The Stacks project

Proof. We will use without further mention that $X_0$ is quasi-compact (Topology, Lemma 5.12.9) and $\pi _0(X)$ is profinite (Topology, Lemma 5.23.9). Picture

If (2) holds, the continuous bijective map $f : X_0 \to \pi _0(X)$ is a homeomorphism by Topology, Lemma 5.17.8. Given an open covering $X = \bigcup U_ i$, we get an open covering $\pi _0(X) = \bigcup f(X_0 \cap U_ i)$. By Topology, Lemma 5.22.4 we can find a finite open covering of the form $\pi _0(X) = \coprod V_ j$ which refines this covering. Since $X_0 \to \pi _0(X)$ is bijective each connected component of $X$ has a unique closed point, whence is equal to the set of points specializing to this closed point. Hence $\pi ^{-1}(V_ j)$ is the set of points specializing to the points of $f^{-1}(V_ j)$. Now, if $f^{-1}(V_ j) \subset X_0 \cap U_ i \subset U_ i$, then it follows that $\pi ^{-1}(V_ j) \subset U_ i$ (because the open set $U_ i$ is closed under generalizations). In this way we see that the open covering $X = \coprod \pi ^{-1}(V_ j)$ refines the covering we started out with. In this way we see that (2) implies (1).

Assume (1). Let $x, y \in X$ be closed points. Then we have the open covering $X = (X \setminus \{ x\} ) \cup (X \setminus \{ y\} )$. It follows from (1) that there exists a disjoint union decomposition $X = U \amalg V$ with $U$ and $V$ open (and closed) and $x \in U$ and $y \in V$. In particular we see that every connected component of $X$ has at most one closed point. By Topology, Lemma 5.12.8 every connected component (being closed) also does have a closed point. Thus $X_0 \to \pi _0(X)$ is bijective. In this way we see that (1) implies (2).

Assume $X_0$ is closed in $X$ and every point specializes to a unique point of $X_0$. Then $X_0$ is a spectral space (Topology, Lemma 5.23.5) consisting of closed points, hence profinite (Topology, Lemma 5.23.8). Let $x, y \in X_0$ be distinct. By Topology, Lemma 5.22.4 we can find a disjoint union decomposition $X_0 = U_0 \amalg V_0$ with $U_0$ and $V_0$ open and closed and $x \in U_0$ and $y \in V_0$. Let $U \subset X$, resp. $V \subset X$ be the set of points specializing to $U_0$, resp. $V_0$. Observe that $X = U \amalg V$. By Topology, Lemma 5.24.7 we see that $U$ is an intersection of quasi-compact open subsets. Hence $U$ is closed in the constructible topology. Since $U$ is closed under specialization, we see that $U$ is closed by Topology, Lemma 5.23.6. By symmetry $V$ is closed and hence $U$ and $V$ are both open and closed. This proves that $x, y$ are not in the same connected component of $X$. In other words, $X_0 \to \pi _0(X)$ is injective. The map is also surjective by Topology, Lemma 5.12.8 and the fact that connected components are closed. In this way we see that the final condition implies (2). $\square$

Proof. The subset $Y_0 \subset Y$ of closed points is closed because $Y_0 = X_0 \cap Y$. Since $X$ is $w$-local, every $y \in Y$ specializes to a unique point of $X_0$. This specialization is in $Y$, and hence also in $Y_0$, because $\overline{\{ y\} }\subset Y$. In conclusion, $Y$ is $w$-local. $\square$

Proof. Note that $Y$ is a closed subspace of $X \times T$ as $\pi _0(X)$ is a profinite space hence Hausdorff (use Topology, Lemmas 5.23.9 and 5.3.4). Since $X \times T$ is spectral (Topology, Lemma 5.23.10) it follows that $Y$ is spectral (Topology, Lemma 5.23.5). Let $Y \to \pi _0(Y) \to T$ be the canonical factorization (Topology, Lemma 5.7.9). It is clear that $\pi _0(Y) \to T$ is surjective. The fibres of $Y \to T$ are homeomorphic to the fibres of $X \to \pi _0(X)$. Hence these fibres are connected. It follows that $\pi _0(Y) \to T$ is injective. We conclude that $\pi _0(Y) \to T$ is a homeomorphism by Topology, Lemma 5.17.8.

Next, assume that $X$ is w-local and let $X_0 \subset X$ be the set of closed points. The inverse image $Y_0 \subset Y$ of $X_0$ in $Y$ maps bijectively onto $T$ as $X_0 \to \pi _0(X)$ is a bijection by Lemma 61.2.1. Moreover, $Y_0$ is quasi-compact as a closed subset of the spectral space $Y$. Hence $Y_0 \to \pi _0(Y) = T$ is a homeomorphism by Topology, Lemma 5.17.8. It follows that all points of $Y_0$ are closed in $Y$. Conversely, if $y \in Y$ is a closed point, then it is closed in the fibre of $Y \to \pi _0(Y) = T$ and hence its image $x$ in $X$ is closed in the (homeomorphic) fibre of $X \to \pi _0(X)$. This implies $x \in X_0$ and hence $y \in Y_0$. Thus $Y_0$ is the collection of closed points of $Y$ and for each $y \in Y_0$ the set of generalizations of $y$ is the fibre of $Y \to \pi _0(Y)$. The lemma follows. $\square$