Lemma 29.7.10. Let $S$ be a scheme. Let $X$, $Y$ be schemes over $S$. Let $f, g : X \to Y$ be morphisms of schemes over $S$. Let $U \subset X$ be an open subscheme such that $f|_ U = g|_ U$. If the scheme theoretic closure of $U$ in $X$ is $X$ and $Y \to S$ is separated, then $f = g$.

**Proof.**
Follows from the definitions and Schemes, Lemma 26.21.5.
$\square$

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