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Consider the chain complex given by

$\{0\}\longrightarrow \Delta _{2}(X;\mathbb {Z} )\longrightarrow \Delta _{1}(X;\mathbb {Z} )\longrightarrow \Delta _{0}(X;\mathbb {Z} )\longrightarrow \{0\}\,.$

Using the decomposition given in the diagram, we have

$\Delta _{2}(X;\mathbb {Z} )=\mathbb {Z} \langle \alpha ,\beta \rangle \,,$

$\Delta _{1}(X;\mathbb {Z} )=\mathbb {Z} \langle A,B,C,D,E,F\rangle \,,$

$\Delta _{0}(X;\mathbb {Z} )=\mathbb {Z} \langle a,b,c,d\rangle \,.$

The boundary operators are given as follows:

$\partial _{3}:\{0\}\longrightarrow \Delta _{2}(X;\mathbb {Z} )\,,$

$\partial _{2}:\Delta _{2}(X;\mathbb {Z} )\longrightarrow \Delta _{1}(X;\mathbb {Z} )\,,$

$\partial _{1}:\Delta _{1}(X;\mathbb {Z} )\longrightarrow \Delta _{0}(X;\mathbb {Z} )\,,$

$\partial _{0}:\Delta _{0}(X;\mathbb {Z} )\longrightarrow \{0\}\,.$

The homology groups are, by definition, given by $H_{i}(X;\mathbb {Z} )\cong \ker(\partial _{i})/{\text{im}}(\partial _{i+1})$ for all $0\leq i\leq 2$. The homology groups are well defined since $\partial ^{2}\equiv 0$ and so ${\text{im}}(\partial _{i+1})\subseteq \ker(\partial _{i})$ for all $0\leq i\leq 2$.

Let us first consider $\partial _{3}$. Trivially, ${\text{im}}(\partial _{3})=\ker(\partial _{3})=\{0\}.$ Next consider $\partial _{2}$:

Failed to parse (syntax error): {\displaystyle \partial_2\alpha = A+C+D+B \, , \ \ \ \partial_2\beta = -(B + E + D + F) \, .}

As elements of $\mathbb {Z} \langle A,B,C,D,E,F\rangle$ both $\partial _{2}\alpha$ and $\partial _{2}\beta$ are linearly independent, and so ${\text{im}}(\partial _{2})\cong \mathbb {Z} ^{2}$, while $\ker(\partial _{2})=\{0\}$.

Next we consider $\partial _{1}$. We have the following:

$\partial _{1}A=d-a\,,\ \ \ \partial _{1}B=a-b\,,\ \ \ \partial _{1}C=b-c\,,$

$\partial _{1}D=c-d\,,\ \ \ \partial _{1}E=b-c\,,\ \ \ \partial _{1}F=d-a\,.$

Clearly $\partial _{1}E=\partial _{1}C$ and $\partial _{1}F=\partial _{1}A$. Thus neither $\partial _{1}E$ nor $\partial _{1}F$ contribute towards the rank of $\partial _{1}$, and may be discounted from further consideration. Consider the system of equation $\partial _{1}A=d-a$, $\partial _{1}B=a-b$, $\partial _{1}C=b-c$ and $\partial _{1}D=c-d$. These may be re-written in matrix notation as follows:

$\partial _{1}\left(\left[{\begin{array}{c}A\\B\\C\\D\end{array}}\right]\right)=\left[{\begin{array}{cccc}-1&0&0&1\\1&-1&0&0\\0&1&-1&0\\0&0&1&-1\end{array}}\right]\left[{\begin{array}{c}a\\b\\c\\d\end{array}}\right].$

The 4-by-4 matrix on the right hand side has rank four, meaning that $\partial _{1}$ must also have rank three. It follows that ${\text{im}}(\partial _{1})\cong \mathbb {Z} ^{3}$ while $\ker(\partial _{1})\cong \mathbb {Z} ^{3}$.

Finally, we consider $\partial _{0}$. Since $\partial _{0}:\Delta _{0}(X;\mathbb {Z} )\longrightarrow \{0\}$, it follows that ${\text{im}}(\partial _{0})\cong \{0\}$ while $\ker(\partial _{0})=\Delta _{0}(X;\mathbb {Z} )\cong \mathbb {Z} ^{4}$.

Using the aforementioned definitions of the homology groups, we have

$H_{2}(X;\mathbb {Z} ):=\{0\}/\{0\}\cong \{0\}\,,\ \ \ H_{1}(X;\mathbb {Z} ):=\mathbb {Z} ^{3}/\mathbb {Z} ^{2}\cong \mathbb {Z} \,,\ \ \ H_{0}(X;\mathbb {Z} ):=\mathbb {Z} ^{4}/\mathbb {Z} ^{3}\cong \mathbb {Z} \,.$

As a trivial consequence, the Euler characteristic:

$\chi (X)=\dim(H_{0})-\dim(H_{1})+\dim(H_{2})=1-1+0=0\,.$