Proof. Let $a, b : U_1 \to U_2$ be two morphisms between schemes étale over $S$. Assume the base changes of $a$ and $b$ to $X$ agree. We have to show that $a = b$. By Proposition 41.6.3 it suffices to show that $a$ and $b$ agree on points and residue fields. This is clear because for every $u \in U_1$ we can find a point $v \in X \times _ S U_1$ mapping to $u$. $\square$
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