Prove that $f(\complement_{E}(A)) = \complement_{E'} (f(A))$

Question

I have a problem which is related to mappings and sets as follows:

Given two sets E and E', and a mapping $f: E \rightarrow E'$. Prove that $f$ is bijective if and only if

$\forall A \in \mathfrak{B}(E')$

$$f(\complement_{E}(A)) = \complement_{E'} (f(A))$$

I feel so confused because I just know that $f^{-1}(\complement_{E}(A)) = \complement_{E'} (f^{-1}(A))$. I have tried to utilize the properties of bijection but I failed.

Answer 1

I'll just deal with the reverse implication, that is we assume

$$ f(A)^c = f(A^c)$$

for all $A$ subsets of $E$.

Taking one non-empty subset $B$, we have

$$E' = f(B) \cup f(B)^c = f(B) \cup f(B^c) = f(E),$$

hence $f$ is surjective.

Assume $f(x) = f(y)$ and $x\not= y$, that is $f$ is not injective. We have

$$ f(y) \in f(\{x\}^c) = f(\{x \})^c$$

which implies $$ f(y)\not= f(x), $$

contradiction. So $f$ must be injective too.

(See also many image/preimage facts listed here.)