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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.

Hoang Nam
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  • If $f$ is bijective then you have $f=(f^{-1})^{-1}$ and hence $f(A^c)=(f^{-1})^{-1}(A^c)=[(f^{-1})^{-1}(A)]^c=[f(A)]^c$. Conversely, you have injectivity and surjectivity immediately from $f(A^c)=[f(A)]^c$. – user10354138 Jul 13 '20 at 02:38
  • @user10354138 I don't think it is right because as you can see that C in the left hand side is $\complement_{E}$ but in the right hand side is $\complement_{E'}$, In your answer, how do you show that? – Hoang Nam Jul 13 '20 at 02:42
  • Moreover, to imply surjective and injective properties conversely from the equation you said is not easy – Hoang Nam Jul 13 '20 at 02:45
  • What is $\mathfrak{B}(E')$? – ir7 Jul 13 '20 at 03:29
  • @ir7 It is just a symbol which means that A is a subset of E' – Hoang Nam Jul 13 '20 at 03:39
  • @ir7 I think he means the powerset $\mathcal{P}(E)$ of $E$. – user10354138 Jul 13 '20 at 06:28

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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.)

ir7
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  • Assume is not injective and show that we get a contradiction (using our original hypothesis which is reflected in $f({x}^c) = f({x })^c$). So, it must be injective. – ir7 Jul 13 '20 at 04:18
  • How about to prove from $f$ is bijective? Can I imply directly $f(\complement_{E}(A)) = \complement_{E'} (f(A))$ from $f$ is bijective? – Hoang Nam Jul 13 '20 at 04:26
  • Apply $f^{-1}$ on both sides. You should end up with $A^c = A^c$ (using the property you said you already know). – ir7 Jul 13 '20 at 04:30