Definition (HHK).

Let $K$ be a field. Let $X$ be a $K$ -variety and $G$ a linear algebraic group over $K$ . We say that $G$ acts strongly transitively on $X$ if $G$ acts on $X$ and, for any field extension $L/K$ , either $X(L)=\emptyset$ or $G(L)$ acts transitively on $X(L)$ .

Definition.

Let $K$ be a field.

1. (1) [Kaplansky] The $u$ -invariant of $K$ , denoted by $u(K)$ , is the maximal dimension of anisotropic quadratic forms over $K$ . We say that $u(K)=\infty$ if there exist anisotropic quadratic forms over $K$ of arbitrarily large dimension.

2. (2) [HHK] The strong $u$ -invariant of $K$ , denoted by $u_{s}(K)$ , is the smallest real number  $m$ such that:

   • ∙ $u(E)\leqslant m$ for all finite field extensions $E/K$ ;

   • ∙ $\frac{1}{2}u(E)\leqslant m$ for all finitely generated field extensions $E/K$ of transcendence degree 1.

   We say that $u_{s}(K)=\infty$ if there exist such field extensions $E$ of arbitrarily large $u$ -invariant.

Setting 1. Let $R_{i},i=1,2$ , be an integral domain endowed with a non-archimedean submultiplicative norm $|\cdot |_{R_{i}}$ , with respect to which it is complete. Set $F_{i}=\operatorname{Frac}R_{i},i=1,2$ . Let $F$ be an infinite field embedded in both $F_{1}$ and $F_{2}$ . Let $F_{0}$ be a complete ultrametric field with non-trivial valuation such that there exist bounded morphisms  $R_{i}{\hookrightarrow}F_{0}$ , $i=1,2$ . Suppose that the image of  $F_{1}$ is dense in  $F_{0}$ . Let  $A_{i}$ be an  $R_{i}$ -module such that  $A_{i}\subseteq F_{i}$ . Suppose that  $A_{i}$ is finitely generated as an  $R_{i}$ -module, i.e. that there exists a surjective  $R_{i}$ -linear morphism  $\unicode[STIX]{x1D711}_{i}:R_{i}^{n_{i}}{\twoheadrightarrow}A_{i}$ for some positive integer  $n_{i},i=1,2$ . Let us endow  $A_{i}$ with the quotient semi-norm induced from  $\unicode[STIX]{x1D711}_{i}$ . Assume that  $A_{i}$ is complete and the morphism  $A_{i}{\hookrightarrow}F_{0}$ is bounded for  $i=1,2$ . We remark that this implies that the semi-norm on $A_{i}$ is a norm. Suppose that the induced map  $\unicode[STIX]{x1D70B}:A_{1}\oplus A_{2}\rightarrow F_{0}$ is surjective. Finally, suppose that the norm of  $F_{0}$ is equivalent to the quotient norm induced by the surjective morphism  $\unicode[STIX]{x1D70B}:A_{1}\oplus A_{2}{\twoheadrightarrow}F_{0}$ , where  $A_{1}\oplus A_{2}$ is endowed with the usual max semi-norm  $|\cdot |_{\max }$ .

Proof. Since ${\mathcal{O}}(X)$ is an integral domain, $\operatorname{Frac}{\mathcal{O}}(X)\subseteq \mathscr{M}(X)$ by the definition of  $\mathscr{M}$ . Let $f\in \mathscr{M}(X)$ . The sheaf $f{\mathcal{O}}\,\cap \,{\mathcal{O}}\subseteq \mathscr{M}$ is non-zero and coherent, so, by Kiehl’s theorem, it has a non-zero global section $x$ . Then there exists $y\in {\mathcal{O}}(X)\backslash \{0\}$ for which $f=x/y\in \operatorname{Frac}{\mathcal{O}}(X)$ .◻

Proof. Let us start by remarking that the set of poles of a meromorphic function is a divisor and as such consists of only rigid points. This implies that a meromorphic function cannot have a pole on any non-rigid point (including $\unicode[STIX]{x1D702}$ ), which is why it makes sense to evaluate it at $\unicode[STIX]{x1D702}$ .

That $\{\unicode[STIX]{x1D702}\}$ is an affinoid domain of $U$ (respectively $V$ ) can be checked directly from the definition of an affinoid domain. By the Gerritzen–Grauert theorem (see [Reference TemkinTem05]), we obtain that it is a rational domain in $U$ (respectively $V$ ). Then, by the easy implication of [Reference BerkovichBer90, Corollary 2.2.10], the meromorphic functions on $U$ (respectively $V$ ) with no poles in $\{\unicode[STIX]{x1D702}\}$ are dense in ${\mathcal{O}}(\{\unicode[STIX]{x1D702}\})$ . Seeing as $\unicode[STIX]{x1D702}$ is a type 3 point, ${\mathcal{O}}(\{\unicode[STIX]{x1D702}\})=\mathscr{M}(\{\unicode[STIX]{x1D702}\})={\mathcal{H}}(\unicode[STIX]{x1D702})$ : the completed residue field of  $\unicode[STIX]{x1D702}$ . Finally, this implies that the image with respect to the restriction morphism of $\mathscr{M}(U)$ (respectively $\mathscr{M}(V)$ ) in $\mathscr{M}(\{\unicode[STIX]{x1D702}\})$ is dense.◻

Proof. By construction, the rings $R_{U},R_{V},F_{W}$ are all complete with respect to non-archimedean semi-norms. Since ${\mathcal{O}}(W)={\mathcal{H}}(\unicode[STIX]{x1D702})$ , the normed ring $F_{W}$ is a complete ultrametric non-trivially valued field. As $U,V$ , and $W$ are integral, by Lemma 1.2,  $\mathscr{M}(U)=F_{U}$ ,  $\mathscr{M}(V)=F_{V}$ , and $\mathscr{M}(W)=F_{W}$ . This shows the existence of embeddings of $F_{C}$ into $F_{U},F_{V}$ , and $F_{W}$ . The restriction morphisms $R_{U},R_{V}\rightarrow F_{W}$ are bounded by construction. From Lemma 1.3, $F_{U},F_{V}$ have dense images in $F_{W}$ .

Notice that for $Z\in \{U,V,W\}$ , ${\mathcal{O}}(Z){\hookrightarrow}{\mathcal{O}}(D)(Z){\hookrightarrow}\mathscr{M}(Z)$ . In particular, this means that ${\mathcal{O}}(D)(W)={\mathcal{O}}(W)=\mathscr{M}(W)$ . Since ${\mathcal{O}}(D)$ is a coherent sheaf, $A_{U}$ (respectively  $A_{V}$ ) is a finitely generated $R_{U}$ -module (respectively $R_{V}$ -module). The completeness of  $A_{U}$ (respectively $A_{V}$ ) follows from the fact that ideals of affinoid algebras are closed. The morphism  ${\mathcal{O}}(D)(U)=A_{U}{\hookrightarrow}F_{W}={\mathcal{O}}(D)(W)$ is the restriction morphism of the sheaf ${\mathcal{O}}(D)$ , so it is bounded. The same is true for $A_{V}{\hookrightarrow}F_{W}$ .

