The Stacks project

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. $\square$