)$ but the other inclusion is correct only if $f$ is injective. You don't need the bijectivity but at least a map which is injective. In your proof, $x\notin A$ doesn't imply that $f(x)\notin f(A)$. Indeed, take $f(x)=x^2$ and take $A=[0,1]$. You have that $f(A)=[0,1]$. You see that $f(-1)\in f(A)$ but $-1\notin A$.
