The statement (1) can be checked locally. Let $\sigma$ be a cone of $\Sigma$. Then

$$X_{\sigma,H}=\operatorname{Spec}(H[M_{\sigma}])=\operatorname{Spec}(K[M_{\sigma}]\underset{K}{\otimes}H)=\operatorname{Spec}(H)\underset{K}{\times}X_{\sigma,K}.$$

This proves the first assertion. The commutativity of the diagram follows from the fact that the map $X_{\Sigma,H}^{{\text{\rm an}}}\to X^{{\text{\rm an}}}_{\Sigma,K}$ is given by the restriction of seminorms.

The statement (2) can also be checked locally. Let $\Lambda$ be a polyhedron of $\Pi$. Let $\Lambda^{\prime}=e_{K^{\prime}/K}\Lambda$. Then it is clear that

$$K^{\circ}[{\mathcal{X}}_{\Lambda}]\underset{K^{\circ}}{\otimes}H^{\circ}\subset H^{\circ}[{\mathcal{X}}_{\Lambda^{\prime}}].$$

Since the right-hand side ring is integrally closed, the integral closure of the left side ring is contained in the right side ring. Therefore we need to prove that $H[{\widetilde{M}}_{\Lambda^{\prime}}]$ is integral over the left side ring. Let $(a,l)\in{\widetilde{M}}_{\Lambda^{\prime}}$. Thus $(e_{H/K}a,l)\in{\widetilde{M}}_{\Lambda}$. Then the monomial $\chi^{a}\varpi^{\prime}{}^{l}\in H^{\circ}[{\mathcal{X}}_{\Lambda^{\prime}}]$ satisfies

$$(\chi^{a}\varpi^{\prime}{}^{l})^{e_{H/K}}=(\chi^{e_{H/K}a}\varpi^{l})\in K^{\circ}[{\mathcal{X}}_{\Lambda}]\underset{K^{\circ}}{\otimes}H^{\circ}.$$

Hence $\chi^{a}\varpi^{\prime}{}^{l}$ is integral over $K^{\circ}[{\mathcal{X}}_{\Lambda}]\underset{K^{\circ}}{\otimes}H^{\circ}.$ Since these monomials generate $H^{\circ}[{\mathcal{X}}_{\Lambda^{\prime}}]$, we obtain the result.

To prove (3), let $p^{\prime}\in X^{{\text{\rm an}}}_{0,H}$ and let $p\in X^{{\text{\rm an}}}_{0,K}$ be the corresponding point. Then ${\operatorname{val}}_{K}(p)=\frac{{\operatorname{val}}_{H}(p^{\prime})}{e_{H/K}}$. Therefore, if we write $u={\operatorname{val}}_{K}(p)$ and $u^{\prime}=\frac{{\operatorname{val}}_{H}(p^{\prime})}{e_{H/K}}$, we have

$$\psi_{\|\cdot\|^{\prime}}(u^{\prime})=\frac{1}{\lambda_{H}}\log\|s(p^{\prime})\|^{\prime}=\frac{e_{H/K}}{\lambda_{K}}\log\|s(p)\|=e_{H/K}\psi(u)=e_{H/K}\psi(u^{\prime}/e_{H/K}).$$

Finally, statement (4) follows directly from the definition of $\theta_{\Sigma}$ because the horizontal arrow is given by the restriction of seminorms. ∎