We follow the proof of [Gub10, Proposition 5.7] but in order to establish the result for an arbitrary non-archimedean field $K$ (not necessarily algebraically closed), we use [Gub13, Proposition 6.22] instead of [Gub07, Proposition 4.4]. Let $\tau$ be an open face of $\mathfrak{D}$. We prove first that $R:=\red(p_{\mathfrak{X}^{\prime\prime}}^{-1}(\tau))$ is a stratum of $\tilde{\mathfrak{X}}^{\prime\prime}$. There is a unique stratum $S$ of $\tilde{\mathfrak{X}}$ such that $\tau$ is contained in the interior of $\Delta_{S}$. Let $\mathfrak{U}$ be a formal open subset of $\mathfrak{X}$ such that $S$ is the distinguished stratum of $\mathfrak{U}$ (Proposition 2.5). As strata are compatible with localization we may assume $\mathfrak{X}=\mathfrak{U}$. Let $\psi_{1}^{\prime}:\mathfrak{X}^{\prime\prime}\rightarrow\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}$ be the base change of the composition of the étale map $\psi:\mathfrak{X}\rightarrow\mathfrak{X}(\boldsymbol{n},\boldsymbol{a},m)$ with the projection on the first factor $\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})$. By [Gub13, Proposition 6.22] the first part of the proposition holds for $\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}$. Let $T$ be the stratum of $\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}$ corresponding to $\tau$, i.e.

(2.1)

$\displaystyle\tau=p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}(\red^{-1}(T))$

and

(2.2)

$\displaystyle T=\red(p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}^{-1}(\tau)).$

where $p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}:\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime an}\rightarrow\Delta_{S}$ is the retraction map. We prove $R=\tilde{\psi}_{1}^{\prime-1}(T)$. First we observe that

$$\red(({\psi^{\prime}}_{1}^{\textup{an}})^{-1}(p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}^{-1}(\tau)))=\tilde{\psi}_{1}^{\prime-1}(\red(p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}^{-1}(\tau))).$$

The inclusion $\subseteq$ is clear because $\red\circ{\psi^{\prime}}_{1}^{\textup{an}}=\tilde{\psi}^{\prime}_{1}\circ\red$. The other inclusion follows from this fact and an application of [Gub13, Proposition 6.22]. For details we refer to the proof of [Gub10, Proposition 5.7]. We conclude

$$R=\red(p_{\mathfrak{X}^{\prime\prime}}^{-1}(\tau))=\red(({\psi^{\prime}}_{1}^{\textup{an}})^{-1}(p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}^{-1}(\tau)))=\tilde{\psi}_{1}^{\prime-1}(\red(p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}^{-1}(\tau)))\overset{(\ref{T=})}{=}\tilde{\psi}_{1}^{\prime-1}(T).$$

By [Ber99, Lemma 2.2] $R$ is a strata subset. To see that $R$ is indeed a stratum it is enough to show that $R$ is irreducible. But this follows from

$$\tilde{\psi}_{1}^{\prime-1}(T)=(T\times\tilde{\mathfrak{X}}(m))\times_{\tilde{\mathfrak{X}}(\boldsymbol{n},\boldsymbol{a},m)^{\prime}}\tilde{\mathfrak{X}}^{\prime\prime}\cong(T\times\tilde{\mathfrak{X}}(m))\times_{\{\tilde{0}\}\times\tilde{\mathfrak{X}}(m)}\tilde{\psi}^{-1}(\{\tilde{0}\}\times\tilde{\mathfrak{X}}(m))\cong T\times S,$$

where the latter is irreducible by [Gro65, Corollaire 4.5.8 (i)]. As the open faces of $\mathfrak{D}$ cover $\Delta$, every stratum of $\tilde{\mathfrak{X}}^{\prime\prime}$ is obtained this way. It remains to prove that we can recover $\tau$ from $R$. First note that

$$p_{\mathfrak{X}^{\prime\prime}}(({\psi^{\prime}}_{1}^{\textup{an}})^{-1}(\red^{-1}(T)))=p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}(\red^{-1}(T)).$$

