Lemma 87.9.10. Let S be a scheme. Let A, B be weakly admissible topological rings over S. Any morphism f : \text{Spf}(B) \to \text{Spf}(A) of affine formal algebraic spaces over S is equal to \text{Spf}(f^\sharp ) for a unique continuous S-algebra map f^\sharp : A \to B.
Proof. Let f : \text{Spf}(B) \to \text{Spf}(A) be as in the lemma. Let J \subset B be a weak ideal of definition. By Lemma 87.9.4 there exists a weak ideal of definition I \subset A such that \mathop{\mathrm{Spec}}(B/J) \to \text{Spf}(B) \to \text{Spf}(A) factors through \mathop{\mathrm{Spec}}(A/I). By Schemes, Lemma 26.6.4 we obtain an S-algebra map A/I \to B/J. These maps are compatible for varying J and define the map f^\sharp : A \to B. This map is continuous because for every weak ideal of definition J \subset B there is a weak ideal of definition I \subset A such that f^\sharp (I) \subset J. The equality f = \text{Spf}(f^\sharp ) holds by our choice of the ring maps A/I \to B/J which make up f^\sharp . \square
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