Choose $0<\delta<1/2$ so that the $2\delta$-realization of $\operatorname{bsd}(T,T)$ satisfies the conclusion of Lemma 3.5 for $K=\partial\mathcal{V}$ and $N=p_{2}(N_{2}(\partial\mathcal{V}))$.

By construction of $\mathfrak{X}^{\delta}$, the set of values on one of the polyhedra $\mathfrak{S}$${}_{\epsilon}+\mathbb{R}_{\geq\epsilon/2}\tau$ is contained in the set of values on its boundary which is contained in $\tau$. Statement (1) follows from $\langle$$\mathfrak{S}$$,\tau\rangle=1$.

For (2), let $n\in V_{w}\backslash N_{2}(\partial\mathcal{V})$. Then $p_{2}(n)$ belongs to a cell $($$\mathfrak{T}$$,w)$ of the $2\delta$-realization of $\operatorname{bsd}(T,T)$, so that $\mathfrak{X}^{\delta}(n)=w$. Also, say, $n\in\overline{U}_{v}$. If we parameterize $\mathcal{F}_{n}(t)$ such that $\mathcal{F}_{n}(0)=n$ (and $\dot{\mathcal{F}}_{n}(t)=\mathfrak{X}(\mathcal{F}_{n}(t))$), then, by (1), $\langle v,\mathcal{F}_{n}(t)\rangle=\lambda(v)+t$. So $\mathcal{F}_{n}(t)-t\widehat{\tau}$ stays in the hyperplane $\langle v,\cdot\rangle=1$, where $\tau=\operatorname{carrier}_{T}($$\mathfrak{T}$$)$.

For $t>0$, $\mathcal{F}_{n}(t)=n+tw$. For $-\delta<t<0$, let $\ell\in(\mathbb{R}^{d})^{*}$ be a linear functional which takes the values $0$ on $w$, and $1$ on the opposite side of $\mathfrak{T}$. Then $\ell(p_{2}(n))<1-2\delta$, and $\frac{d}{dt}\ell(p_{2}(\mathcal{F}_{n}(t)-t\widehat{\tau}))=\ell(w-\widehat{\tau})=1$. So in this time range, $\mathcal{F}_{n}(t)=n+tw$ as well. For $t<-\delta$, $\mathcal{F}_{n}(t)$ belongs to $\Delta^{\vee}_{\lambda^{\delta}}$ by (1). ∎