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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