Continuous Open Maps transfer Bases

Last updated Nov 1, 2022

Let $X, Y$ be Topological Spaces and let $X$ have Topological Basis $\mathcal{B}$. Suppose there exists continuous Open Map $F:X \to Y$. Then $F(\mathcal{B}) := {F(B) \subset Y : B \in \mathcal{B}}$ is a Topological Basis for $Y$.

First note $F(\mathcal{B})$ is a collection of Open sets because $F$ is an Open Map. Let $U \subset Y$ be Open. Because $F$ is continuous, $F^{-1}(U)$ is Open. Thus there exists $\mathcal{B}{U} \subset \mathcal{B}$ so that $\bigcup\limits{B \in \mathcal{B}{U}}B = F^{-1}(U)$. Then $\bigcup\limits{B \in \mathcal{B}_{U}}F(B) = U$. Since $U$ was arbitrary, $F(\mathcal{B})$ is a Topological Basis for $Y$. $\blacksquare$