TITLE: Prove/Disprove: Let $f : X \rightarrow Y$ be a function. Do we have $f(A^c) = (f(A))^c ?$
QUESTION [0 upvotes]: let $X$  and $Y$  be two  sets  where $A \subset X $. Let $f : X \rightarrow  Y$ be  a  function. Do we  have  $f(A^c) = (f(A))^c ?$ 
My attempt : I think yes, I take $ A= [0, \infty)$ , $X=Y= \mathbb{R}$ and $f(x) = x$ Now $ f( \mathbb{R} / A ) = (f(A))^c$
I thinks  im not on the right track
Any hints/solution will be appreciated

REPLY [1 votes]: Let $X=\Bbb R$, $Y=[0,+\infty)$, $f(x)=x^2$, $A = Y$ then $f[A^c]=(0, +\infty), f[A]=Y$, so $(f[A])^c = \emptyset$. 

Theorem: Suppose $f:X \to Y$ obeys $\forall A \subseteq X: f[A^\complement]= f[A]^\complement$. Then $f$ is a bijection.

First not that $f$ must be onto: $$f[X]=f[\emptyset^\complement] = f[\emptyset]^\complement= \emptyset^\complement= Y$$ by plugging in $A=\emptyset$.
Suppose $x_1 \neq x_2$ exist in $X$ such that $f(x_1)=f(x_2)$. Call the common image point $p$. Then $p \in f[\{x_1\}^\complement]$, as witnessed by $x_2$ while $p \notin f[\{x_1\}]^\complement = \{p\}^\complement$ contradicting the property for $A=\{x_1\}$.
So any non-bijection will give a counterexample, and for a bijection $f$, the property $\forall A \subseteq X: f[A^\complement]= f[A]^\complement$ does hold. So the property is equivalent to being a bijection.
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