The Stacks project

Proof. Suppose we are given a covering $S = \bigcup _{i \in I} U_ i$ as in (3). First, let $U \subset S$ be any affine open. For any $u \in U$ we can find an index $i(u) \in I$ such that $u \in U_{i(u)}$. As standard opens form a basis for the topology on $U_{i(u)}$ we can find $W_ u \subset U \cap U_{i(u)}$ which is standard open in $U_{i(u)}$. By compactness we can find finitely many points $u_1, \ldots , u_ n \in U$ such that $U = \bigcup _{j = 1}^ n W_{u_ j}$. For each $j$ write $f^{-1}U_{i(u_ j)} = \bigcup _{k \in K_ j} V_{jk}$ as a finite union of affine opens. Since $W_{u_ j} \subset U_{i(u_ j)}$ is a standard open we see that $f^{-1}(W_{u_ j}) \cap V_{jk}$ is a standard open of $V_{jk}$, see Algebra, Lemma 10.17.4. Hence $f^{-1}(W_{u_ j}) \cap V_{jk}$ is affine, and so $f^{-1}(W_{u_ j})$ is a finite union of affines. This proves that the inverse image of any affine open is a finite union of affine opens.

Next, assume that the inverse image of every affine open is a finite union of affine opens. Let $K \subset S$ be any quasi-compact open. Since $S$ has a basis of the topology consisting of affine opens we see that $K$ is a finite union of affine opens. Hence the inverse image of $K$ is a finite union of affine opens. Hence $f$ is quasi-compact.

Finally, assume that $f$ is quasi-compact. In this case the argument of the previous paragraph shows that the inverse image of any affine is a finite union of affine opens. $\square$

Proof. Omitted. $\square$

Proof. This follows from the definitions and Topology, Lemma 5.12.2. $\square$

Proof. Follows from the definitions and Topology, Lemma 5.12.3. $\square$

Proof. We have (1) $\Rightarrow$ (2) by Topology, Lemma 5.19.2. Assume (2). Let $U \subset S$ be an affine open. It suffices to prove that $f(X) \cap U$ is closed. Since $U \cap f(X)$ is stable under specializations in $U$, we have reduced to the case where $S$ is affine. Because $f$ is quasi-compact we deduce that $X = f^{-1}(S)$ is quasi-compact as $S$ is affine. Thus we may write $X = \bigcup _{i = 1}^ n U_ i$ with $U_ i \subset X$ open affine. Say $S = \mathop{\mathrm{Spec}}(R)$ and $U_ i = \mathop{\mathrm{Spec}}(A_ i)$ for some $R$-algebra $A_ i$. Then $f(X) = \mathop{\mathrm{Im}}(\mathop{\mathrm{Spec}}(A_1 \times \ldots \times A_ n) \to \mathop{\mathrm{Spec}}(R))$. Thus the lemma follows from Algebra, Lemma 10.41.5. $\square$

Proof. According to Topology, Lemma 5.19.7 if $f$ is closed then specializations lift along $f$. Conversely, suppose that specializations lift along $f$. Let $Z \subset X$ be a closed subset. We may think of $Z$ as a scheme with the reduced induced scheme structure, see Definition 26.12.5. Since $Z \subset X$ is closed the restriction of $f$ to $Z$ is still quasi-compact. Moreover specializations lift along $Z \to S$ as well, see Topology, Lemma 5.19.5. Hence it suffices to prove $f(X)$ is closed if specializations lift along $f$. In particular $f(X)$ is stable under specializations, see Topology, Lemma 5.19.6. Thus $f(X)$ is closed by Lemma 26.19.7. $\square$