The Stacks project

5.8 Irreducible components

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

  1. 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 $X = Z_2$.

  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.

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

  2. Any irreducible component of $X$ is closed.

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

  4. 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. $\square$

Lemma 5.8.4. Let $X$ be a topological space and suppose $X = \bigcup _{i = 1, \ldots , n} X_ i$ where each $X_ i$ is an irreducible closed subset of $X$ and no $X_ i$ is contained in the union of the other members. Then each $X_ i$ is an irreducible component of $X$ and each irreducible component of $X$ is one of the $X_ i$.

Proof. Let $Y$ be an irreducible component of $X$. Write $Y = \bigcup _{i = 1, \ldots , n} (Y \cap X_ i)$ and note that each $Y \cap X_ i$ is closed in $Y$ since $X_ i$ is closed in $X$. By irreducibility of $Y$ we see that $Y$ is equal to one of the $Y \cap X_ i$, i.e., $Y \subset X_ i$. By maximality of irreducible components we get $Y = X_ i$.

Conversely, take one of the $X_ i$ and expand it to an irreducible component $Y$, which we have already shown is one of the $X_ j$. So $X_ i \subset X_ j$ and since the original union does not have redundant members, $X_ i = X_ j$, which is an irreducible component. $\square$

Lemma 5.8.5. Let $f : X \to Y$ be a surjective, continuous map of topological spaces. If $X$ has a finite number, say $n$, of irreducible components, then $Y$ has $\leq n$ irreducible components.

Proof. Say $X_1, \ldots , X_ n$ are the irreducible components of $X$. By Lemmas 5.8.2 and 5.8.3 the closure $Y_ i \subset Y$ of $f(X_ i)$ is irreducible. Since $f$ is surjective, we see that $Y$ is the union of the $Y_ i$. We may choose a minimal subset $I \subset \{ 1, \ldots , n\} $ such that $Y = \bigcup _{i \in I} Y_ i$. Then we may apply Lemma 5.8.4 to see that the $Y_ i$ for $i \in I$ are the irreducible components of $Y$. $\square$

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 5.8.6. Let $X$ be a topological space.

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

  2. 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.

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

  4. 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 5.8.7. Let $X$ be a topological space and let $Y\subset X$.

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

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

  3. 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 5.8.8. Let $X$ be a topological space and let $(X_ i)_{i\in I}$ be a covering of $X$.

  1. 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 $i\in I$.

  2. 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 $i\in I$.

  3. 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 5.8.7. 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 5.8.7. 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 5.8.9. 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 5.8.10. 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 5.8.11. Let $X$ and $Y$ be as in Example 5.8.9 and Example 5.8.10. Then, $X\amalg Y$ is neither Kolmogorov nor quasi-sober.

Example 5.8.12. 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 5.8.13. 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 5.8.14. 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 5.8.15. 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 5.8.14. Hence we win. $\square$

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

reference

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. $\square$

Lemma 5.8.17. Let $X$ be a connected topological space with a finite number of irreducible components $X_1, \ldots , X_ n$. If $n > 1$ there is an $1 \leq j \leq n$ such that $X' = \bigcup _{i \not= j} X_ i$ is connected.

Proof. This is a graph theory problem. Let $\Gamma $ be the graph with vertices $V = \{ 1, \ldots , n\} $ and an edge between $i$ and $j$ if and only if $X_ i \cap X_ j$ is nonempty. Connectedness of $X$ means that $\Gamma $ is connected. Our problem is to find $1 \leq j \leq n$ such that $\Gamma \setminus \{ j\} $ is still connected. You can do this by choosing $j, j' \in E$ with maximal distance and then $j$ works (choose a leaf!). Details omitted. $\square$


Comments (9)

Comment #1691 by Zeshen Gu on

Hi,

Possibly there is a typo:

In proof of Lemma 5.7.3, "suppose that T′=Z1∪Z2 is a union of two closed subset of T", should the last be replaced by ?

Comment #3379 by Kazuki Masugi on

In the proof of the Lemma5.8.5(1), "x,y \in X" should be "x,y \in Y"

Comment #3380 by Kazuki Masugi on

And in the proof of the Lemma5.8.5(2), "Hence there exists with . It follows ." should be "Hence there exists with . It follows ."

Comment #4296 by Goodluckthere on

Possible typo: In the proof of the third part of lemma 004.W you take to be a set of irreducible sets but later you consider to be a set of indices of elements in .

Comment #4361 by Victor on

Possible tiny typo. In the aforementioned proof of the Lemma5.8.5(2) it should actually be "Hence there exists x with " instead of "Hence there exists x with ". Nothing else need to be changed. ( will in fact be in but we'd need to establish first)

Comment #4461 by on

@Goodluckthere: do not understand your comment. Sorry.

@Victor: Very good, thanks. Fixed here.

Comment #9537 by George on

To expand on Goodluckthere's comment (which I believe to be correct), in 004W we introduce "the set of irreducible subsets (implicitly with in some indexing set ) with ", and we go on to say consistent with this. But then we introduce a partial order on as if it's the set of the indices (not subsets) corresponding to and talk about instead of and so on. So we have switched to treating as the set of irreducible subsets rather than the implicit "", say. Really there's no need to introduce a partial order on the indices, you can just order the set of subsets by inclusion to accomplish the same thing.

Perhaps a minor point but I also picked up on it while reading.


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