The fact that the seminorm $\theta_{\sigma}(\gamma)$ extends the norm of $K$ is clear. Let now $f=\sum_{m}\alpha_{m}\chi^{m}$ and $g=\sum_{l}\beta_{l}\chi^{l}$ and write $fg=\sum_{k}\varepsilon_{k}\chi^{k}$ with $\varepsilon_{k}=\sum_{m+l=k}\alpha_{m}\beta_{l}$. Then, since the absolute value of $K$ is ultrametric,

Let $M_{f}=\{m\in M_{\sigma}|\sup_{m^{\prime}}(|\alpha_{m^{\prime}}|\gamma(m^{\prime}))=|\alpha_{m}|\gamma(m)\}$. We define $M_{g}$ analogously. Let $r$ be a vertex of the Minkowski sum $\operatorname{conv}(M_{f})+\operatorname{conv}(M_{g})$. Then there is a unique decomposition $r=m_{r}+l_{r}$ with $m_{r}\in M_{f}$ and $l_{r}\in M_{g}$. Hence $\varepsilon_{r}=\alpha_{m_{r}}\beta_{l_{r}}$. Thus

Thus $\theta_{\sigma}(\gamma)(fg)=\theta_{\sigma}(\gamma)(f)\theta_{\sigma}(\gamma)(g)$. Hence, it is a multiplicative.

We show next that the map $\theta_{\sigma}$ is continuous. The topology of $X^{{\text{\rm an}}}_{\sigma}$ is the coarsest topology that makes the functions $p\to|f(p)|$ continuous for all $f\in K[M_{\sigma}]$. Thus to show that $\theta_{\sigma}$ is continuous it is enough to show that the map $\gamma\to|f(\theta_{\sigma}(\gamma))|$ is continuous on $X_{\sigma}(\mathbb{R}_{\geq 0})=\operatorname{Hom}_{\operatorname{sg}}(M_{\sigma},\mathbb{R}_{\geq 0})$. The topology of $X_{\sigma}(\mathbb{R}_{\geq 0})$ is the coarsest topology such that, for each $m\in M_{\sigma}$, the map $\gamma\to\gamma(m)$ is continuous. Since, for $f=\sum_{m\in M_{\sigma}}\alpha_{m}\chi^{m}$, we have that

we obtain that $\theta_{\sigma}$ is continuous. Since each $\theta_{\sigma}$ is a section of $\rho_{\sigma}$, they are injective.

The fact that the maps $\theta_{\sigma}$ glue together to give a continuous map $\theta_{\Sigma}$ and that $\theta_{\Sigma}$ is a section of $\rho_{\Sigma}$ follows easily from the definitions. This implies in particular that $\theta_{\Sigma}$ is injective. When $\Sigma$ is complete, since $X_{\Sigma}(\mathbb{R}_{\geq 0})$ is compact and $X^{{\text{\rm an}}}_{\Sigma}$ is Hausdorff, the map $\theta_{\Sigma}$ is proper. We deduce that the map $\theta_{\Sigma}$ is proper in general, by using the same argument that shows that the function $\rho_{\Sigma}$ is proper.

The last assertion is clear from the definition of $\theta_{\sigma}(\gamma)$. ∎