Lemma 35.13.6. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a fpqc covering. Suppose that for each $i$ we have an open subset $W_ i \subset T_ i$ such that for all $i, j \in I$ we have $\text{pr}_0^{-1}(W_ i) = \text{pr}_1^{-1}(W_ j)$ as open subsets of $T_ i \times _ T T_ j$. Then there exists a unique open subset $W \subset T$ such that $W_ i = f_ i^{-1}(W)$ for each $i$.

Proof. Apply Lemma 35.13.1 to the map $\coprod _{i \in I} T_ i \to T$. It implies there exists a subset $W \subset T$ such that $W_ i = f_ i^{-1}(W)$ for each $i$, namely $W = \bigcup f_ i(W_ i)$. To see that $W$ is open we may work Zariski locally on $T$. Hence we may assume that $T$ is affine. Using Topologies, Definition 34.9.1 we may choose a standard fpqc covering $\{ g_ j : V_ j \to T\} _{j \in J}$ which refines $\{ T_ i \to T\} _{i \in I}$. Let $\alpha : J \to I$ and $h_ j : V_ j \to T_{\alpha (j)}$ be as in Sites, Definition 7.8.1. Then $g_ j^{-1}(W) = h_ j^{-1}(W_{\alpha (j)})$. Thus we may assume $\{ f_ i : T_ i \to T\}$ is a standard fpqc covering. In this case we may apply Morphisms, Lemma 29.25.12 to the morphism $\coprod T_ i \to T$ to conclude that $W$ is open. $\square$

Comment #7530 by WhatJiaranEatsTonight on

Could you please tell me how to reduce it to the case that $\{f_i:T_i\to T\}$ is a standard fpqc covering?

Comment #7533 by on

In the proof we only do this reduction when we already have $W$ as a subset and $T$ is affine. By Definition 34.9.1 the covering can be refined by a standard one, say $\{V_j \to T\}$ and $V_j$ maps to $T_{i(j)}$ over $T$. Then the pullback of $W$ to $V_j$ is also the pullback of $W_{i(j)}$ and hence open. Maybe this should be added?

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