If $U\,\cup \,V$ is not the entire $C$ , it is an affinoid domain thereof (see [Reference DucrosDuc, Théorème 6.1.3]). By Tate’s acyclicity theorem [Reference BerkovichBer90, ch. 2, Proposition 2.2.5],

$$\begin{eqnarray}0\rightarrow H^{0}(U\,\cup \,V,{\mathcal{O}}(D))\rightarrow H^{0}(U,{\mathcal{O}}(D))\oplus H^{0}(V,{\mathcal{O}}(D))\rightarrow H^{0}(U\,\cap \,V,{\mathcal{O}}(D))\rightarrow 0\end{eqnarray}$$

is an exact admissible sequence, from which we obtain the surjective admissible morphism  $A_{U}\oplus A_{V}{\twoheadrightarrow}{\mathcal{O}}(D)(W)=F_{W}$ .

Suppose that $U\,\cup \,V=C$ . Since $C$ is then compact and integral, by [Reference DucrosDuc, Théorèmes 6.1.3 and 3.7.2], it is either an affinoid domain (a case we dealt with in the paragraph above) or a projective curve. If $C$ is projective, by [Reference LiuLiu02, § 7.5, Proposition 5.5] for large enough $n$ , $H^{1}(U\,\cup \,V,{\mathcal{O}}(D))=0$ . The Mayer–Vietoris exact sequence now produces a bounded surjective morphism  $A_{U}\oplus A_{V}{\twoheadrightarrow}{\mathcal{O}}(D)(W)=F_{W}$ . Admissibility follows from Banach’s open mapping theorem if $k$ is not non-trivially valued (for a proof, see [Reference BourbakiBou53]) and by a change of basis followed by the open mapping theorem if it is (see [Reference BerkovichBer90, ch. 2, Proposition 2.1.2(ii)]).◻

Proof. (1) See the paragraph above the statement.

(2) Suppose that $c\neq 0$ . Let $D$ be any real number such that $D>1$ . For any $c\in F_{0}$ , there exist $a\in A_{1},b\in A_{2}$ (depending on $D$ ) such that $\unicode[STIX]{x1D70B}(a+b)=c$ and $\max (|a|_{A_{1}},|b|_{A_{2}})\leqslant D\cdot |c|_{F_{0}}$ . Otherwise, for any $x\in A_{1},y\in A_{2}$ for which $\unicode[STIX]{x1D70B}(x+y)=c$ , one would have  $|x\,+\,y|_{\max }=\max (|x|_{A_{1}},|y|_{A_{2}})>D\cdot |c|_{F_{0}}$ . Then

$$\begin{eqnarray}|c|_{F_{0}}=\inf _{\substack{ x\in A_{1},y\in A_{2} \\ \unicode[STIX]{x1D70B}(x+y)=c}}|x+y|_{\max }\geqslant D\cdot |c|_{F_{0}},\end{eqnarray}$$

which is impossible if $c\neq 0$ . Thus, there exist $a$ and $b$ as above and, for  $d=D^{-1}\in (0,1)$ , one obtains $d\cdot \max (|a|_{A_{1}},|b|_{A_{2}})\leqslant |c|_{F_{0}}$ .

If $c=0$ , the statement is true regardless of the choice of $d$ .◻

Proof. The rational function $f$ can be written as $(f_{1},\ldots ,f_{n})$ , where the $f_{i}$ are elements of $F_{0}[T_{1},\ldots ,T_{n},S_{1},\ldots ,S_{n}]_{(T_{1},\ldots ,T_{n},S_{1},\ldots ,S_{n})}$ . Furthermore, since $f_{i}(0,0)=0$ , they belong to the maximal ideal of this ring. Lemmas 2.1 and 2.3 of [Reference Harbater, Hartmann and KrashenHHK09] remain true in our setting without any changes to their proofs (this is where the condition $f(x,0)=f(0,x)=x$ is crucial). They tell us that:

1. (1) we can see these rational functions as elements of $F_{0}[[T_{1},\ldots ,T_{n},S_{1},\ldots ,S_{n}]]$ ;

2. (2) there exists $M\geqslant 1$ such that

   $$\begin{eqnarray}f_{i}=S_{i}+T_{i}+\mathop{\sum }_{|(l,m)|\geqslant 2}c_{l,m}^{i}\text{}\underline{T}^{l}\text{}\underline{S}^{m}\in F_{0}[[T_{1},\ldots ,T_{n},S_{1},\ldots ,S_{n}]],\end{eqnarray}$$
   with $|c_{l,m}^{i}|_{F_{0}}\leqslant M^{|(l,m)|}$ for $i=1,2,\ldots ,n$ and $(l,m)\in \mathbb{N}^{2n}$ , where $|(l,m)|$ is the sum of the coordinates of $(l,m)$ . (We remark that since $f_{i}(x,0)=f_{i}(0,x)=x$ for any $x$ for which $(0,x),(x,0)\in \widetilde{S}$ , we can even assume that $l,m$ are both non-zero.)

Since $\widetilde{S}$ is open, $\widetilde{S}(F_{0})$ is Zariski open in $\mathbb{A}^{2n}(F_{0})$ and so it is open in $F_{0}^{2n}$ in the topology induced by the max norm (which is finer than the Zariski one). Seeing as $0\in \widetilde{S}(F_{0})$ , there exists $\unicode[STIX]{x1D6FF}>0$ such that for any $(x,y)\in F_{0}^{2n}$ with $|(x,y)|_{F_{0}}<\unicode[STIX]{x1D6FF}$ , one has $(x,y)\in \widetilde{S}(F_{0})$ and  $f(x,y)$ is defined.

Let us fix the constant $d$ given by Lemma 1.8. Let $0<\unicode[STIX]{x1D700}^{\prime }\leqslant \min \{1/2M,d^{2}/M^{4},\unicode[STIX]{x1D6FF}/2\}$ . Set $\unicode[STIX]{x1D700}=d\unicode[STIX]{x1D700}^{\prime }$ . Since $\unicode[STIX]{x1D700}<\unicode[STIX]{x1D700}^{\prime }<\min (1/M,\unicode[STIX]{x1D6FF}/2),$ by [Reference Harbater, Hartmann and KrashenHHK09, Lemma 2.1], for any $(x,y)\in \widetilde{S}(F_{0})$ with $|(x,y)|_{F_{0}}\leqslant \unicode[STIX]{x1D700}^{\prime }$ , $f(x,y)$ is well defined and the series by which $f_{i}(x,y)$ is given is convergent in $F_{0}$ , $i=1,2,\ldots ,n$ .

