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

  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. $\square$


Comments (0)

There are also:

  • 6 comment(s) on Section 5.5: Bases

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0D5Q. Beware of the difference between the letter 'O' and the digit '0'.