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In his paper Etale cohomology for non-Archimedean analytic space (IHES), Berkovich explained how to glue $k$-analytic spaces (Prop. 1.3.3) and show its uniqueness using the Prop 1.3.2 (gluing morphism).

In Prop 1.3.3, it has two cases that we can glue the $k$-analytic spaces $X_i$ along $X_{ij}\subset X_i$.

(a) all $X_{ij}$ are open in $X_i$ (b) all $_{ij}$ are closed in $X_i$ and for each $i$, $X_{ij}$ is non empty for finitely many $j$.

Q1. Do we need, in (b) of Prop 1.3.3, the condition that $X_{ij}$ are closed in $X_i$? I'm not sure where he used this condition in the construction.

Prop 1.3.2 says if $X$ is $k$-analytic space with covering $\{X_i\}$ by analytic domains that is quasi-net (i.e. for each $x\in X$ there are $X_{i_1},\cdots,X_{i_n}$ such that $x\in \cap X_{i_k}$ and $\cup X_{i_k}$ is neighborhood at $x$), then we can glue the morphisms from $X_i$ to $Y$ that are compatible as the morphism from $X$ to $Y$. Here the quasi-net condition is necessary for example to make sure the continuity of glued morphism. He used this to prove the uniqueness of glued analytic spaces.

Q2. When we glue $\{X_i\}$ as in Prop.1.3.3, does the covering $\{X_i\}$ of $X$ (glued analytic space) becomes quasi-net?(so that we can apply the Prop 1.3.2)

I think this can fail when the $X_{ij}$ are not open, as the quotient map $\prod X_i \to X$ is not a open map.

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  • $\begingroup$ If $X_{ij}$'s are not closed then the quotient topology might be weird, so that $X$ might not be even Hausdorff. You need some sort of compatibility between topologies of $X_{i}$. For your second question in the case of 1.3.3(b), indeed it forms a quasinet, because $x\in X$ can be contained in only finitely many $X_{i}$'s. Namely pick one $i$ such that $x\in X_{i}$, then any $x\in X_{j}$ would mean $x\in X_{ij}$ so there are possibly only finitely many $X_{j}$'s that can contain $x$. It is a neighborhood because $\cup \left(X_{j}-\cup X_{jk}\right)\subset\cup X_{j}$ is open. $\endgroup$
    – GTA
    Commented Feb 19, 2020 at 4:45
  • $\begingroup$ I think the last question on the relation with adic spaces should be asked separately. $\endgroup$
    – Jakob Werner
    Commented Feb 19, 2020 at 7:44
  • $\begingroup$ @GTA I guess you mean X_j - union of X_jk where X_jk's do not contain x and X_jk are closed so X_j - union of X_jk is open and take union of them? I think this is correct and gives me more sense why we require X_ij being closed. $\endgroup$
    – quasi-mathematician
    Commented Feb 19, 2020 at 19:42
  • $\begingroup$ @JakobWerner I see. I removed it now $\endgroup$
    – quasi-mathematician
    Commented Feb 19, 2020 at 19:43

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