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

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

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

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

Proof. Define an equivalence relation $\sim$ on points of $X$ by the rule

$x \sim y \Leftrightarrow (\forall U \in \mathcal{B} : x \in U \Leftrightarrow y \in U)$

Let $Y$ be the set of equivalence classes and $f : X \to Y$ the natural map. Part (2) holds by construction. The assumptions on $\mathcal{B}$ exactly mirror the assumptions in Lemma 5.5.2 on the set of subsets $f(U)$, $U \in \mathcal{B}$. Hence there is a unique topology on $Y$ such that (3) holds. Then (1) is clear as well. $\square$

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