The criterion 1 unfolds the definition of the fact that $z\in\mathfrak{M}$. The criterion 2 is equivalent to the criterion 1, as $\vvvert\mathord{\cdot}\vvvert_{X|x}$ is the quotient algebra norm of $\vvvert\mathord{\cdot}\vvvert_{\widehat{\kappa}(x)}$ for the evaluation map $\mathrm{ev}(x)$. The criterion 3 is equivalent to the criterion 2: if 2 holds, then

$$\forall n\in\mathbb{N},\quad\lvert e_{1}(x)\rvert_{z}\leq C(z)^{\frac{1}{n}}\cdot\vvvert e_{1}^{\otimes n}(x)\vvvert_{X|x}^{\frac{1}{n}},$$

so 3 holds after a limit process for $n\to\infty$. Conversely, if 3 holds, then since $V_{n}(L)(x)$ is spaned by $e_{1}^{\otimes n}(x)$ over $\widehat{\kappa}(x)$, one has

$$\forall n\in\mathbb{N},\quad\lvert s_{n}\rvert_{z}\leq\vvvert s_{n}(x)\vvvert_{(X|x);\mathrm{sp}}\leq\vvvert s_{n}(x)\vvvert_{X|x},$$

so 2 holds by the ultra-metricity of $\lvert\mathord{\cdot}\rvert_{z}$ and the orthogonality of $\vvvert\mathord{\cdot}\vvvert_{X|x}$ for $V_{n}(L)$’s. ∎

For any $x\in X^{\mathrm{an}}$ and any $e_{1}(x)\in V_{1}(L)(x)$, we have

$$\begin{split}\vvvert e_{1}(x)\vvvert_{(X|x);\mathrm{sp}}&=\lim_{\begin{subarray}{c}n\to\infty\end{subarray}}\vvvert e_{1}^{\otimes n}(x)\vvvert_{X|x}^{\frac{1}{n}}\\ &=\lim_{\begin{subarray}{c}n\to\infty\end{subarray}}\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})(e_{1}(x))(x)=\mathcal{P}(\vvvert\mathord{\cdot}\vvvert)(e_{1}(x))(x).\end{split}$$

∎

By Proposition 2.94, the morphism $p(\boldsymbol{0})$ of schemes of finite type over $\spec k$ induces a continuous map betweeen the topological space of their analytification:

$$p(\boldsymbol{0})^{\mathrm{an}}:Tot(L^{\vee})^{\mathrm{an}}\rightarrow(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}.$$

Since

$$p(\boldsymbol{0}):Tot(L^{\vee})\setminus\mathbb{O}\rightarrow\spec(V_{{\scriptscriptstyle\bullet}}(L))\setminus\boldsymbol{0}$$

is an isomorphism of schemes of finite type, its analytification induces a homeomorphism by Proposition 2.94

$$p(\boldsymbol{0})^{\mathrm{an}}:Tot(L^{\vee})^{\mathrm{an}}\setminus\mathbb{O}^{\mathrm{an}}\rightarrow(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}\setminus\boldsymbol{0}^{\mathrm{an}}.$$

∎

Starting with the continuous map in Lemma 3.19, we can determine the pre-image of $\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L,\vvvert\mathord{\cdot}\vvvert))$: let $z\in(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}\setminus\boldsymbol{0}^{\mathrm{an}}$ be a point and $(x,e^{\vee}(x))$ be its unique pre-image under $p(\boldsymbol{0})^{\mathrm{an}}$, where $x\in X^{\mathrm{an}}$ and $e^{\vee}(x)\in L^{\vee}(x)$. By Lemma 3, if we fix a non-zero element $e_{1}(x)\in L(x)$, the point $z$ lies in $\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L,\vvvert\mathord{\cdot}\vvvert))$ if and only if

$$\lvert e_{1}(x)\rvert_{z}\leq\lvert e_{1}(x)\rvert_{\mathcal{P}(\vvvert\mathord{\cdot}\vvvert)}(x).$$

This condition is equivalent to

$$\lvert e^{\vee}(e_{1})(x)\rvert\leq\lvert e_{1}(x)\rvert_{\mathcal{P}(\vvvert\mathord{\cdot}\vvvert)}(x).$$

