= Solution
Choose an <orientation> for each <simplex>. The chain <group> $C_k$ is the <free abelian group> on the oriented $k$-simplices, with <orientation> reversal changing sign. On an ordered <simplex> set
$$
\partial_k[v_0,\ldots,v_k]=\sum_{i=0}^k(-1)^i[v_0,\ldots,\widehat v_i,\ldots,v_k].
$$
Its face maps respect the identifications in a <Delta-complex>. In $\partial_{k-1}\partial_k$, deleting any pair of vertices occurs in the two possible orders with signs $(-1)^{i+j-1}$ and $(-1)^{i+j}$, which cancel. Hence $\partial^2=0$. The groups of <homological cycles> and boundaries are $Z_k=\ker\partial_k$ and $B_k=\operatorname{im}\partial_{k+1}$; the identity just proved gives $B_k\subseteq Z_k$, so $H_k=Z_k/B_k$ is defined. Mod-two <homology> uses the same <chain complex> with coefficients reduced modulo two.
Here the edge identifications put all six vertices into one vertex class. Let the three boundary edge classes be $a,b,c$ in cyclic order, and orient the internal diagonals as $d=[v_1,v_3]$, $e=[v_3,v_5]$, $f=[v_5,v_1]$. With the four face orientations listed in the source, the boundaries are
$$
\partial\sigma_1=2a-d,\quad\partial\sigma_2=2b-e,\quad\partial\sigma_3=2c-f,\quad\partial\sigma_4=d+e+f.
$$
Thus $C_2=\mathbb Z^4$, $C_1=\mathbb Z^6$, $C_0=\mathbb Z$, and $\partial_1=0$. The <Euler characteristic> is
$$
\boxed{\chi=1-6+4=-1}.
$$
The face-boundary map is injective over the integers: coefficients of $a,b,c$ force the first three face coefficients to vanish, after which the diagonal coefficients force the fourth to vanish. Its cokernel eliminates $d,e,f$ and leaves generators $a,b,c$ with the single relation $2(a+b+c)=0$. Since $a+b+c$ is primitive in $\mathbb Z^3$, this cokernel is $\mathbb Z^2\oplus\mathbb Z/2$. Therefore
$$
\boxed{H_k=\begin{cases}\mathbb Z,&k=0,\\\mathbb Z^2\oplus\mathbb Z/2,&k=1,\\0,&k\ge2.\end{cases}}
$$
Modulo two the first three face-boundary columns are $d,e,f$, and the last is their sum. The rank is three, the kernel is one-dimensional, and the edge quotient has dimension three. Consequently
$$
\boxed{H_k(-;\mathbb Z/2)=\begin{cases}\mathbb Z/2,&k=0,2,\\(\mathbb Z/2)^3,&k=1,\\0,&k\ge3.\end{cases}}
$$
The top mod-two class is the sum of all four faces. Integral <homology> has no such top class, reflecting nonorientability.
\Image[/past-exam-of-the-topology-and-groups-and-algebraic-topology-course-of-the-university-of-oxford/2007-hexagon-chain-complex.png]
{title=The hexagon complex and its face boundaries}
{description=Paired boundary edges have labels a, b, c. The three internal diagonal classes are d, e, f; all displayed vertices become one vertex after identification.}
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