Proposition 4.2.5. Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^{X}$ be a filter base. If $\emptyset \not\in f^{-1}(\fB) = \bracs{f^{-1}(E)| E \in \fB}$, then $f^{-1}(\fB)$ is also a filter base.
Proof. (FB1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f^{-1}(G) \subset f^{-1}(E \cap F) = f^{-1}(E) \cap f^{-1}(F)$.
(FB2): By assumption, $\emptyset \not \in f^{-1}(\fB)$.$\square$