The inclusion $\subseteq$ is clear because $p_{\mathfrak{X}^{\prime\prime}}=p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}\circ{\psi^{\prime}}_{1}^{\textup{an}}$. For the other inclusion, let $x\in p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}(\red^{-1}(T))=\tau$. As the sets $\red^{-1}(T^{\prime})$ with $T^{\prime}$ varying over the strata of $\tilde{\mathfrak{X}}(\boldsymbol{n},\boldsymbol{a})^{\prime}$ cover ${\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}^{\textup{an}}$ and using [Gub13, Proposition 6.22] and the fact the $p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}$ restricts to the identity on $\Delta$ we deduce $x\in\red^{-1}(T)$. Hence $x$ is an element of the left hand side which proves the equality claimed in the display. Now the rest is an easy calculation:

	$\displaystyle p_{\mathfrak{X}^{\prime\prime}}(\red^{-1}(R))$	$\displaystyle=p_{\mathfrak{X}^{\prime\prime}}(\red^{-1}(\tilde{\psi}_{1}^{\prime-1}(T)))$
		$\displaystyle=p_{\mathfrak{X}^{\prime\prime}}(({\psi^{\prime}}_{1}^{\textup{an}})^{-1}(\red^{-1}(T)))$
		$\displaystyle=p_{\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}}(\red^{-1}(T))$
		$\displaystyle\overset{(\ref{tau=})}{=}\tau.$

Finally we want to show that $R$ may be replaced by a nonempty subset $Y$ of $R$. Clearly, the arguments in [Gub10, Proposition 5.7] generalize to the polystable situation, so we presume the claim for $K$ algebraically closed and show how to drop this assumption. Let $\mathbb{C}_{K}$ be the completion of an algebraic closure of $K$. We denote by $\pi:\mathfrak{X}^{\prime\prime}_{\mathbb{C}_{K}}\rightarrow\mathfrak{X}^{\prime\prime}$ the base change of $\mathfrak{X}^{\prime\prime}$ to $\mathbb{C}_{K}^{\circ}$. Let $R^{\prime}$ be the union of the strata of $\tilde{\mathfrak{X}}^{\prime\prime}_{\mathbb{C}_{K}}$ lying over $R$. Then $\pi$ induces a surjection $p_{\mathfrak{X}^{\prime\prime}_{\mathbb{C}_{K}}}(\red^{-1}(R^{\prime}))\twoheadrightarrow p_{\mathfrak{X}^{\prime\prime}}(\red^{-1}(R))$ as the strata in $R^{\prime}$ correspond to open faces lying over $\tau$. Let $Y^{\prime}$ be a lift of $Y$ in $R^{\prime}$. By [Gub10, Proposition 5.7] we have $p_{\mathfrak{X}^{\prime\prime}_{\mathbb{C}_{K}}}(\red^{-1}(Y^{\prime}))=p_{\mathfrak{X}^{\prime\prime}_{\mathbb{C}_{K}}}(\red^{-1}(R^{\prime}))$. Clearly $p_{\mathfrak{X}^{\prime\prime}}(\red^{-1}(Y))\subseteq p_{\mathfrak{X}^{\prime\prime}}(\red^{-1}(R))$ and hence it is enough to show that the restriction of $\pi$ to $p_{\mathfrak{X}^{\prime\prime}_{\mathbb{C}_{K}}}(\red^{-1}(Y^{\prime}))$ factors through $p_{\mathfrak{X}^{\prime\prime}}(\red^{-1}(Y))$. We have the following commutative diagram:

Let $x\in p_{\mathfrak{X}^{\prime\prime}_{\mathbb{C}_{K}}}(\red^{-1}(Y^{\prime}))$ and $y\in\red^{-1}(Y^{\prime})$ with $p_{\mathfrak{X}^{\prime\prime}_{\mathbb{C}_{K}}}(y)=x$ then $\pi(x)=\pi(p_{\mathfrak{X}^{\prime\prime}_{\mathbb{C}_{K}}}(y))=p_{\mathfrak{X}^{\prime\prime}}(\pi(y))\in p_{\mathfrak{X}^{\prime\prime}}(\red^{-1}(Y))$. This proves the claim. ∎