Lemma 87.27.5. Let $S$ be a scheme. Let $X$ be a McQuillan affine formal algebraic space over $S$. Let $f : Y \to X$ be a closed immersion of formal algebraic spaces over $S$. Then $Y$ is a McQuillan affine formal algebraic space and $f$ corresponds to a continuous homomorphism $A \to B$ of weakly admissible topological $S$-algebras which is taut, has closed kernel, and has dense image.
Proof. Write $X = \text{Spf}(A)$ where $A$ is a weakly admissible topological ring. Let $I_\lambda $ be a fundamental system of weakly admissible ideals of definition in $A$. Then $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )$ is a closed subscheme of $\mathop{\mathrm{Spec}}(A/I_\lambda )$ and hence affine (Definition 87.27.1). Say $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$. The ring map $A/I_\lambda \to B_\lambda $ is surjective. Hence the projections
\[ B = \mathop{\mathrm{lim}}\nolimits B_\lambda \longrightarrow B_\lambda \]
are surjective as the compositions $A \to B \to B_\lambda $ are surjective. It follows that $Y$ is McQuillan by Lemma 87.9.6. The ring map $A \to B$ is taut by Lemma 87.19.8. The kernel is closed because $B$ is complete and $A \to B$ is continuous. Finally, as $A \to B_\lambda $ is surjective for all $\lambda $ we see that the image of $A$ in $B$ is dense. $\square$
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