The Stacks project

Lemma 61.2.1. Let $X$ be a spectral space. Let $X_0 \subset X$ be the set of closed points. The following are equivalent

  1. Every open covering of $X$ can be refined by a finite disjoint union decomposition $X = \coprod U_ i$ with $U_ i$ open and closed in $X$.

  2. The composition $X_0 \to X \to \pi _0(X)$ is bijective.

Moreover, if $X_0$ is closed in $X$ and every point of $X$ specializes to a unique point of $X_0$, then these conditions are satisfied.

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

\[ \xymatrix{ X_0 \ar[rd]_ f \ar[r] & X \ar[d]^\pi \\ & \pi _0(X) } \]

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). $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 61.2: Some topology

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0968. Beware of the difference between the letter 'O' and the digit '0'.