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.
$\square$

## 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Your email address will not be published.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.

## Comments (3)

Comment #7530 by WhatJiaranEatsTonight on

Comment #7533 by Johan on

Comment #7660 by Stacks Project on

There are also: