Lemma 3.9.11. Let $f : X \to Y$ be a morphism of schemes. Assume there exists an fpqc covering $\{ g_ j : Y_ j \to Y\} _{j \in J}$ such that $g_ j$ factors through $f$. Then $\text{size}(Y) \leq \text{size}(X)$.

Proof. Let $V \subset Y$ be an affine open. By definition there exist $n \geq 0$ and $a : \{ 1, \ldots , n\} \to J$ and affine opens $V_ i \subset Y_{a(i)}$ such that $V = g_{a(1)}(V_1) \cup \ldots \cup g_{a(n)}(V_ n)$. Denote $h_ j : Y_ j \to X$ a morphism such that $f \circ h_ j = g_ j$. Then $h_{a(1)}(V_1) \cup \ldots \cup h_{a(n)}(V_ n)$ is a quasi-compact subset of $f^{-1}(V)$. Hence we can find a quasi-compact open $W \subset f^{-1}(V)$ which contains $h_{a(i)}(V_ i)$ for $i = 1, \ldots , n$. In particular $V = f(W)$.

On the one hand this shows that the cardinality of the set of affine opens of $Y$ is at most the cardinality of the set $S$ of quasi-compact opens of $X$. Since every quasi-compact open of $X$ is a finite union of affines, we see that the cardinality of this set is at most $\sup |S|^ n = \max (\aleph _0, |S|)$. On the other hand, we have $\mathcal{O}_ Y(V) \subset \prod _{i = 1, \ldots , n} \mathcal{O}_{Y_{a(i)}}(V_ i)$ because $\{ V_ i \to V\}$ is an fpqc covering. Hence $\mathcal{O}_ Y(V) \subset \mathcal{O}_ X(W)$ because $V_ i \to V$ factors through $W$. Again since $W$ has a finite covering by affine opens of $X$ we conclude that $|\mathcal{O}_ Y(V)|$ is bounded by the size of $X$. The lemma now follows from the definition of the size of a scheme. $\square$

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