Lemma 5.5.6 (0D5Q)—The Stacks project

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 $Y$ .

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. $\square$

The Stacks project

Comments (0)

Post a comment