Lemma 32.4.9. In Situation 32.4.5. Suppose we are given an $i$ and a morphism $T \to S_ i$ such that

1. $T \times _{S_ i} S = \emptyset$, and

2. $T$ is quasi-compact.

Then $T \times _{S_ i} S_{i'} = \emptyset$ for all sufficiently large $i'$.

Proof. By Lemma 32.2.3 we see that $T \times _{S_ i} S = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} T \times _{S_ i} S_{i'}$. Hence the result follows from Lemma 32.4.3. $\square$

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).