Remark 98.27.3. In Situation 98.27.1 let $V$ be a locally Noetherian scheme over $S$. Let $(Z, u', \hat x)$ be a triple satisfying (1), (2), and (3) above. We want to explain a way to think about the compatibility condition (4). It will not be mathematically precise as we are going use a fictitious category $\textit{An}_ S$ of analytic spaces over $S$ and a fictitious analytification functor

For example if $Y = \text{Spf}(k[[t]])$ over $S = \mathop{\mathrm{Spec}}(k)$, then $Y^{an}$ should be thought of as an open unit disc. If $Y = \mathop{\mathrm{Spec}}(k)$, then $Y^{an}$ is a single point. The category $\textit{An}_ S$ should have open and closed immersions and we should be able to take the open complement of a closed. Given $Y$ the morphism $Y_{red} \to Y$ should induces a closed immersion $Y_{red}^{an} \to Y^{an}$. We set $Y^{rig} = Y^{an} \setminus Y_{red}^{an}$ equal to its open complement. If $Y$ is an algebraic space and if $Z \subset Y$ is closed, then the morphism $Y_{/Z} \to Y$ should induce an open immersion $Y_{/Z}^{an} \to Y^{an}$ which in turn should induce an open immersion

Also, given a formal modification $g : Y' \to Y$ of locally Noetherian formal algebraic spaces, the induced morphism $g^{rig} : (Y')^{rig} \to Y^{rig}$ should be an isomorphism. Given $\text{An}_ S$ and the analytification functor, we can consider the requirement that

commutes. This makes sense as $g^{rig} : (X'_{T'})^{rig} \to W^{rig}$ is an isomorphism and $U' = X' \setminus T'$. Finally, under some assumptions of faithfulness of the analytification functor, this requirement will be equivalent to the compatibility condition formulated above. We hope this will motivate the reader to think of the compatibility of $u'$ and $\hat x$ as the requirement that some maps be equal, rather than asking for the existence of a certain commutative diagram.

## Comments (0)