Lemma 26.19.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

1. $f : X \to S$ is quasi-compact,

2. the inverse image of every affine open is quasi-compact, and

3. there exists some affine open covering $S = \bigcup _{i \in I} U_ i$ such that $f^{-1}(U_ i)$ is quasi-compact for all $i$.

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. $\square$

Comment #667 by Anfang Zhou on

Typo. It should be "a covering $S = \bigcup_{i \in I} U_i$" in the first line of the proof.

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