Another lemma on intersections of $d$-simplices

Question

Let $d\ge1$. A $d$-simplex $S$ is the convex hull in $\mathbb R^d$ of the vertices $v_0,\dots,v_d\in\mathbb R^d$ where $\{v_1-v_0,\dots,v_d-v_0\}$ is a linearly independent set of $d$ vectors; for any non-empty subset $T$ of vertices, their convex hull is a face of $S$ and it is a facet of $S$ if $|T|=d$.

Let $S$ be a $d$-simplex. Let $m\ge1$. Let $S_1,\dots,S_m$ be $d$-simplices such that for $1\le i<j\le m$, $S_i\cap S_j$ is either empty or a face of $S_i$ and $S_j$. Assume that $S=\bigcup_{i=1}^m S_i$.

Let $i\in\{1,\dots,m\}$. Show that if a facet of $S_i$ is a subset of a facet of $S$, then it is not a face of any other $S_j$ for $j\in\{1,\dots,m\}$; otherwise, it is a face of exactly one other $S_j$ for some $j\in\{1,\dots,m\}$.

You may use, if needed, the solution to A lemma on intersections of $d$-simplices

Answer 1

1. If a facet $A$ of $S_i$ lying in a hyperplane $\alpha$ belongs to a facet of $S$, then both $S,S_i$ lie on the same side $\alpha^+$ of $\alpha$. If $A$ is a facet of another simplex $S_j$, $j\ne i$, then let $p$ be a barycentre of $A$ (or any other relatively interior point of $A$), then for small enough $\varepsilon>0$ the half of a ball of radius $\varepsilon>0$ centered at $p$ and belonging to $\alpha^+$ is contained both in $S_i$ and $S_j$ (indeed, a $d$-simplex is an intersection of $d+1$ half-spaces, and being contained in a simplex is equivalent to being contained in each of these half-spaces. For all of them except $\alpha^+$ the inclusion follows from $\varepsilon$ being small, and for $\alpha^+$ it is granted by the definition of our half-ball). This contradicts to $S_i\cap S_j$ being a common face.

2. Consider our facet $A$ which belongs to hyperplane $\alpha$, $S\subset \alpha^+$, but now $A$ does not belong to a facet of $S$. Let $p$ be an interior point of $A$. If $p$ lies on the boundary of $S$, then $A$ belongs to the same facet of $S$ which contains $p$ (otherwise some points of $\alpha$ close to $p$ lie outside $S$ that contradicts to $S_i\subset S$.) So, $p$ is an interior point of $S$. Denoting by $h$ any vector for which $p+h$ lies in an interior of $\alpha^+$, we note that for small $t>0$ the point $p+th$ lies in $S_i$ and $p-th$ in some $S_j$, $j\ne i$ (this $j$ may a priori depend on $t$). Therefore there exists a boundary point of $S_j$ on the segment between $p-th$ and $p+th$. There exists the same $j$ for which it is the case for arbitrarily small $t$. Then $p$ itself belongs to a boundary of $S_j$. Since $S_i\cap S_j$ is a common face of $S_i$ and $S_j$, this face is $A$.