Let $X$ and $Y$ be sets with $f: X \to Y$, and $E, F \subset Y$, then $E \subset F$ if and only if $f^{-1}(E) \subset f^{-1}(F)$. *Proof*. Suppose that $E \subset F$, then for any $x \in f^{-1}(E)$, $f(x) \in E \subset F$. Therefore $f^{-1}(E) \subset f^{-1}(F)$. Suppose that $f^{-1}(E) \subset f^{-1}(F)$, then for any $y \in E$, $f(f^{-1}(y)) \subset F$ and $E \subset F$.