The Stacks project

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. $\square$

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. $\square$

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. $\square$

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 $Y$. $\square$

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). $\square$

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). $\square$

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. $\square$

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. $\square$

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

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. $\square$

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. $\square$