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. $\square$

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