Lemma 87.5.9. Let $\varphi : A \to B$ be a continuous homomorphism of linearly topologized rings. If $\varphi $ is taut and $A$ is pre-admissible, then $B$ is pre-admissible.
Proof. Let $I \subset A$ be an ideal of definition. Let $I_\lambda \subset A$ be a fundamental system of open ideals. Then the closure $J$ of $IB$ is open and the closures $J_\lambda $ of $I_\lambda B$ are open and form a fundamental system of open ideals of $B$. For every $\lambda $ there is an $n$ such that $I^ n \subset I_\lambda $. Observe that $J^ n$ is contained in the closure of $I^ nB$. Thus $J^ n \subset J_\lambda $ and we conclude $J$ is an ideal of definition. $\square$
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