Let $a=(a_{1},a_{2},\ldots ,a_{n})\in \mathbb{A}^{n}(F_{0})$ be such that $|a|_{F_{0}}\leqslant \unicode[STIX]{x1D700}$ . Let $u_{0}=0\in A_{1}^{n}$ and $v_{0}=0\in A_{2}^{n}$ . Using induction, one constructs sequences $(u_{s})_{s}$ in $A_{1}^{n}$ and $(v_{s})_{s}$ in $A_{2}^{n}$ such that the following conditions are satisfied:

1. (1) $|u_{s}|_{A_{1}},|v_{s}|_{A_{2}}\leqslant \unicode[STIX]{x1D700}^{\prime }$ for all $s\geqslant 0$ ;

2. (2) $|u_{s}-u_{s-1}|_{A_{1}},|v_{s}-v_{s-1}|_{A_{2}}\leqslant \unicode[STIX]{x1D700}^{\prime (s+1)/2}$ for all $s\geqslant 1$ ;

3. (3) $|f(u_{s},v_{s})-a|_{F_{0}}\leqslant d\unicode[STIX]{x1D700}^{\prime (s+2)/2}$ for all $s\geqslant 0$ .

The first terms $u_{0}$ and $v_{0}$ satisfy conditions (1) and (3). We notice that the first condition implies that $|(u_{s},v_{s})|_{F_{0}}\leqslant \unicode[STIX]{x1D700}^{\prime }$ , so $f(u_{s},v_{s})$ is well defined and $f_{i}(u_{s},v_{s})$ is convergent for $s\in \mathbb{N}$ and  $i=1,2,\ldots ,n$ . Suppose that for $j\geqslant 0$ , we have constructed $u_{j}$ and $v_{j}$ satisfying all three conditions above. Then $d_{j}:=a-f(u_{j},v_{j})\in F_{0}^{n}$ is well defined and $|d_{j}|_{F_{0}}\leqslant d\unicode[STIX]{x1D700}^{\prime (j+2)/2}$ . From Lemma 1.8, there exist $u_{j}^{\prime }\in A_{1}^{n}$ and $v_{j}^{\prime }\in A_{2}^{n}$ such that $d_{j}=u_{j}^{\prime }+v_{j}^{\prime }$ and $d\cdot \max (|u_{j}^{\prime }|_{A_{1}},|v_{j}^{\prime }|_{A_{2}})\leqslant |d_{j}|_{F_{0}}\leqslant d\unicode[STIX]{x1D700}^{\prime (j+2)/2}$ .

Set $u_{j+1}=u_{j}+u_{j}^{\prime }$ and $v_{j+1}=v_{j}+v_{j}^{\prime }$ . Then $|u_{j+1}|_{A_{1}}\leqslant \max (\unicode[STIX]{x1D700}^{\prime },\unicode[STIX]{x1D700}^{\prime (j+2)/2})=\unicode[STIX]{x1D700}^{\prime }$ and the same is true for $v_{j+1}$ . Also, $|u_{j+1}-u_{j}|_{A_{1}}=|u_{j}^{\prime }|_{A_{1}}\leqslant \unicode[STIX]{x1D700}^{\prime (j+2)/2}$ and, similarly, $|v_{j+1}-v_{j}|_{A_{2}}\leqslant \unicode[STIX]{x1D700}^{\prime (j+2)/2}$ .

For $r\in \mathbb{N}$ , $i\in \{1,2,\ldots ,r\}$ , and $\unicode[STIX]{x1D6FC}\in F_{0}^{r}$ , let $\unicode[STIX]{x1D6FC}_{i}$ be the $i$ th coordinate of $\unicode[STIX]{x1D6FC}$ . For  $p=(p_{1},p_{2},\ldots ,p_{r})\in \mathbb{N}^{r}$ , set $\unicode[STIX]{x1D6FC}^{p}:=\prod _{i=1}^{r}\unicode[STIX]{x1D6FC}_{i}^{p_{i}}$ . Then, for the third condition,

$$\begin{eqnarray}\displaystyle |f_{i}(u_{j+1},v_{j+1})-a_{i}|_{F_{0}} & = & \displaystyle \biggl|u_{j+1,i}+v_{j+1,i}-a_{i}+\mathop{\sum }_{|(l,m)|\geqslant 2}c_{l,m}^{i}u_{j+1}^{l}v_{j+1}^{m}\biggr|_{F_{0}}\nonumber\\ \displaystyle & = & \displaystyle \biggl|u_{j,i}+v_{j,i}+u_{j,i}^{\prime }+v_{j,i}^{\prime }-a_{i}+\mathop{\sum }_{|(l,m)|\geqslant 2}c_{l,m}^{i}u_{j+1}^{l}v_{j+1}^{m}\biggr|_{F_{0}}\nonumber\\ \displaystyle & = & \displaystyle \biggl|f_{i}(u_{j},v_{j})-a_{i}+u_{j,i}^{\prime }+v_{j,i}^{\prime }+\mathop{\sum }_{|(l,m)|\geqslant 2}c_{l,m}^{i}(u_{j+1}^{l}v_{j+1}^{m}-u_{j}^{l}v_{j}^{m})\biggr|_{F_{0}}\nonumber\\ \displaystyle & = & \displaystyle \biggl|-d_{j,i}+u_{j,i}^{\prime }+v_{j,i}^{\prime }+\mathop{\sum }_{|(l,m)|\geqslant 2}c_{l,m}^{i}(u_{j+1}^{l}v_{j+1}^{m}-u_{j}^{l}v_{j}^{m})\biggr|_{F_{0}}\nonumber\\ \displaystyle & = & \displaystyle \biggl|\mathop{\sum }_{|(l,m)|\geqslant 2}c_{l,m}^{i}(u_{j+1}^{l}v_{j+1}^{m}-u_{j}^{l}v_{j}^{m})\biggr|_{F_{0}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \max _{|(l,m)|\geqslant 2}|c_{l,m}^{i}|_{F_{0}}\cdot |u_{j+1}^{l}v_{j+1}^{m}-u_{j}^{l}v_{j}^{m}|_{F_{0}}.\nonumber\end{eqnarray}$$

