Proving the inclusion $\ev_{\mathcal{X}}(X)\subset\Delta_{\mathcal{X}}$ is a matter of unwinding definitions. The reverse inclusion will follow from (a). Hence (ii) implies (i).

The proof of (ii) is essentially the same as that of [JM11, Proposition 3.1]. It is also closely related to [Ber99, Lemma 5.6] and [Thu07, Corollaire 3.13]. Fix a subset $J\subset I$ with $E_{J}\neq\emptyset$, let $\xi_{J}$ be its generic point and let $\sigma_{J}$ be the corresponding face of $\Delta_{\mathcal{X}}$. It will be enough to show the existence and uniqueness of a continuous map $\emb_{\mathcal{X}}:\sigma_{J}\to X$ satisfying (a) and (b) of Theorem 3.1 for $s\in\sigma_{J}$.

For each $j\in J$ pick a local equation $z_{j}\in\mathcal{O}_{\mathcal{X},\xi_{J}}$ of $E_{j}$, so that $(z_{j})_{j\in J}$ is a regular system of parameters of $\mathcal{O}_{\mathcal{X},\xi_{J}}$ thanks to the SNC condition. Property (a) means that the valuation defined by

$$\val_{\mathcal{X},s}(f):=-\log|f(\emb_{\mathcal{X}}(s)|$$

takes value $s_{j}$ on $z_{j}$. After choosing a field of representatives of $\kappa(\xi_{J})$ in $\mathcal{O}_{\mathcal{X},\xi_{J}}$, Cohen’s theorem yields an isomorphism

(3.2)

$$\widehat{\mathcal{O}}_{\mathcal{X},\xi_{J}}\simeq\kappa(\xi_{J})[[t_{j},j\in J]]$$

sending $z_{j}$ to $t_{j}$. We first deal with the uniqueness of $\emb_{\mathcal{X}}$ on $\sigma_{J}$. Assume thus that $\emb_{\mathcal{X}},\emb^{\prime}_{\mathcal{X}}:\sigma_{J}\to X$ are two continuous maps satisfying (a) and (b) for $s\in\sigma_{J}$. When $s$ belongs to the relative interior $\rel(\sigma_{J})$, the corresponding valuations $\val_{\mathcal{X},s}$, $\val_{\mathcal{X},s}^{\prime}$ have center $\xi_{J}$ on $\mathcal{X}$, hence extend by continuity to $\widehat{\mathcal{O}}_{\mathcal{X},\xi_{J}}$. The isomorphism (3.2) enables us to write any given $f\in\widehat{\mathcal{O}}_{\mathcal{X},\xi_{J}}$ as $f=\sum_{\alpha\in\mathbf{N}^{J}}f_{\alpha}z^{\alpha}$ with $f_{\alpha}\in\widehat{\mathcal{O}}_{\mathcal{X},\xi_{J}}$, in such a way that each non-zero $f_{\alpha}$ is a unit. For any $s\in\rel(\sigma_{J})$ we then have

$$\val_{\mathcal{X},s}(f_{\alpha}z^{\alpha})=\langle s,\alpha\rangle=\val^{\prime}_{\mathcal{X},s}(f_{\alpha}z^{\alpha})$$

for each $\alpha\in\mathbf{N}^{J}$. If $(s_{j})_{j\in J}$ is $\mathbf{Q}$-linearly independent then these numbers are furthermore mutually distinct as $\alpha$ ranges over $\mathbf{N}^{J}$, and the ultrametric property yields

(3.3)

$$\val_{\mathcal{X},s}(f)=\min_{\alpha\in\mathbf{N}^{J}}\langle s,\alpha\rangle=\val^{\prime}_{\mathcal{X},s}(f).$$

We conclude that $\emb_{\mathcal{X}}(s)=\emb^{\prime}_{\mathcal{X}}(s)$ on the dense set of points $s\in\rel(\sigma_{J})$ such that $(s_{j})_{j\in J}$ is $\mathbf{Q}$-linearly independent, hence $\emb_{\mathcal{X}}=\emb_{\mathcal{X}}^{\prime}$ on $\sigma_{J}$ by continuity.

Let us now define $\emb_{\mathcal{X}}$ on $\sigma_{J}$. Recall that a monomial valuation $v$ on the ring of formal power series $\kappa(\xi_{J})[[t_{j},\,j\in J]]$ is a valuation that is uniquely determined by its values on monomials, i.e. by $s_{j}=v(t_{j})$, $j\in J$. Such a valuation acts on

$$g=\sum_{\alpha\in\mathbf{N}^{J}}g_{\alpha}t^{\alpha}\in\kappa(\xi_{J})[[t_{j},\,j\in J]]$$

by

(3.4)

$$v(g)=\min\{\langle s,\alpha\rangle,\,g_{\alpha}\neq 0\}.$$

Using the isomorphism (3.2) we may thus define $\val_{\mathcal{X},s}$ by pulling back the monomial valuation of $\kappa(\xi_{J})[[t_{j},\,j\in J]]$ with value $s_{j}$ on $t_{j}$. The center of $\val_{\mathcal{X},s}$ is then equal to the generic point of $\bigcap_{s_{j}>0}\left\{z_{j}=0\right\}$, i.e. the generic point of $E_{J^{\prime}}$ where $\sigma_{J^{\prime}}$ is the face containing $s$ in its relative interior. The continuity of $s\mapsto\val_{\mathcal{X},s}(f)$ on $\sigma_{J}$ is also easy to see using (3.4). Setting $\emb_{\mathcal{X}}(s)=\exp\left(-\val_{\mathcal{X},s}\right)$ therefore concludes the proof.