= Solution
Label the six vertices $A,B,C,D,E,F$ clockwise, starting at the upper-left vertex. The arrows identify $AB$ with $ED$, $BC$ with $FE$, and $FA$ with $DC$, preserving the indicated directions. Consequently the <quotient space> has exactly two vertices,
$$
P=\{A,C,E\},\qquad Q=\{B,D,F\}.
$$
Orient its three <topological cells> by $a:A\to B$, $b:B\to C$, and $c:F\to A$. The clockwise boundary of the hexagonal <topological cell> traverses
$$
abc^{-1}a^{-1}b^{-1}c.
$$
Every <topological cell> of dimension one therefore occurs once with each sign. The <cellular chain complex>, with integral coefficients, is
$$
0\longrightarrow\mathbb Z\xrightarrow{\partial_2=0}\mathbb Z^3
\xrightarrow{\partial_1}\mathbb Z^2\longrightarrow0,
\qquad
\partial_1=
\begin{pmatrix}-1&1&1\\1&-1&-1\end{pmatrix}.
$$
Here the rows correspond to $P,Q$. The <image> of $\partial_1$ is the primitive subgroup generated by $Q-P$, and its <kernel> is freely generated by $a+b$ and $a+c$. Thus the <homology groups> are
$$
\boxed{H_j(X;\mathbb Z)=
\begin{cases}
\mathbb Z,&j=0,2,\\
\mathbb Z^2,&j=1,\\
0,&j>2.
\end{cases}}
$$
The two vertex classes are important: identifying all six vertices would give an incorrect <cellular chain complex>. The result agrees with the <homology> of a <torus>.
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