## 5.8 Irreducible components

Definition 5.8.1. Let $X$ be a topological space.

We say $X$ is *irreducible*, if $X$ is not empty, and whenever $X = Z_1 \cup Z_2$ with $Z_ i$ closed, we have $X = Z_1$ or $X = Z_2$.

We say $Z \subset X$ is an *irreducible component* of $X$ if $Z$ is a maximal irreducible subset of $X$.

An irreducible space is obviously connected.

Lemma 5.8.2. Let $f : X \to Y$ be a continuous map of topological spaces. If $E \subset X$ is an irreducible subset, then $f(E) \subset Y$ is irreducible as well.

**Proof.**
Clearly we may assume $E = X$ (i.e., $X$ irreducible) and $f(E) = Y$ (i.e., $f$ surjective). First, $Y \not= \emptyset $ since $X \not= \emptyset $. Next, assume $Y = Y_1 \cup Y_2$ with $Y_1$, $Y_2$ closed. Then $X = X_1 \cup X_2$ with $X_ i = f^{-1}(Y_ i)$ closed in $X$. By assumption on $X$, we must have $X = X_1$ or $X = X_2$, hence $Y = Y_1$ or $Y = Y_2$ since $f$ is surjective.
$\square$

Lemma 5.8.3. Let $X$ be a topological space.

If $T \subset X$ is irreducible so is its closure in $X$.

Any irreducible component of $X$ is closed.

Any irreducible subset of $X$ is contained in an irreducible component of $X$.

Every point of $X$ is contained in some irreducible component of $X$, in other words, $X$ is the union of its irreducible components.

**Proof.**
Let $\overline{T}$ be the closure of the irreducible subset $T$. If $\overline{T} = Z_1 \cup Z_2$ with $Z_ i \subset \overline{T}$ closed, then $T = (T\cap Z_1) \cup (T \cap Z_2)$ and hence $T$ equals one of the two, say $T = Z_1 \cap T$. Thus clearly $\overline{T} \subset Z_1$. This proves (1). Part (2) follows immediately from (1) and the definition of irreducible components.

Let $T \subset X$ be irreducible. Consider the set $A$ of irreducible subsets $T \subset T_\alpha \subset X$. Note that $A$ is nonempty since $T \in A$. There is a partial ordering on $A$ coming from inclusion: $\alpha \leq \alpha ' \Leftrightarrow T_\alpha \subset T_{\alpha '}$. Choose a maximal totally ordered subset $A' \subset A$, and let $T' = \bigcup _{\alpha \in A'} T_\alpha $. We claim that $T'$ is irreducible. Namely, suppose that $T' = Z_1 \cup Z_2$ is a union of two closed subsets of $T'$. For each $\alpha \in A'$ we have either $T_\alpha \subset Z_1$ or $T_\alpha \subset Z_2$, by irreducibility of $T_\alpha $. Suppose that for some $\alpha _0 \in A'$ we have $T_{\alpha _0} \not\subset Z_1$ (say, if not we're done anyway). Then, since $A'$ is totally ordered we see immediately that $T_\alpha \subset Z_2$ for all $\alpha \in A'$. Hence $T' = Z_2$. This proves (3). Part (4) is an immediate consequence of (3) as a singleton space is irreducible.
$\square$

A singleton is irreducible. Thus if $x \in X$ is a point then the closure $\overline{\{ x\} }$ is an irreducible closed subset of $X$.

Definition 5.8.4. Let $X$ be a topological space.

Let $Z \subset X$ be an irreducible closed subset. A *generic point* of $Z$ is a point $\xi \in Z$ such that $Z = \overline{\{ \xi \} }$.

The space $X$ is called *Kolmogorov*, if for every $x, x' \in X$, $x \not= x'$ there exists a closed subset of $X$ which contains exactly one of the two points.

The space $X$ is called *quasi-sober* if every irreducible closed subset has a generic point.

The space $X$ is called *sober* if every irreducible closed subset has a unique generic point.

