The Stacks project

Proof. Let $F$ be a closed subset of $X$, if $y\in F$ then $\{ y\} \subset F$, so $\overline{\{ y\} } \subset \overline{F} = F$ as $F$ is closed. Thus for all specialization $x$ of $y$, we have $x\in F$.

Let $x, y\in X$ such that $x\in \overline{\{ y\} }$ and let $T$ be a subset of $X$. Saying that $T$ is stable under specialization means that $y\in T$ implies $x\in T$ and reciprocally saying that $T$ is stable under generalization means that $x\in T$ implies $y\in T$. Therefore (3) is proven using contraposition.

The second property follows from (1) and (3) by considering the complement. $\square$

Proof. Suppose that $T$ is stable under specialization, then for all $y\in T$ we have $\overline{\{ y\} } \subset T$. Thus $T = \bigcup _{y\in T} \overline{\{ y\} }$ which is an union of closed subsets of $X$. Reciprocally, suppose that $T = \bigcup _{i\in I}F_ i$ where $F_ i$ are closed subsets of $X$. If $y\in T$ then there exists $i\in I$ such that $y\in F_ i$. As $F_ i$ is closed, we have $\overline{\{ y\} } \subset F_ i \subset T$, which proves that $T$ is stable under specialization. $\square$

Proof. Let $z'\leadsto z$ be a specialization in $Z$ and let $x' \in X$ such as $g\circ f (x') = z'$. Then because specializations lift along $g$, there exists a specialization $f(x') \leadsto y$ of $f(x')$ in $Y$ such that $g(y) = z$. Likewise, because specializations lift along $f$, there exists a specialization $x' \leadsto x$ of $x'$ in $X$ such that $f(x) = y$. It provides a specialization $x' \leadsto x$ of $x'$ in $X$ such that $g\circ f(x) = z$. In other words, specialization lift along $g\circ f$. $\square$

Proof. Let $y' \leadsto y$ be a specialization in $Y$ where $y'\in f(T)$ and let $x'\in T$ such that $f(x') = y'$. Because specialization lift along $f$, there exists a specialization $x'\leadsto x$ of $x'$ in $X$ such that $f(x) = y$. But $T$ is stable under specialization so $x\in T$ and then $y \in f(T)$. Therefore $f(T)$ is stable under specialization.

The proof of (2) is identical, using that generalizations lift along $f$. $\square$

Proof. Assume $f$ is closed. Let $y' \leadsto y$ in $Y$ and any $x'\in X$ with $f(x') = y'$ be given. Consider the closed subset $T = \overline{\{ x'\} }$ of $X$. Then $f(T) \subset Y$ is a closed subset, and $y' \in f(T)$. Hence also $y \in f(T)$. Hence $y = f(x)$ with $x \in T$, i.e., $x' \leadsto x$.

Assume $f$ is open, $X$ Noetherian, every irreducible closed subset of $X$ has a generic point, and $Y$ is Kolmogorov. Let $y' \leadsto y$ in $Y$ and any $x \in X$ with $f(x) = y$ be given. Consider $T = f^{-1}(\{ y'\} ) \subset X$. Take an open neighbourhood $x \in U \subset X$ of $x$. Then $f(U) \subset Y$ is open and $y \in f(U)$. Hence also $y' \in f(U)$. In other words, $T \cap U \not= \emptyset$. This proves that $x \in \overline{T}$. Since $X$ is Noetherian, $T$ is Noetherian (Lemma 5.9.2). Hence it has a decomposition $T = T_1 \cup \ldots \cup T_ n$ into irreducible components. Then correspondingly $\overline{T} = \overline{T_1} \cup \ldots \cup \overline{T_ n}$. By the above $x \in \overline{T_ i}$ for some $i$. By assumption there exists a generic point $x' \in \overline{T_ i}$, and we see that $x' \leadsto x$. As $x' \in \overline{T}$ we see that $f(x') \in \overline{\{ y'\} }$. Note that $f(\overline{T_ i}) = f(\overline{\{ x'\} }) \subset \overline{\{ f(x')\} }$. If $f(x') \not= y'$, then since $Y$ is Kolmogorov $f(x')$ is not a generic point of the irreducible closed subset $\overline{\{ y'\} }$ and the inclusion $\overline{\{ f(x')\} } \subset \overline{\{ y'\} }$ is strict, i.e., $y' \not\in f(\overline{T_ i})$. This contradicts the fact that $f(T_ i) = \{ y'\}$. Hence $f(x') = y'$ and we win. $\square$

