Lemma 5.5.6. Let $X$ be a topological space. Let $\mathcal{B}$ be a collection of opens of $X$. Assume $X = \bigcup _{U \in \mathcal{B}} U$ and for $U, V \in \mathcal{B}$ we have $U \cap V = \bigcup _{W \in \mathcal{B}, W \subset U \cap V} W$. Then there is a continuous map $f : X \to Y$ of topological spaces such that

for $U \in \mathcal{B}$ the image $f(U)$ is open,

for $U \in \mathcal{B}$ we have $f^{-1}(f(U)) = U$, and

the opens $f(U)$, $U \in \mathcal{B}$ form a basis for the topology on $Y$.

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