On the other hand,

$$\begin{eqnarray}\displaystyle u_{j+1}^{l}v_{j+1}^{m}-u_{j}^{l}v_{j}^{m} & = & \displaystyle (u_{j}+u_{j}^{\prime })^{l}(v_{j}+v_{j}^{\prime })^{m}-u_{j}^{l}v_{j}^{m}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\substack{ 0\leqslant \unicode[STIX]{x1D6FD}\leqslant l \\ 0\leqslant \unicode[STIX]{x1D6FE}\leqslant m}}A_{\unicode[STIX]{x1D6FD}}B_{\unicode[STIX]{x1D6FE}}u_{j}^{\unicode[STIX]{x1D6FD}}u_{j}^{\prime l-\unicode[STIX]{x1D6FD}}v_{j}^{\unicode[STIX]{x1D6FE}}v_{j}^{\prime m-\unicode[STIX]{x1D6FE}}-u_{j}^{l}v_{j}^{m}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{0\leqslant \unicode[STIX]{x1D6FC}\leqslant (l,m)}\;\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FD}+\unicode[STIX]{x1D6FE}=\unicode[STIX]{x1D6FC} \\ 0\leqslant \unicode[STIX]{x1D6FD}\leqslant l \\ 0\leqslant \unicode[STIX]{x1D6FE}\leqslant m}}A_{\unicode[STIX]{x1D6FD}}B_{\unicode[STIX]{x1D6FE}}u_{j}^{\unicode[STIX]{x1D6FD}}u_{j}^{\prime l-\unicode[STIX]{x1D6FD}}v_{j}^{\unicode[STIX]{x1D6FE}}v_{j}^{\prime m-\unicode[STIX]{x1D6FE}}-u_{j}^{l}v_{j}^{m},\nonumber\end{eqnarray}$$

where $A_{\unicode[STIX]{x1D6FD}},B_{\unicode[STIX]{x1D6FE}}$ are integers (implying that they are of norm at most one on $F_{0}$ ). Thus, $u_{j+1}^{l}v_{j+1}^{m}-u_{j}^{m}v_{j}^{l}$ is a finite sum of monomials of degree $|(l,m)|$ in the variables $u_{j,i},u_{j,i}^{\prime },v_{j,i},v_{j,i}^{\prime },i=1,2,\ldots ,n$ , where the degree in $u_{j,i},v_{j,i}$ is strictly smaller than $|(l,m)|$ . Finally, since the norm is multiplicative and non-archimedean,

$$\begin{eqnarray}\displaystyle |u_{j+1}^{l}v_{j+1}^{m}-u_{j}^{l}v_{j}^{m}|_{F_{0}} & {\leqslant} & \displaystyle \max _{\substack{ 0\leqslant \unicode[STIX]{x1D6FD}+\unicode[STIX]{x1D6FE}<(l,m) \\ 0\leqslant \unicode[STIX]{x1D6FD}\leqslant l,0\leqslant \unicode[STIX]{x1D6FE}\leqslant m}}|u_{j}^{\unicode[STIX]{x1D6FD}}|_{F_{0}}|v_{j}^{\unicode[STIX]{x1D6FE}}|_{F_{0}}|u_{j}^{\prime l-\unicode[STIX]{x1D6FD}}|_{F_{0}}|v_{j}^{\prime m-\unicode[STIX]{x1D6FE}}|_{F_{0}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \max _{\substack{ 0\leqslant \unicode[STIX]{x1D6FD}+\unicode[STIX]{x1D6FE}<(l,m) \\ 0\leqslant \unicode[STIX]{x1D6FD}\leqslant l,0\leqslant \unicode[STIX]{x1D6FE}\leqslant m}}\unicode[STIX]{x1D700}^{\prime |(\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE})|}(\unicode[STIX]{x1D700}^{\prime (j+2)/2})^{|(l,m)|-|(\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE})|},\nonumber\end{eqnarray}$$

so $|u_{j+1}^{l}v_{j+1}^{m}-u_{j}^{l}v_{j}^{m}|_{F_{0}}\leqslant \max _{0\leqslant \unicode[STIX]{x1D703}<|(l,m)|}\unicode[STIX]{x1D700}^{\prime \unicode[STIX]{x1D703}}\cdot (\unicode[STIX]{x1D700}^{\prime (j+2)/2})^{|(l,m)|-\unicode[STIX]{x1D703}}$ . This, combined with

$$\begin{eqnarray}|c_{l,m}^{i}|_{F_{0}}\leqslant M^{|(l,m)|},\end{eqnarray}$$

implies that

$$\begin{eqnarray}\displaystyle |f_{i}(u_{j+1},v_{j+1})-a_{i}|_{F_{0}} & {\leqslant} & \displaystyle \max _{\substack{ |(l,m)|\geqslant 2 \\ 0\leqslant \unicode[STIX]{x1D703}<|(l,m)|}}M^{|(l,m)|}\unicode[STIX]{x1D700}^{\prime \unicode[STIX]{x1D703}}\cdot (\unicode[STIX]{x1D700}^{\prime (j+2)/2})^{|(l,m)|-\unicode[STIX]{x1D703}}\nonumber\\ \displaystyle & = & \displaystyle \max _{\substack{ |(l,m)|\geqslant 2 \\ 0\leqslant \unicode[STIX]{x1D703}<|(l,m)|}}(M\unicode[STIX]{x1D700}^{\prime })^{\unicode[STIX]{x1D703}}\cdot (M\unicode[STIX]{x1D700}^{\prime (j+2)/2})^{|(l,m)|-\unicode[STIX]{x1D703}}.\nonumber\end{eqnarray}$$

Since $\unicode[STIX]{x1D700}^{\prime }\geqslant \unicode[STIX]{x1D700}^{\prime (j+2)/2}$ and $M\unicode[STIX]{x1D700}^{\prime }<1$ , one obtains

$$\begin{eqnarray}|f_{i}(u_{j+1},v_{j+1})-a_{i}|_{F_{0}}\leqslant \max _{|(l,m)|\geqslant 2}(M\unicode[STIX]{x1D700}^{\prime })^{|(l,m)|-1}\cdot (M\unicode[STIX]{x1D700}^{\prime (j+2)/2})\leqslant M\unicode[STIX]{x1D700}^{\prime }\cdot M\unicode[STIX]{x1D700}^{\prime (j+2)/2}.\end{eqnarray}$$

At the same time, $M^{2}\cdot \unicode[STIX]{x1D700}^{\prime 1+((j+2)/2)}=((M^{2}/d)\unicode[STIX]{x1D700}^{\prime 1/2})\,d\unicode[STIX]{x1D700}^{\prime (j+3)/2}\leqslant d\unicode[STIX]{x1D700}^{\prime (j+3)/2}$ , which concludes the induction argument.