A topological space $X$ is Kolmogorov, quasi-sober, resp. sober if and only if the map $x\mapsto \overline{\{ x\} }$ from $X$ to the set of irreducible closed subsets of $X$ is injective, surjective, resp. bijective. Hence we see that a topological space is sober if and only if it is quasi-sober and Kolmogorov.

Lemma 5.8.5. Let $X$ be a topological space and let $Y\subset X$.

If $X$ is Kolmogorov then so is $Y$.

Suppose $Y$ is locally closed in $X$. If $X$ is quasi-sober then so is $Y$.

Suppose $Y$ is locally closed in $X$. If $X$ is sober then so is $Y$.

**Proof.**
Proof of (1). Suppose $X$ is Kolmogorov. Let $x,y\in Y$ with $x\neq y$. Then $\overline{\overline{\{ x\} }\cap Y}=\overline{\{ x\} }\neq \overline{\{ y\} }= \overline{\overline{\{ y\} }\cap Y}$. Hence $\overline{\{ x\} }\cap Y\neq \overline{\{ y\} }\cap Y$. This shows that $Y$ is Kolmogorov.

Proof of (2). Suppose $X$ is quasi-sober. It suffices to consider the cases $Y$ is closed and $Y$ is open. First, suppose $Y$ is closed. Let $Z$ be an irreducible closed subset of $Y$. Then $Z$ is an irreducible closed subset of $X$. Hence there exists $x \in Z$ with $\overline{\{ x\} } = Z$. It follows $\overline{\{ x\} } \cap Y = Z$. This shows $Y$ is quasi-sober. Second, suppose $Y$ is open. Let $Z$ be an irreducible closed subset of $Y$. Then $\overline{Z}$ is an irreducible closed subset of $X$. Hence there exists $x \in \overline{Z}$ with $\overline{\{ x\} }=\overline{Z}$. If $x\notin Y$ we get the contradiction $Z=Z\cap Y\subset \overline{Z}\cap Y=\overline{\{ x\} }\cap Y=\emptyset $. Therefore $x\in Y$. It follows $Z=\overline{Z}\cap Y=\overline{\{ x\} }\cap Y$. This shows $Y$ is quasi-sober.

Proof of (3). Immediately from (1) and (2).
$\square$

Lemma 5.8.6. Let $X$ be a topological space and let $(X_ i)_{i\in I}$ be a covering of $X$.

Suppose $X_ i$ is locally closed in $X$ for every $i\in I$. Then, $X$ is Kolmogorov if and only if $X_ i$ is Kolmogorov for every $i\in I$.

Suppose $X_ i$ is open in $X$ for every $i\in I$. Then, $X$ is quasi-sober if and only if $X_ i$ is quasi-sober for every $i\in I$.

Suppose $X_ i$ is open in $X$ for every $i\in I$. Then, $X$ is sober if and only if $X_ i$ is sober for every $i\in I$.

**Proof.**
Proof of (1). If $X$ is Kolmogorov then so is $X_ i$ for every $i\in I$ by Lemma 5.8.5. Suppose $X_ i$ is Kolmogorov for every $i\in I$. Let $x,y\in X$ with $\overline{\{ x\} }=\overline{\{ y\} }$. There exists $i\in I$ with $x\in X_ i$. There exists an open subset $U\subset X$ such that $X_ i$ is a closed subset of $U$. If $y\notin U$ we get the contradiction $x\in \overline{\{ x\} }\cap U=\overline{\{ y\} }\cap U=\emptyset $. Hence $y\in U$. It follows $y\in \overline{\{ y\} }\cap U=\overline{\{ x\} }\cap U\subset X_ i$. This shows $y\in X_ i$. It follows $\overline{\{ x\} }\cap X_ i=\overline{\{ y\} }\cap X_ i$. Since $X_ i$ is Kolmogorov we get $x=y$. This shows $X$ is Kolmogorov.

