ScalingStacks

Proof. By Proposition 3.26, $\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi|_{Y}})$ is equal to $\phi|_{Y}$, so it is continuous. Hence for any $\epsilon>0$, by Proposition 3.23, one has

$$\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi_{X|Y}))=\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi|_{Y}))\subseteq\text{Int}^{\mathrm{top}}_{V_{{\scriptscriptstyle\bullet}}}(\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi|_{Y}(\epsilon)))).$$

By Proposition 2.88, the topology on $\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi(\epsilon)_{X|Y}^{\mathrm{aff}}))$ coincides the induced topology from $\spec(V_{{\scriptscriptstyle\bullet}}(L|_{Y}))^{\mathrm{an}}$, hence the set $\text{Int}^{\mathrm{top}}_{V_{{\scriptscriptstyle\bullet}}}(\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi|_{Y}(\epsilon))))$ which is open in $\spec(V_{{\scriptscriptstyle\bullet}}(L|_{Y}))^{\mathrm{an}}$ is also open in $\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi(\epsilon)_{X|Y}^{\mathrm{aff}}))$. Therefore this set is contained in $\text{Int}^{\mathrm{top}}_{\mathfrak{M}}(\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi|_{Y}(\epsilon))))$. ∎