The second property of the sequences $(u_{s})_{s},(v_{s})_{s}$ tells us that they are Cauchy (hence convergent) in the complete spaces $A_{1}^{n},A_{2}^{n}$ , respectively. Let $u\in A_{1}^{n}$ and $v\in A_{2}^{n}$ be the corresponding limits. The first property implies that $|(u,v)|_{F_{0}}\leqslant \unicode[STIX]{x1D700}^{\prime }<\unicode[STIX]{x1D6FF}$ , so $(u,v)\in \widetilde{S}(F_{0})$ and $f(u,v)$ is well defined. Lastly, the third property implies that $f(u,v)=a$ .◻

Proof. We recall that the projective line $\mathbb{P}_{k}^{1,\text{an}}$ is uniquely path connected and can be obtained by adding a rational point $\infty$ to $\mathbb{A}_{k}^{1,\text{an}}$ . For any two points $a,b\in \mathbb{P}_{k}^{1,\text{an}}$ , we denote by $[a,b]$ the unique path connecting them.

Let $A$ be a connected component of $\mathbb{P}_{k}^{1,\text{an}}\backslash \{\unicode[STIX]{x1D702}\}$ that does not contain $\infty$ . In particular, $A\subseteq \mathbb{A}_{k}^{1,\text{an}}$ . Let $\unicode[STIX]{x1D702}_{0}$ be any rigid point of $A$ . There exists a unique irreducible polynomial $P\in k[T]$ such that $\unicode[STIX]{x1D702}_{0}=\unicode[STIX]{x1D702}_{P,0}$ . Then $\unicode[STIX]{x1D702}\in [\unicode[STIX]{x1D702}_{P,0},\infty ]$ .

Let $\unicode[STIX]{x1D711}$ be the finite morphism $\mathbb{P}_{k}^{1,\text{an}}\rightarrow \mathbb{P}_{k}^{1,\text{an}}$ determined by the map $k[T]\rightarrow k[T]$ , $T\mapsto P(T).$ Seeing as $\unicode[STIX]{x1D711}(\unicode[STIX]{x1D702}_{P,0})=\unicode[STIX]{x1D702}_{T,0}$ and $\unicode[STIX]{x1D711}(\infty )=\infty$ , $[\unicode[STIX]{x1D702}_{P,0},\infty ]$ is mapped by $\unicode[STIX]{x1D711}$ to $[\unicode[STIX]{x1D702}_{T,0},\infty ]$ . Set $\unicode[STIX]{x1D702}^{\prime }=\unicode[STIX]{x1D711}(\unicode[STIX]{x1D702})$ . The path connecting $\unicode[STIX]{x1D702}_{T,0}$ to $\infty$ in $\mathbb{P}_{k}^{1,\text{an}}$ is $\{\unicode[STIX]{x1D702}_{T,s}:s\in \mathbb{R}_{{\geqslant}0}\}\,\cup \,\{\infty \}$ . For any $s\geqslant 0,|T|_{\unicode[STIX]{x1D702}_{T,s}}=s$ and, if $\unicode[STIX]{x1D702}_{T,s}$ is a type 3 point, then it is the only one in $\mathbb{P}_{k}^{1,\text{an}}$ for which $|T|=s$ . Furthermore, $\unicode[STIX]{x1D702}_{T,s}$ is a type 2 (respectively type 3) point if and only if $s\in \sqrt{|k^{\times }|}$ (respectively $s\not \in \sqrt{|k^{\times }|}$ ).

Thus, there exists $r>0$ such that $\unicode[STIX]{x1D702}^{\prime }=\unicode[STIX]{x1D702}_{T,r}$ . Since $\unicode[STIX]{x1D711}(\unicode[STIX]{x1D702})=\unicode[STIX]{x1D702}_{T,r}$ , by construction, $\unicode[STIX]{x1D702}=\unicode[STIX]{x1D702}_{P,r}$ and $|P|_{\unicode[STIX]{x1D702}_{P,r}}=r$ . Seeing as a finite morphism preserves the type of the point (i.e.  $\unicode[STIX]{x1D702}_{T,r}$ is a type 2 (respectively type 3) point if and only if $\unicode[STIX]{x1D702}_{P,r}$ is so), we obtain (1) and the first part of (2).

To prove the second part of (2), we need to show that if $r\not \in \sqrt{|k^{\times }|}$ , $\unicode[STIX]{x1D702}_{P,r}$ is the only point in $\mathbb{A}_{k}^{1,\text{an}}$ for which $|P|=r$ . Since $P$ is irreducible, by [Reference DucrosDuc, 3.4.24.3], $|P|$ is strictly increasing in $[\unicode[STIX]{x1D702}_{P,0},\infty )$ and locally constant elsewhere. Hence, $\unicode[STIX]{x1D702}_{P,r}$ is the only point in $[\unicode[STIX]{x1D702}_{P,0},\infty )$ for which $|P|=r$ and, since it is a type 3 point (i.e.  $\mathbb{A}_{k}^{1,\text{an}}$ has exactly two connected components), it is the only such point in $\mathbb{A}_{k}^{1,\text{an}}$ .◻

Proof. If $V$ is strictly affinoid, this is [Reference VázquezVáz16, Lemma 2.3]. The proof can be extended to the general case by replacing classical reduction with Temkin’s graded reduction (see [Reference TemkinTem04, Propositions 3.3 and 3.4]).◻

Proof. By [Reference BerkovichBer90, Corollary 2.5.13(ii)], $\unicode[STIX]{x2202}_{B}(V/C)=\unicode[STIX]{x2202}V$ . By [Reference BerkovichBer93, Proposition 1.5.5(ii)], since $C$ is boundaryless, $\unicode[STIX]{x2202}_{B}(V/C)=\unicode[STIX]{x2202}_{B}(V)$ . Finally, in view of Lemma 2.4, $\unicode[STIX]{x2202}V=\unicode[STIX]{x2202}_{B}(V/C)=\unicode[STIX]{x1D6E4}(V)$ .◻

Proof. Let $x\in C$ . Seeing as any curve is a good Berkovich space (i.e. all points have a neighborhood that is an affinoid domain), we may assume that $C$ is an affinoid domain. Let $U$ be an open neighborhood of $x$ in $C$ . There exists an open neighborhood of $x$ in $U$ given by $\{|f_{i}|<r_{i},|g_{j}|>s_{j}:i=1,2,\ldots ,n,j=1,2,\ldots ,m\}$ , where $f_{i},g_{j}$ are analytic functions on $C$ and $r_{i},s_{j}\in \mathbb{R}_{{>}0}$ .

