Two properties on $f^{-1}(A)$ For any function $f : X \rightarrow Y$ and any subset A of Y, define
$$f^{-1}(A) = \{x \in X: f(x) \in A\}$$ Let $A^c$ denote the complement of A in Y. For subsets $A_1,A_2$ of Y, consider the following statements:
(i) $ f^{-1} (A^c_1 \bigcap A^c_2) = (f^{-1}(A_1))^c \bigcup (f^{-1}(A_2))^c  $
(ii) If $ f^{-1} (A_1) = f^{-1} (A_2)$ then $A_1 = A_2 $
Then which of the above statements are always true?
My effort: The first statement can not be true unless $A_1 = A_2$. So that's not always true. For the 2nd statement, let x = $ f^{-1}(A_1) = f^{-1}(A_2)$, then $f(x) = A_1 = A_2$. Since f is a function, f can not have duplicate values of f(x) for the same value of x. That's what I read in books. E.g. the function $y^2 = x$, is actually two functions, $y=+\sqrt x$ and $y=-\sqrt x$, since, for same x, there are two values of y. So, my answer is, (ii) is always true, (i) is not always true.
But the answer given is, neither (i) nor (ii) is always true.  Any pointers on where my understanding is incorrect, is highly appreciated.
 A: On $X=\{1,2\}$ let $f(1)=f(2)=0$ viewed as a  function into $Y=\mathbb R$. Then $f^{-1}(\{0,1\})=f^{-1}(\{0,2\})$ but $\{0,1\} \neq \{0,2\}$
A: Not necessarily $A_1=A_2$.
Actually, when $A_1\subset A_2$, the second statement could be true. Indeed, if $f(x)\in A_1 \subset A_2$, then $f(x)\in A_2$.

Little note: Your writing $f(x)=A_1=A_2$ is strange since $f(x)$ is a value while $A$ is a set.
A: You're getting confused with the notation $f^{-1}$... which is confusing!
Unfortunately, $f^{-1}$ is used to denote two different things:

*

*If $f$ is a bijection between $X$ and $Y$, for $y \in Y$ $f^{-1}(y)$ is the inverse image of $y$ under the map $f^{-1}$.

*$f^{-1}$ is also used to denote the inverse image of a subset $B \subseteq Y$. $f^{-1}(B) = \{x \in X: f(x) \in B\}$. To avoid this confusion, it may be interesting to denote $\{x \in X: f(x) \in B\}$ by $f^{-1}[B]$ instead of $f^{-1}(B)$.

Bottom line your main error is when you wrote $x=f^{-1}(A_1) = f^{-1}(A_2)$. $f^{-1}(A_1),f^{-1}(A_2)$ are subsets of $X$, not elements of $X$. And as pointed in other answers, $f(f^{-1}(A))$ may be different of $A$. Though you always have $f(f^{-1}(A)) \subseteq A$.
