Assume that $f : X \rightarrow Y $ is surjective. Show that $f(A^c) = (f(A))^c \ \forall A\subset X$ iff $f$ is also injective.

So I tried starting with the right implication $\Rightarrow$

Since $f$ is surjective we know that every $ y\in Y$ corresponds to some $x\in X$. Then we want to prove that this $x \in X $ is at most one.

I came to the conclusion that since $f(A^c) = (f(A))^c \ \Leftrightarrow f(A) \cap (f(A))^c = \emptyset \ \Rightarrow \not\exists x \in A $ s.t $f(x) \in (f(A))^c$ and $\not\exists x \in A^c $ s.t $f(x) \in f(A)$. Now i want to prove that if there $\exists (f(x_1),\ f(x_2)) \in A$ s.t $ f(x_1)=f(x_2)$ then $x_1=x_2$ and of course $\exists (f(x_1),\ f(x_2)) \in A^c$ s.t $ f(x_1)=f(x_2)$ then $x_1=x_2$

But here I have troubles.

  • 1
    $\begingroup$ Hint: Remember you are assuming $f(A^C) = (f(A))^C$ for all subsets of $X$. This includes the case where $A = \{x_1\}$. $\endgroup$ – Browning Jan 29 '16 at 22:41

If $f(A^c) = (f(A))^c \ \forall A\subset X,$ in particular, given $x\in X$ it is $f(\{x\}^c) = \{f(x)\}^c.$ That is, if $y\not \in \{x\}$ ($y\ne x$) then $f(y)\in \{f(x)\}^c$ (that is, $f(y)\ne f(x)$). This shows that $f$ is injective.

The converse implication must be easy as $f$ is bijective.

  • $\begingroup$ "$y \not \in \{x\}$"then shouldnt $y \in \{x\}^c$ and hence $f(y) \in \{f(\{x\}^c)$? But if $y_1, \ y_2 \not \in \{ x \}$ then $f(y_1), \ f(y_2) \in \{f(x)\}^c$ How does one know that $f(y_1) \not = f(y_2)$ ? $\endgroup$ – Olba12 Jan 29 '16 at 22:54
  • $\begingroup$ We are showing that for any $x$ there is no $y$ with $f(x)=f(y).$ Note that $x$ can be any element of $X.$ Thus we can conclude that $f$ in injective. $\endgroup$ – mfl Jan 29 '16 at 22:57
  • $\begingroup$ "$y \not \in \{x\}$"then shouldnt $y \in \{x\}^c$ and hence $f(y) \in \{f(\{x\}^c)$? $\endgroup$ – Olba12 Jan 29 '16 at 22:59
  • $\begingroup$ Yes. And thus means $f(y)\ne f(x).$ Since this holds for any $x,$ $f$ must be injective. $\endgroup$ – mfl Jan 29 '16 at 23:01
  • $\begingroup$ Okey, but you wrote $f(y) \in \{f(x)\}^c$ , I think it should be $f(y) \in f(\{x\}^c)$ $\endgroup$ – Olba12 Jan 29 '16 at 23:02

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