= 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 <sequence>
$$
\cdots\to H_n(M\cap N)\xrightarrow{(i_*,-j_*)}
H_n(M)\oplus H_n(N)\to H_n(X)\to H_{n-1}(M\cap N)\to\cdots.
$$
The 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 $c_i*K$ 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 $K$ gives
$$
L\simeq\Sigma K\vee\Sigma K.
$$
This 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. Therefore
$$
\boxed{\widetilde H_n(L)\cong
\widetilde H_{n-1}(K)\oplus\widetilde H_{n-1}(K)}
$$
for every $n$, with the convention that negative reduced homology vanishes. Since $K$ is nonempty, $L$ is connected, so explicitly
$$
H_0(L)\cong\mathbb Z,\qquad
H_1(L)\cong\widetilde H_0(K)^{\oplus2},
$$
$$
H_n(L)\cong H_{n-1}(K)^{\oplus2}\quad(n\geq2).
$$
Back to article page