Let $r_{i}^{\prime },s_{j}^{\prime }\in \mathbb{R}_{{>}0}\backslash \sqrt{|k^{\times }|}$ be such that $r_{i}^{\prime }<r_{i},s_{j}^{\prime }>s_{j}$ and $|f_{i}(x)|<r_{i}^{\prime },|g_{j}(x)|>s_{j}^{\prime }$ for all $i$ and $j$ . Set $V=\{|f_{i}|\leqslant r_{i}^{\prime },|g_{j}|\geqslant s_{j}^{\prime }\}$ . It is an affinoid domain of $C$ and a neighborhood of $x$ contained in $U$ .

As $\{|f_{i}|<r_{i}^{\prime },|g_{j}^{\prime }|>s_{j}\}$ is open, it is contained in $\operatorname{Int}(V)$ , so $\unicode[STIX]{x2202}V\subseteq \bigcup _{i=1}^{n}\{|f_{i}|=r_{i}^{\prime }\}\,\cup \,\bigcup _{j=1}^{m}\{|g_{j}|=s_{j}^{\prime }\}$ . Let $y\in \bigcup _{i=1}^{n}\{|f_{i}|=r_{i}^{\prime }\}\,\cup \,\bigcup _{j=1}^{m}\{|g_{j}|=s_{j}^{\prime }\}$ . Since there exists an analytic function $f$ on $C$ such that $|f(y)|\not \in \sqrt{|k^{\times }|}$ , the point $y$ is of type 3, implying that the boundary of $V$ contains only type 3 points.◻

Proof. Let $Z$ be any effective Cartier divisor on $\mathbb{P}_{k}^{1,\text{an}}$ so that $H^{1}(\mathbb{P}_{k}^{1,\text{an}},{\mathcal{O}}(Z))=0$ . We will use induction on the cardinality $n$ of a nice cover. If $n=2$ , then this is Corollary 1.10. Suppose that the result is true for some $n-1$ . If ${\mathcal{D}}=\{U_{1},U_{2},\ldots ,U_{n}\}$ , since $\bigcup _{i=1}^{n}U_{i}$ is connected, from Lemma 2.20, there exist $n-1$ elements of ${\mathcal{U}}$ whose union remains connected. For simplicity of notation, suppose them to be the elements of ${\mathcal{D}}^{\prime }=\{U_{1},U_{2},\ldots ,U_{n-1}\}$ . By Lemma 2.8, $\bigcup _{i=1}^{n-1}U_{i}\,\cap \,U_{n}$ is single type 3 point, so $U_{n}$ intersects exactly one of the elements of ${\mathcal{D}}^{\prime }$ . To simplify the notation, suppose it to be $U_{n-1}$ . Set $\{\unicode[STIX]{x1D702}\}=U_{n-1}\,\cap \,U_{n}$ , so that $S_{{\mathcal{D}}}=S_{{\mathcal{D}}^{\prime }}\,\cup \,\{\unicode[STIX]{x1D702}\}$ .

Let $(g_{s})_{s\in S_{{\mathcal{D}}}}$ be any element of $\prod _{s\in S_{{\mathcal{D}}}}G(\mathscr{M}(\{s\}))$ . By the induction hypothesis, for $(g_{s})_{s\in S_{{\mathcal{D}}^{\prime }}}\in \prod _{s\in S_{{\mathcal{D}}^{\prime }}}G(\mathscr{M}(\{s\}))$ , there exists $(g_{U})_{U\in {\mathcal{D}}^{\prime }}\in \prod _{U\in {\mathcal{D}}^{\prime }}G(\mathscr{M}(U))$ satisfying the conditions of the statement.

• ∙ Suppose that $T_{{\mathcal{D}}}(U_{n})=0$ . By Corollary 1.10, there exist $a\in G(\mathscr{M}(U_{n}))$ and $b\in G(\mathscr{M}(\bigcup _{i=1}^{n-1}U_{i}))$ such that $g_{\unicode[STIX]{x1D702}}\cdot g_{U_{n-1}}=a\cdot b$ in $G(\mathscr{M}(\{\unicode[STIX]{x1D702}\}))$ . For any $i\neq n$ , set $g_{U_{i}}^{\prime }=g_{U_{i}}\cdot b^{-1}$ in $G(\mathscr{M}(U_{i}))$ . Also, set $g_{U_{n}}^{\prime }=a$ in $G(\mathscr{M}(U_{n}))$ .

• ∙ Suppose that $T_{{\mathcal{D}}}(U_{n})=1$ . By Corollary 1.10, there exist $c\in G(\mathscr{M}(\bigcup _{i=1}^{n-1}U_{i}))$ and $d\in G(\mathscr{M}(U_{n}))$ such that $g_{U_{n-1}}^{-1}\cdot g_{\unicode[STIX]{x1D702}}=c\cdot d$ in $G(\mathscr{M}(\{\unicode[STIX]{x1D702}\}))$ . For any $i\neq n$ , set $g_{U_{i}}^{\prime }=g_{U_{i}}\cdot c$ in $G(\mathscr{M}(U_{i}))$ . Also, set $g_{U_{n}}^{\prime }=d^{-1}$ in $G(\mathscr{M}(U_{n}))$ .

The family $(g_{U_{i}}^{\prime })_{i=1}^{n}\in \prod _{i=1}^{n}G(\mathscr{M}(U_{i}))$ satisfies the conditions of the statement for $(g_{s})_{s\in S_{{\mathcal{D}}}}$ .◻

Proof. By Theorem 2.6, we may assume that the cover ${\mathcal{V}}$ only contains elements with finite boundary consisting of only type 3 points. Since $Y$ is compact, we may also assume that ${\mathcal{V}}$ is finite.

Let $f:Y\rightarrow \mathbb{D}$ be a finite surjective morphism we obtain from Noether’s normalization lemma, where $\mathbb{D}$ is the closed unit disc in $\mathbb{P}_{k}^{1,\text{an}}$ . Set $S=f(\bigcup _{V\in {\mathcal{V}}}\unicode[STIX]{x2202}V)$ . It is a finite set of type 3 points. By Lemma 2.14, there exists a nice cover ${\mathcal{D}}$ of $\mathbb{D}$ such that $S_{{\mathcal{D}}}=S$ . Let   $T_{{\mathcal{D}}}$ be a parity function for ${\mathcal{D}}$ (it exists by Lemma 2.19). From Proposition 2.21, the connected components of $f^{-1}(Z^{\prime })$ for all $Z^{\prime }\in {\mathcal{D}}$ form a nice cover ${\mathcal{U}}$ of $Y$ such that $f^{-1}(S_{{\mathcal{D}}})=S_{{\mathcal{U}}}$ and $T_{{\mathcal{D}}}$ induces a parity function $T_{{\mathcal{U}}}$ for ${\mathcal{U}}$ .

