 Solution

Let $K_n$ be the abstract simplex on the vertex setthus every subset of these vertices is a simplex. Let $L_{n-1}$ be its boundary subcomplex of a simplex, consisting of all proper faces, so the only simplex of $K_n$ absent from $L_{n-1}$ is the full $n$-simplex $[v_0\cdots v_n]$.

The geometric realization of a simplicial complex $|K_n|$ is the usual solid geometric $n$-simplex, which is homeomorphic to the closed disk $D^n$. Under such a homeomorphism its topological boundary is $|L_{n-1}|$, and