The Stacks project

[Page 68 ff, EGA1-second]

Lemma 5.8.16. Let $X$ be a topological space. There is a canonical continuous map

\[ c : X \longrightarrow X' \]

from $X$ to a sober topological space $X'$ which is universal among continuous maps from $X$ to sober topological spaces. Moreover, the assignment $U' \mapsto c^{-1}(U')$ is a bijection between opens of $X'$ and $X$ which commutes with finite intersections and arbitrary unions. The image $c(X)$ is a Kolmogorov topological space and the map $c : X \to c(X)$ is universal for maps of $X$ into Kolmogorov spaces.

Proof. Let $X'$ be the set of irreducible closed subsets of $X$ and let

\[ c : X \to X', \quad x \mapsto \overline{\{ x\} }. \]

For $U \subset X$ open, let $U' \subset X'$ denote the set of irreducible closed subsets of $X$ which meet $U$. Then $c^{-1}(U') = U$. In particular, if $U_1 \not= U_2$ are open in $X$, then $U'_1 \not= U_2'$. Hence $c$ induces a bijection between the subsets of $X'$ of the form $U'$ and the opens of $X$.

Let $U_1, U_2$ be open in $X$. Suppose that $Z \in U'_1$ and $Z \in U'_2$. Then $Z \cap U_1$ and $Z \cap U_2$ are nonempty open subsets of the irreducible space $Z$ and hence $Z \cap U_1 \cap U_2$ is nonempty. Thus $(U_1 \cap U_2)' = U'_1 \cap U'_2$. The rule $U \mapsto U'$ is also compatible with arbitrary unions (details omitted). Thus it is clear that the collection of $U'$ form a topology on $X'$ and that we have a bijection as stated in the lemma.

Next we show that $X'$ is sober. Let $T \subset X'$ be an irreducible closed subset. Let $U \subset X$ be the open such that $X' \setminus T = U'$. Then $Z = X \setminus U$ is irreducible because of the properties of the bijection of the lemma. We claim that $Z \in T$ is the unique generic point. Namely, any open of the form $V' \subset X'$ which does not contain $Z$ must come from an open $V \subset X$ which misses $Z$, i.e., is contained in $U$.

Finally, we check the universal property. Let $f : X \to Y$ be a continuous map to a sober topological space. Then we let $f' : X' \to Y$ be the map which sends the irreducible closed $Z \subset X$ to the unique generic point of $\overline{f(Z)}$. It follows immediately that $f' \circ c = f$ as maps of sets, and the properties of $c$ imply that $f'$ is continuous. We omit the verification that the continuous map $f'$ is unique. We also omit the proof of the statements on Kolmogorov spaces. $\square$

Comments (4)

Comment #8247 by Elías Guisado on

For another reference, everything except the last part is also proven in EGA, 1971 ed., (2.9.2).

Comment #8892 by on

Sorry, but I could not find this in EGA part I. Can you be more precise with the reference?

Comment #8904 by on

@#8892 I mean the new edition of EGA I published by Springer in 1971, Chapter 0, Proposition (2.9.2), p. 68. The statement that is a bijection between open sets of and is in the same page, (

There are also:

  • 8 comment(s) on Section 5.8: Irreducible components

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 0A2N. Beware of the difference between the letter 'O' and the digit '0'.