The simplicial homology of the tetrahedron boundary

Let $K$ be the boundary of a tetrahedron. It has $4$ vertices, $6$ edges, and $4$ triangular faces. The complex is connected, hence $H_0^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)\cong \mathbb{Z}$.

Orient the faces as $[v_0,v_1,v_2]$, $[v_0,v_1,v_3]$, $[v_0,v_2,v_3]$, and $[v_1,v_2,v_3]$. The alternating sum $[v_1,v_2,v_3]-[v_0,v_2,v_3]+[v_0,v_1,v_3]-[v_0,v_1,v_2]$ is a $2$-cycle because every edge appears twice with opposite signs. Since $C_3(K)=0$, this gives a nonzero class in $H_2^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)$.

The three face boundaries $\mathrm{\partial }[v_0,v_1,v_2],\mathrm{\partial }[v_0,v_1,v_3],\mathrm{\partial }[v_0,v_2,v_3]$ are linearly independent in $C_1(K)$ because the edges $[v_1,v_2]$, $[v_1,v_3]$, and $[v_2,v_3]$ occur in only one of them. Thus $\mathrm{rank}{B}_1(K)\ge 3$. On the other hand, $\mathrm{rank}{C}_1(K)=6$ and $\mathrm{rank}{\mathrm{\partial }}_1=3$ because $H_0^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)\cong \mathbb{Z}$ and $C_0(K)\cong {\mathbb{Z}}^4$, so $\mathrm{rank}{Z}_1(K)=6-3=3$. Hence $B_1(K)=Z_1(K)$ and $H_1^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)=0$.

Now $\mathrm{rank}{\mathrm{\partial }}_2=\mathrm{rank}{B}_1(K)=3$, so $\mathrm{rank}{Z}_2(K)=4-3=1$. Since $Z_2(K)\subseteq C_2(K)\cong {\mathbb{Z}}^4$ and already contains a nonzero cycle, it follows that $H_2^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)=Z_2(K)\cong \mathbb{Z}$. Thus $H_0^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)\cong \mathbb{Z},H_1^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)=0,H_2^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)\cong \mathbb{Z},$ and $H_n^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)=0$ for $n\ge 3$.