Proof of (2). If $X$ is quasi-sober then so is $X_ i$ for every $i\in I$ by Lemma 5.8.5. Suppose $X_ i$ is quasi-sober for every $i\in I$. Let $Y$ be an irreducible closed subset of $X$. As $Y\neq \emptyset $ there exists $i\in I$ with $X_ i\cap Y\neq \emptyset $. As $X_ i$ is open in $X$ it follows $X_ i\cap Y$ is non-empty and open in $Y$, hence irreducible and dense in $Y$. Thus $X_ i\cap Y$ is an irreducible closed subset of $X_ i$. As $X_ i$ is quasi-sober there exists $x\in X_ i\cap Y$ with $X_ i\cap Y=\overline{\{ x\} }\cap X_ i\subset \overline{\{ x\} }$. Since $X_ i\cap Y$ is dense in $Y$ and $Y$ is closed in $X$ it follows $Y=\overline{X_ i\cap Y}\cap Y\subset \overline{X_ i\cap Y}\subset \overline{\{ x\} }\subset Y$. Therefore $Y=\overline{\{ x\} }$. This shows $X$ is quasi-sober.

Proof of (3). Immediately from (1) and (2).
$\square$

Example 5.8.7. Let $X$ be an indiscrete space of cardinality at least $2$. Then $X$ is quasi-sober but not Kolmogorov. Moreover, the family of its singletons is a covering of $X$ by discrete and hence Kolmogorov spaces.

Example 5.8.8. Let $Y$ be an infinite set, furnished with the topology whose closed sets are $Y$ and the finite subsets of $Y$. Then $Y$ is Kolmogorov but not quasi-sober. However, the family of its singletons (which are its irreducible components) is a covering by discrete and hence sober spaces.

Example 5.8.9. Let $X$ and $Y$ be as in Example 5.8.7 and Example 5.8.8. Then, $X\amalg Y$ is neither Kolmogorov nor quasi-sober.

Example 5.8.10. Let $Z$ be an infinite set and let $z\in Z$. We furnish $Z$ with the topology whose closed sets are $Z$ and the finite subsets of $Z\setminus \{ z\} $. Then $Z$ is sober but its subspace $Z\setminus \{ z\} $ is not quasi-sober.

Example 5.8.11. Recall that a topological space $X$ is Hausdorff iff for every distinct pair of points $x, y \in X$ there exist disjoint opens $U, V \subset X$ such that $x \in U$, $y \in V$. In this case $X$ is irreducible if and only if $X$ is a singleton. Similarly, any subset of $X$ is irreducible if and only if it is a singleton. Hence a Hausdorff space is sober.

Lemma 5.8.12. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that (a) $Y$ is irreducible, (b) $f$ is open, and (c) there exists a dense collection of points $y \in Y$ such that $f^{-1}(y)$ is irreducible. Then $X$ is irreducible.

**Proof.**
Suppose $X = Z_1 \cup Z_2$ with $Z_ i$ closed. Consider the open sets $U_1 = Z_1 \setminus Z_2 = X \setminus Z_2$ and $U_2 = Z_2 \setminus Z_1 = X \setminus Z_1$. To get a contradiction assume that $U_1$ and $U_2$ are both nonempty. By (b) we see that $f(U_ i)$ is open. By (a) we have $Y$ irreducible and hence $f(U_1) \cap f(U_2) \not= \emptyset $. By (c) there is a point $y$ which corresponds to a point of this intersection such that the fibre $X_ y = f^{-1}(y)$ is irreducible. Then $X_ y \cap U_1$ and $X_ y \cap U_2$ are nonempty disjoint open subsets of $X_ y$ which is a contradiction.
$\square$

Lemma 5.8.13. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that (a) $f$ is open, and (b) for every $y \in Y$ the fibre $f^{-1}(y)$ is irreducible. Then $f$ induces a bijection between irreducible components.

**Proof.**
We point out that assumption (b) implies that $f$ is surjective (see Definition 5.8.1). Let $T \subset Y$ be an irreducible component. Note that $T$ is closed, see Lemma 5.8.3. The lemma follows if we show that $f^{-1}(T)$ is irreducible because any irreducible subset of $X$ maps into an irreducible component of $Y$ by Lemma 5.8.2. Note that $f^{-1}(T) \to T$ satisfies the assumptions of Lemma 5.8.12. Hence we win.
$\square$

The construction of the following lemma is sometimes called the “soberification”.

Lemma 5.8.14. 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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