 Solution

The simplicial Mayer–Vietoris theorem says that if a simplicial complex $X=M\cup N$ for subcomplexes $M,N$, there is a natural long exact sequenceThe reduced version extends through dimension zero.

For every simplex $\sigma$ of $K$, all faces of $\sigma \cup \{c_i\}$ are either faces of $\sigma$ or have the form $\tau \cup \{c_i\}$ with $\tau$ a face of $\sigma$, so they lie in $L$. Each $c_i\ast K$ is a cone and hence a simplicial complex. For distinct $i,j$, the rays in the last two coordinates through $c_i$ and $c_j$ are not positive multiples. Thus the two cones meet exactly in $K$, so their simplices have common faces only in $K$. Hence $L$ is a simplicial complex.

Topologically, $L$ is the union of three cones on $K$ along their common base. The first two cones form the suspension $\Sigma K$. Attaching the third contractible cone along the equatorial copy of $K$ givesThis also follows by applying reduced Mayer--Vietoris twice: every cone has zero reduced homology, and the two inclusion maps from the common base are null-homotopic inside the cones. Thereforefor every $n$, with the convention that negative reduced homology vanishes. Since $K$ is nonempty, $L$ is connected, so explicitly