Lemma 87.5.5. Let $A \to B$ and $B \to C$ be continuous homomorphisms of linearly topologized rings. If $A \to C$ is taut, then $B \to C$ is taut.
Proof. Let $J \subset B$ be an open ideal with inverse image $I \subset A$. Then the closure of $JC$ contains the closure of $IC$. Hence this closure is open as $A \to C$ is taut. Let $I_\lambda $ be a fundamental system of open ideals of $A$. Let $K_\lambda $ be the closure of $I_\lambda C$. Since $A \to C$ is taut, these form a fundamental system of open ideals of $C$. Denote $J_\lambda \subset B$ the inverse image of $K_\lambda $. Then the closure of $J_\lambda C$ is $K_\lambda $. Hence we see that the closures of the ideals $JC$, where $J$ runs over the open ideals of $B$ form a fundamental system of open ideals of $C$. $\square$
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