= Solution
Use the orientations
$$
t=[v_1v_2v_3],
\qquad
e_{ij}=[v_iv_j]\quad(i<j).
$$
The <simplicial chain complex> has
$$
C_2\cong\mathbb Z,
\qquad C_1\cong\mathbb Z^5,
\qquad C_0\cong\mathbb Z^4,
$$
with
$$
\partial_2t=e_{23}-e_{13}+e_{12},
\qquad
\partial_1e_{ij}=v_j-v_i.
$$
The map $\partial_2$ is injective, so $H_2(K)=0$. The complex is connected, so $H_0(K)\cong\mathbb Z$ and $\operatorname{rank}\partial_1=3$. Hence $\ker\partial_1$ has rank $5-3=2$; quotienting by the rank-one primitive subgroup $\operatorname{im}\partial_2$ gives
$$
H_1(K)\cong\mathbb Z.
$$
There are no higher simplices. Therefore
$$
H_n(K)\cong
\begin{cases}
\mathbb Z,&n=0,1,\\
0,&n\geq2.
\end{cases}
$$
The result agrees with the <Euler characteristic> $4-5+1=0$.
Back to article page
