Inclusions of closed sets

Question

Let $X, Y$ be compact Hausdorff spaces, $f : X \to Y$ a continuous function, $A \subseteq X$ closed and $B \subseteq Y$ closed. My conjecture is that $$ \overline{f[A]} \cap B = \emptyset \quad\text{ iff }\quad A \cap f^{-1}[B] = \emptyset. $$ Here $\overline{f[A]}$ is the closure of the direct image of $A$ under $f$. Is there a nice way to prove this?

Answer 1

Since $A$ is a closed subset of a compact set, it is itself compact, so its image by $f$ is compact (thus closed). So your statement really reads $$ f(A)\cap B = \emptyset \quad \text{iff}\quad A\cap f^{-1}(B) =\emptyset. $$ By defining $$ I = \{ (a,b)\in A\times B \text{ such that } f(a)=b\} $$ one has $$f(A) \cap B = \{ b \in B \text{ such that } \exists a \in A \text{ such that } (a,b)\in I\}$$ and $$A \cap f^{-1}(B) = \{ a \in A \text{ such that } \exists b \in B \text{ such that } (a,b)\in I\}.$$ So both sides of your statement are equivalent to $I=\emptyset$.