Lemma 86.9.10. 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 86.9.4 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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