Hence there exists a continuous surjective map

$$p(\boldsymbol{0})^{\mathrm{an}}:\overline{\mathbb{D}}^{\vee}(L,\mathcal{P}(\vvvert\mathord{\cdot}\vvvert),0)\rightarrow(\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L,\vvvert\mathord{\cdot}\vvvert))).$$

If we remove $\mathbb{O}^{\mathrm{an}}$ and $\boldsymbol{0}^{\mathrm{an}}$ from the domain and image, the restricted map is indeed a homeomorphism. ∎

The assertion is clear if $e_{1}(x)=0$. For $e_{1}(x)\neq 0$, let $e_{n}(x)=e_{1}^{\otimes n}(x)$, note that

$$\lvert e_{1}^{\vee}(x)\rvert_{\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=(\lvert e_{n}^{\vee}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}})^{\frac{1}{n}}.$$

By Corollary 3.15, one has

$$\lvert e_{n}^{\vee}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=\max\Big\{\lvert e_{n}^{\vee}(x)(s_{n,j})\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert_{n}^{-1}\Big\}$$

so

$$\lvert e_{1}^{\vee}(x)\rvert_{\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=\max\Big\{\lvert e_{n}^{\vee}(x)(s_{n,j})\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert_{n}^{-1}\Big\}^{\frac{1}{n}}.$$

Tautologically, one has

$$s_{n,j}(z)=(e_{1}^{\otimes n})^{\vee}(x)(s_{n,j})=e_{n}^{\vee}(x)(s_{n,j}),$$

so the criterion holds. By these defining equations, it is easy to see that the image of the open dual unit disc bundle is an open set. ∎

As $\phi$ is continuous, $\mathbb{D}^{\vee}(L,\phi)\setminus\mathbb{O}^{\mathrm{an}}$ is an open subset of $Tot(L^{\vee})^{\mathrm{an}}\setminus\mathbb{O}^{\mathrm{an}}$. By Lemma 3.19, under the map $p(\boldsymbol{0})^{\mathrm{an}}$, the image of $\mathbb{D}^{\vee}(L,\phi)\setminus\mathbb{O}^{\mathrm{an}}$ is an open subset of $(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}\setminus\boldsymbol{0}^{\mathrm{an}}$, so it is also an open subset of $(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}$. It suffices to treat $\boldsymbol{0}^{\mathrm{an}}$ which is the image of $\mathbb{O}^{\mathrm{an}}$.

As $L$ is ample, there exist $n\in\mathbb{N}$ such that $L^{\otimes n}$ is globally generated. Let $\{s_{n,j}\}_{j\in\{0,\dots,d_{n}\}}$ be a basis and let $\psi$ be the Fubini-Study metric associated with some ultrametric norm $\lVert\mathord{\cdot}\rVert_{n}$ for which this basis is orthogonal. As both $\frac{1}{n}\psi$ and $\phi$ are continuous and $X^{\mathrm{an}}$ is compact, there exist $\alpha\in\mathbb{R}$ such that

$$\forall x\in X^{\mathrm{an}},\ \frac{1}{n}\psi(\alpha)(x)\leq\phi(x),$$

so

$$p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\frac{1}{n}\psi(\alpha)))\subseteq p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\phi)).$$

the left hand side is an open subset of $(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}$ by Lemma 3.21. Then

$$p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\phi))=p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\phi)\setminus\mathbb{O}^{\mathrm{an}})\cup p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\frac{1}{n}\psi(\alpha)))$$

is an open set in $(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}$. ∎

By Proposition 3.20, the left hand side $\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L,\phi))$ is identified with $p(\boldsymbol{0})^{\mathrm{an}}(\overline{\mathbb{D}}^{\vee}(L,\phi))$, which is contained in the open subset $p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\phi(\epsilon)))$. This open subset is contained in $p(\boldsymbol{0})^{\mathrm{an}}(\overline{\mathbb{D}}^{\vee}(L,\phi(\epsilon)))$ which is identified with $\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L,\phi(\epsilon)))$, hence this open subset is contained in the topological interior of the later, the right hand side. ∎