Let us show that ${\mathcal{U}}$ refines ${\mathcal{V}}$ . Suppose, by contradiction, that $Z\in {\mathcal{U}}$ is such that there does not exist an element of ${\mathcal{V}}$ containing it. Then there must exist some $a\in \bigcup _{V\in {\mathcal{V}}}\unicode[STIX]{x2202}V\subseteq S_{{\mathcal{U}}}$ such that $a\in \operatorname{Int}(Z)$ . Since $a\in S_{{\mathcal{U}}}$ , there exists $U\in {\mathcal{U}}$ such that $a\in \unicode[STIX]{x2202}U$ . But, then, $Z\,\cap \,U\neq \unicode[STIX]{x2202}Z\,\cap \,\unicode[STIX]{x2202}U$ , which contradicts the fact that ${\mathcal{U}}$ is a nice cover of $Y$ . Consequently, ${\mathcal{U}}$ must refine ${\mathcal{V}}$ .

Set $G^{\prime }={\mathcal{R}}_{K/\mathscr{M}(\mathbb{D})}(G)$ : the restriction of scalars from $K$ to $\mathscr{M}(\mathbb{D})$ of $G$ . It is still a connected rational linear algebraic group (see [Reference MilneMil72, § 1]).

Recall that for any $s\in S_{{\mathcal{D}}}$ , one obtains $G^{\prime }(\mathscr{M}(\{s\}))=G(\mathscr{M}(\{s\})\otimes _{\mathscr{M}(\mathbb{D})}\mathscr{M}(Y))$ . By the lemma above, $G^{\prime }(\mathscr{M}(\{s\}))=\prod _{x\in f^{-1}(s)}G(\mathscr{M}(\{x\}))$ .

Consequently, $(g_{y})_{y\in S_{{\mathcal{U}}}}\in \prod _{y\in S_{{\mathcal{U}}}}G(\mathscr{M}(\{y\}))$ determines uniquely an element $(h_{s})_{s\in S_{{\mathcal{D}}}}$ of $\prod _{s\in S_{{\mathcal{D}}}}G^{\prime }(\mathscr{M}(\{s\}))$ . By Proposition 3.2, there exists $(h_{Z})_{Z\in {\mathcal{D}}}\in \prod _{Z\in {\mathcal{D}}}G^{\prime }(\mathscr{M}(Z))$ such that if for two different $Z_{0},Z_{1}\in {\mathcal{D}}$ with $T_{{\mathcal{D}}}(Z_{0})=0$ , $s\in Z_{0}\,\cap \,Z_{1}$ , then $h_{s}=h_{Z_{0}}\cdot h_{Z_{1}}^{-1}$ in $G^{\prime }(\mathscr{M}(\{s\}))$ .

For any $Z\in {\mathcal{D}}$ , let $Z_{1},Z_{2},\ldots ,Z_{r}$ be the connected components of $f^{-1}(Z)$ . The application $\mathscr{M}(Z)\otimes _{\mathscr{M}(\mathbb{D})}\mathscr{M}(Y)\rightarrow \prod _{i=1}^{r}\mathscr{M}(Z_{i})$ maps $h_{Z}$ to an element $(g_{Z_{1}},g_{Z_{2}},\ldots ,g_{Z_{r}})$ of $\prod _{i=1}^{r}G(\mathscr{M}(Z_{i}))$ . Thus, for any $U\in {\mathcal{U}}$ , we have an element $g_{U}\in G(\mathscr{M}(U))$ . It remains to show that given different $U_{0},U_{1}\in {\mathcal{U}}$ with $T_{{\mathcal{U}}}(U_{0})=0$ such that $y\in U_{0}\,\cap \,U_{1}$ for some $y\in S_{{\mathcal{U}}}$ , we have $g_{y}=g_{U_{0}}\cdot g_{U_{1}}^{-1}$ in $G(\mathscr{M}(\{y\}))$ . This is a consequence of the relation between $T_{{\mathcal{D}}}$ and $T_{{\mathcal{U}}}$ and of the commutativity of the following diagram for any $Z\in {\mathcal{D}}$ and any $s\in Z$ of type 3.

Proof. Since $Y$ is irreducible and normal, ${\mathcal{O}}_{x}$ is a domain for all $x\in Y$ and $\mathscr{M}_{x}=\operatorname{Frac}{\mathcal{O}}_{x}$ .

Seeing as $K{\hookrightarrow}\mathscr{M}_{x}$ for all $x\in Y$ , the implication ‘ $\Rightarrow$ ’ is true.

Suppose that $X(\mathscr{M}_{x})\neq \emptyset$ for all $x\in Y$ . Then there exists an open cover ${\mathcal{V}}$ of $C$ such that for any $V\in {\mathcal{V}}$ , $X(\mathscr{M}(V))\neq \emptyset$ . Let ${\mathcal{U}}$ be a nice refinement of ${\mathcal{V}}$ given by Proposition 3.3 and $T_{{\mathcal{U}}}$ its associated parity function. We remark that for any $U\in {\mathcal{U}}$ , we have $X(\mathscr{M}(U))\neq \emptyset$ .

For $U\in {\mathcal{U}}$ , let $x_{U}\in X(\mathscr{M}(U))$ . For any $y\in S_{{\mathcal{U}}}$ , there exists exactly one element $U_{i}\in {\mathcal{U}}$ with $T_{{\mathcal{U}}}(U_{i})=i$ , $i=0,1$ , containing it. From the transitivity of the action of $G$ , there exists $g_{y}\in G(\mathscr{M}(\{y\}))$ such that $x_{U_{0}}=g_{y}\cdot x_{U_{1}}$ in $G(\mathscr{M}(\{y\}))$ . This gives us an element $(g_{y})_{y\in S_{{\mathcal{U}}}}\in \prod _{y\in S_{{\mathcal{U}}}}G(\mathscr{M}(\{y\}))$ . By Proposition 3.3, there exists $(g_{U})_{U\in {\mathcal{U}}}\in \prod _{U\in {\mathcal{U}}}G(\mathscr{M}(U))$ satisfying: for any different $U^{\prime },U^{\prime \prime }\in {\mathcal{U}}$ containing some point $y\in S_{{\mathcal{U}}}$ such that $T_{{\mathcal{U}}}(U^{\prime })=0$ (implying that $T_{{\mathcal{U}}}(U^{\prime \prime })=1$ ), $g_{y}=g_{U^{\prime }}\cdot g_{U^{\prime \prime }}^{-1}$ in $G(\mathscr{M}\{y\})$ .

