Lemma 26.23.7. Let $j : X \to Y$ be a morphism of schemes. If
$j$ is injective on points, and
for any $x \in X$ the ring map $j^\sharp _ x : \mathcal{O}_{Y, j(x)} \to \mathcal{O}_{X, x}$ is surjective,
then $j$ is a monomorphism.
Proof. Let $a, b : Z \to X$ be two morphisms of schemes such that $j \circ a = j \circ b$. Then (1) implies $a = b$ as underlying maps of topological spaces. For any $z \in Z$ we have $a^\sharp _ z \circ j^\sharp _{a(z)} = b^\sharp _ z \circ j^\sharp _{b(z)}$ as maps $\mathcal{O}_{Y, j(a(z))} \to \mathcal{O}_{Z, z}$. The surjectivity of the maps $j^\sharp _ x$ forces $a^\sharp _ z = b^\sharp _ z$, $\forall z \in Z$. This implies that $a^\sharp = b^\sharp $. Hence we conclude $a = b$ as morphisms of schemes as desired. $\square$
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