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Denoting $Q_E:B\to B/E$ the quotient map, $Q_EB$$Q_EA$ is surjective. Then there exists $r>0$ such that $Q_E(B-\lambda)$$Q_E(A-\lambda)$ is surjective, hence $B=(A-\lambda)(B)+E$, for $|\lambda|<r$.
Denoting $Q_E:B\to B/E$ the quotient map, $Q_EB$ is surjective. Then there exists $r>0$ such that $Q_E(B-\lambda)$ is surjective, hence $B=(A-\lambda)(B)+E$, for $|\lambda|<r$.
Denoting $Q_E:B\to B/E$ the quotient map, $Q_EA$ is surjective. Then there exists $r>0$ such that $Q_E(A-\lambda)$ is surjective, hence $B=(A-\lambda)(B)+E$, for $|\lambda|<r$.
Denoting $Q_E:B\to B/E$ the quotient map, $Q_EB$ is surjective. Then there exists $r>0$ such that $Q_E(B-\lambda)$ is surjective, hence $B=(A-\lambda)(B)+E$, for $|\lambda|<r$.
