To show that $A \to B$ is surjective, we view it as a map of $A$-modules and apply Lemma 10.19.1 (6). We conclude it suffices to show that $A/\mathfrak m_ A \to B/\mathfrak m_ AB$ is surjective. As $A/\mathfrak m_ A = B/\mathfrak m_ B$ it suffices to show that $\mathfrak m_ AB \to \mathfrak m_ B$ is surjective. View $\mathfrak m_ AB \to \mathfrak m_ B$ as a map of $B$-modules and apply Lemma 10.19.1 (6). We conclude it suffices to see that $\mathfrak m_ AB/\mathfrak m_ A\mathfrak m_ B \to \mathfrak m_ B/\mathfrak m_ B^2$ is surjective. This follows from assumption (4).
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