ScalingStacks

Proof. 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. ∎