Suppose that $\varphi : A \to B$ is a continuous map between adic topological rings. One says $\varphi$ is adic if there exists an ideal of definition $I \subset A$ such that the topology on $B$ is $I$-adic. However, this is not a good notion unless we assume $A$ has a finitely generated ideal of definition. In this case, the condition is equivalent to $\varphi$ being taut, see Lemma 85.4.15.

Let $P$ be the property of morphisms $\varphi : A \to B$ of $\textit{WAdm}^{adic*}$ defined by

$P(\varphi )=\text{}\varphi \text{ is adic''}=\text{}\varphi \text{ is taut''}$

(see above for the equivalence). Since $\textit{WAdm}^{adic*}$ is a full subcategory of $\textit{WAdm}^{count}$ it follows trivially from Lemma 85.16.6 that $P$ is a local property on morphisms of $\textit{WAdm}^{adic*}$, see Remark 85.16.4. Combining Lemmas 85.16.3 and 85.16.7 we obtain the result stated in the next paragraph.

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of locally adic* formal algebraic spaces over $S$. Then the following are equivalent

1. $f$ is representable by algebraic spaces (in other words, the equivalent conditions of Lemma 85.14.4 hold),

2. for every commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y }$

with $U$ and $V$ affine formal algebraic spaces, $U \to X$ and $V \to Y$ representable by algebraic spaces and étale, the morphism $U \to V$ corresponds to an adic map in $\textit{WAdm}^{adic*}$ (in other words, the equivalent conditions of Lemma 85.16.3 hold with $P$ as above).

In this situation we will sometimes say that $f$ is an adic morphism. Here it is understood that this notion is only defined for morphisms between formal algebraic spaces which are locally adic*.

Definition 85.17.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$. Assume $X$ and $Y$ are locally adic*. We say $f$ is an adic morphism if $f$ is representable by algebraic spaces. See discussion above.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).