For any $U\in {\mathcal{U}}$ , set $x_{U}^{\prime }=g_{U}^{-1}\cdot x_{U}\in X(\mathscr{M}(U))$ . We have constructed a meromorphic function over $U$ for any $U\in {\mathcal{U}}$ . Let us show that they are compatible, i.e. that they coincide on the intersections of the elements of ${\mathcal{U}}$ . Let $D,E\in {\mathcal{U}}$ be such that $D\,\cap \,E\neq \emptyset$ . Suppose that $T_{{\mathcal{U}}}(D)=0$ . For any $s\in D\,\cap \,E$ , $x_{E}^{\prime }=g_{E}^{-1}\cdot x_{E}=g_{D}^{-1}(g_{D}g_{E}^{-1})\cdot x_{E}=g_{D}^{-1}g_{s}\cdot x_{E}=g_{D}^{-1}x_{D}=x_{D}^{\prime }$ in $X(\mathscr{M}(\{s\}))$ . Consequently, $x_{E}^{\prime }=x_{D}^{\prime }$ in $X(\mathscr{M}(E\,\cap \,D))$ .

Compatibility of these meromorphic functions implies that they can be patched to give a meromorphic function on the entire $Y$ . Thus, $X(K)=X(\mathscr{M}(Y))\neq \emptyset$ .

The second version of this local–global principle is a direct consequence of the first one. ◻

Proof. Suppose that there exists an affinoid neighborhood $N_{0}$ of $e$ in $E$ such that $e\in \unicode[STIX]{x1D6E4}(N_{0})$ . By [Reference BerkovichBer90, Proposition 2.5.20], $\unicode[STIX]{x1D6E4}(N_{0})\subseteq \unicode[STIX]{x2202}_{B}(N_{0}/E)\,\cup \,(\unicode[STIX]{x1D6E4}(E)\,\cap \,N_{0})$ . Since $\unicode[STIX]{x2202}_{B}(N/E)$ is the topological boundary of $N_{0}$ in $E$ (see [Reference BerkovichBer90, Corollary 2.5.13(ii)]), we obtain that $e\not \in \unicode[STIX]{x2202}_{B}(N_{0}/E)$ , implying $e\in \unicode[STIX]{x1D6E4}(E)\,\cap \,N_{0}\subseteq \unicode[STIX]{x1D6E4}(E)$ .

On the other hand, if $e\in \unicode[STIX]{x1D6E4}(E)$ for any affinoid neighborhood $N$ of $e$ in $E$ , since $\unicode[STIX]{x1D6E4}(E)\,\cap \,N\subseteq \unicode[STIX]{x1D6E4}(N)$ (see [Reference BerkovichBer90, Proposition 2.5.20]), we obtain $e\in \unicode[STIX]{x1D6E4}(N)$ .◻

Proof. Let $p$ denote the characteristic exponent of $k$ . Then, by [Reference DucrosDuc09, Theorem 6.10], there exists $n$ such that $Y^{\prime }:=(Y\times k^{1/p^{n}})_{\text{red}}$ is geometrically reduced. Since $k^{1/p^{n}}/k$ is a purely inseparable field extension, the map $f:Y^{\prime }\rightarrow Y$ is a homeomorphism. As $Y^{\prime }$ is geometrically reduced, the set of its smooth points is a non-empty Zariski-open subset, i.e. the complement of a set of rigid points. Consequently, since $y^{\prime }:=f^{-1}(y)$ is non-rigid, it is smooth in $Y^{\prime }$ . We remark also that by [Reference DucrosDuc, Theorem 4.2.14], the image (respectively preimage) of a connected affinoid domain is a connected analytic domain and thus, by [Reference DucrosDuc, Théorème 6.1.3], a connected affinoid domain. Finally, for any affinoid domain $U$ of $Y^{\prime }$ , we have that $\unicode[STIX]{x1D6E4}(U)=f^{-1}(\unicode[STIX]{x1D6E4}(f(U)))$ : by [Reference BerkovichBer90, Proposition 2.5.8(iii) and Corollary 2.5.13(i)], this is true for finite morphisms and taking the reduction of an affinoid space does not change its Shilov boundary. Set $Z^{\prime }=f^{-1}(Z)$ . It suffices to prove the statement for $Y^{\prime },y^{\prime },Z^{\prime }$ .

(1) By [Reference DucrosDuc, Théorème 4.5.4], $y^{\prime }$ has an affinoid neighborhood $A^{\prime }$ in $Y^{\prime }$ with only type 3 points in its boundary that is a virtual annulus. Thus, $A^{\prime }$ has at most two connected components at the neighborhood of $y^{\prime }$ and it is connected there if and only if $y^{\prime }\in \unicode[STIX]{x1D6E4}(A^{\prime })$ .

Finally, $Y^{\prime }$ has at most two connected components at the neighborhood of $y^{\prime }$ and, by Lemma 3.6, it is connected there if and only if $y^{\prime }\in \unicode[STIX]{x1D6E4}(Y^{\prime })$ .

(2) Suppose furthermore that $y^{\prime }\in \unicode[STIX]{x1D6E4}(Y^{\prime })$ , implying $y^{\prime }\in \unicode[STIX]{x1D6E4}(A^{\prime })$ . Set $\unicode[STIX]{x1D6E4}(A^{\prime })=\{y^{\prime },z^{\prime }\}$ , where $z^{\prime }$ is a type 3 point. Then $\unicode[STIX]{x2202}A^{\prime }=\{z^{\prime }\}$ and, by [Reference DucrosDuc, Proposition 4.3.14 and Théorème 6.1.3], $B^{\prime }:=(Y^{\prime }\backslash A^{\prime })\,\cup \,\{z^{\prime }\}$ is an affinoid domain. We have $A^{\prime }\,\cup \,B^{\prime }=Y^{\prime },A^{\prime }\,\cap \,B^{\prime }=\{z^{\prime }\}$ (which implies that $B^{\prime }$ is connected). Finally, by shrinking $A^{\prime }$ if necessary, we can always assume that $z^{\prime }\not \in \unicode[STIX]{x1D6E4}(Y^{\prime })$ and, since $\unicode[STIX]{x1D6E4}(Y^{\prime })\,\cap \,A^{\prime }\subseteq \unicode[STIX]{x1D6E4}(A^{\prime })$ , this implies that $\unicode[STIX]{x1D6E4}(Y^{\prime })\,\cap \,A^{\prime }=\{y^{\prime }\}$ .

(3) If $y^{\prime }\not \in \unicode[STIX]{x1D6E4}(Y^{\prime })$ , then $y^{\prime }\not \in \unicode[STIX]{x1D6E4}(A^{\prime })$ and, for the non-trivially valued field $k^{1/p^{n}}$ , the statement follows from the fact that $A^{\prime }$ is a closed virtual annulus.◻