Proof. Properties (3) and (5) imply that a point $x$ corresponds to an finite equivalence class $\{ u_1, \ldots , u_ n\} \subset U$ of the equivalence relation. Suppose that $x' \in X$ is a second point corresponding to the equivalence class $\{ u'_1, \ldots , u'_ m\} \subset U$. Suppose that $u_ i \leadsto u'_ j$ for some $i, j$. Then for any $r' \in R$ with $s(r') = u'_ j$ by (4) we can find $r \leadsto r'$ with $s(r) = u_ i$. Hence $t(r) \leadsto t(r')$. Since $\{ u'_1, \ldots , u'_ m\} = t(s^{-1}(\{ u'_ j\} ))$ we conclude that every element of $\{ u'_1, \ldots , u'_ m\}$ is the specialization of an element of $\{ u_1, \ldots , u_ n\}$. Thus $\overline{\{ u_1\} } \cup \ldots \cup \overline{\{ u_ n\} }$ is a union of equivalence classes, hence of the form $\pi ^{-1}(Z)$ for some subset $Z \subset X$. By (1) we see that $Z$ is closed in $X$ and in fact $Z = \overline{\{ x\} }$ because $\pi (\overline{\{ u_ i\} }) \subset \overline{\{ x\} }$ for each $i$. In other words, $x \leadsto x'$ if and only if some lift of $x$ in $U$ specializes to some lift of $x'$ in $U$, if and only if every lift of $x'$ in $U$ is a specialization of some lift of $x$ in $U$.

Suppose that both $x \leadsto x'$ and $x' \leadsto x$. Say $x$ corresponds to $\{ u_1, \ldots , u_ n\}$ and $x'$ corresponds to $\{ u'_1, \ldots , u'_ m\}$ as above. Then, by the results of the preceding paragraph, we can find a sequence

which must repeat, hence by (2) we conclude that $\{ u_1, \ldots , u_ n\} = \{ u'_1, \ldots , u'_ m\}$, i.e., $x = x'$. Thus $X$ is Kolmogorov. $\square$

Proof. Assume specializations lift along $f$. Let $Z_0 \subset Z_1 \subset \ldots Z_ e \subset Y$ be a chain of irreducible closed subsets of $X$. Let $\xi _ e \in X$ be a point mapping to the generic point of $Z_ e$. By assumption there exists a specialization $\xi _ e \leadsto \xi _{e - 1}$ in $X$ such that $\xi _{e - 1}$ maps to the generic point of $Z_{e - 1}$. Continuing in this manner we find a sequence of specializations

with $\xi _ i$ mapping to the generic point of $Z_ i$. This clearly implies the sequence of irreducible closed subsets

is a chain of length $e$ in $X$. The case when generalizations lift along $f$ is similar. $\square$

Proof. Let $E$ be constructible and stable under generalization. Let $Y \subset X$ be an irreducible closed subset with generic point $\xi \in Y$. If $E \cap Y$ is nonempty, then it contains $\xi$ (by stability under generalization) and hence is dense in $Y$, hence it contains a nonempty open of $Y$, see Lemma 5.16.3. Thus $E$ is open by Lemma 5.16.5. This proves (2). To prove (1) apply (2) to the complement of $E$ in $X$. $\square$