# Show $f^{-1}(A^c)=(f^{-1}(A))^c$ [duplicate]

Let $f: X \to Y$, and $A\subseteq Y$. Show that $f^{-1}(A^c)=(f^{-1}(A))^c$

I know how to prove that $f^{-1}(A^c)\subseteq(f^{-1}(A))^c$, but stuck on proving $(f^{-1}(A))^c\subseteq f^{-1}(A^c)$. Could someone help with this step please? Thanks.

Suppose $x\in (f^{-1}(A))^c$. Then $x\in X$ and $x\notin f^{-1}(A)$. Thus $f(x)\notin A$, and of course $f(x)\in Y$, so $f(x)\in A^c$. Therefore $x\in f^{-1}(A^c)$.
• Is it always true that if $f(x) \in A$, then $x \in f^{-1}(A)$, and $f(x) \notin A$ implies $x \notin f^{-1}(A)$, without assuming injection or surjection? – user57891 Jan 30 '16 at 6:29
• Yes, both are true, with no assumptions about $f$. Confirming that they're true is just a "definition chase", and you can easily verify them. – BrianO Jan 30 '16 at 6:30
Suppose $x \in (f^{-1}(A))^c$. This means $f(x) \notin A$. But we know $f(x) \in Y$. Hence it must be true that $f(x) \in A^c$. That is, $x \in f^{-1}(A^c)$.