We compute the simplicial homology of the complex described in the text.

Fix notation to refer to the vertices of each tetrahedron in the complex: $v^i_j$ will refer to the $j$'th vertex on the $i$'th tetrahedron -- throughout this problem, $i$ used as an index ranging from $1$ to $n$ will be understood to be taken mod $n$. We perform the indicated identifications and examine equivalence classes of faces: We have $n$ 3-simplices (the tetrahedra):

$$\mathcal{T}_1 = \set{[v^1_0 v^1_1 v^1_2 v^1_3]}, \dots, \mathcal{T}_n = \set{[v^n_0 v^n_1 v^n_2 v^n_3]}$$

For faces, we first identify the vertical faces with the next neighboring vertical face, in a cyclic fashion. This gives

$$[v^1_1 v^1_2 v^1_3] \sim [v^2_0 v^2_2 v^2_3], [v^2_1 v^2_2 v^2_3] \sim [v^3_0 v^3_2 v^3_3], \dots, [v^n_1 v^n_2 v^n_3] \sim [v^1_0 v^1_2 v^1_3]$$

Then we identify the bottom face of each tetrahedron with the top of the next tetrahedron around, giving:

$$[v^1_0 v^1_1 v^1_2] \sim [v^2_0 v^2_1 v^2_3], [v^2_0 v^2_1 v^2_2] \sim [v^3_0 v^3_1 v^3_3], \dots, [v^n_0 v^n_1 v^n_2] \sim [v^1_0 v^1_1 v^1_3]$$

So, we have $2n$ faces:

$$\begin{align*} & A_i = \set{[v^i_1 v^i_2 v^i_3], [v^{i+1}_0 v^{i+1}_2 v^{i+1}_3]} \text{ for each } 1 \leq i \leq n \\ & B_i = \set{[v^i_0 v^i_1 v^i_2], [v^{i+1}_0 v^{i+1}_1 v^{i+1}_3]} \text{ for each } 1 \leq i \leq n \end{align*}$$

Examining carefully the identifications this induces, we have $n+2$ edges:

$$\begin{align*} & p = \setbuilder{[v^i_2 v^i_3]}{1 \leq i \leq n} && \text{(the central pole)} \\ & b = \setbuilder{[v^i_0 v^i_1]}{1 \leq i \leq n} && \text{(the outer rims)} \\ & f_i = \set{[v^i_1 v^i_2], [v^{i+1}_1 v^{i+1}_3], [v^{i+1}_0 v^{i+1}_2], [v^{i+2}_0 v^{i+2}_3]} \text{ for each } 1 \leq i \leq n \end{align*}$$

And only $2$ vertices:

$$\begin{align*} & v = \setbuilder{[v^i_0]}{1 \leq i \leq n} \cup \setbuilder{[v^i_1]}{1 \leq i \leq n} \\ & w = \setbuilder{[v^i_2]}{1 \leq i \leq n} \cup \setbuilder{[v^i_3]}{1 \leq i \leq n} \end{align*}$$

So we have $\Delta_0 \cong \mathbb{Z}^2$, $\Delta_1 \cong \mathbb{Z}^{n+2}$, $\Delta_2 \cong \mathbb{Z}^{2n}$, $\Delta_3 \cong \mathbb{Z}^n$. We calculate the homology groups:

For $\delta_0$, we have $\im \delta_0 = \set{0}$ and $\ker \delta_0 = \Delta_0$.

For $\delta_1$, we have:

$$\begin{align*} & \delta_1(p) = \delta_1([v^0_2 v^0_3]) = [v^0_3] - [v^0_2] = w - w = 0 \\ & \delta_1(b) = \delta_1([v^0_0 v^0_1]) = [v^0_1] - [v^0_0] = v - v = 0 \\ & \delta_1(f_i) = \delta_1([v^i_1 v^i_2]) = [v^i_2] - [v^i_1] = w - v \end{align*}$$

For $\delta_2$, we have:

$$\begin{align*} & \delta_2(A_i) = \delta_2([v^i_1 v^i_2 v^i_3]) = [v^i_2 v^i_3] - [v^i_1 v^i_3] + [v^i_1 v^i_2] = (p) - (f_{i-1}) + (f_i) = p + f_i - f_{i-1}\\ & \delta_2(B_i) = \delta_2([v^i_0 v^i_1 v^i_2]) = [v^i_1 v^i_2] - [v^i_0 v^i_2] + [v^i_0 v^i_1] = (f_i) - (f_{i-1}) + (b) = b + f_i - f_{i-1} \end{align*}$$

For $\delta_3$, we have:

$$\begin{align*} & \delta_3(\mathcal{T}_i) = \delta_3([v^i_0 v^i_1 v^i_2 v^i_3]) = [v^i_1 v^i_2 v^i_3] - [v^i_0 v^i_2 v^i_3] + [v^i_0 v^i_1 v^i_3] - [v^i_0 v^i_1 v^i_2] = A_i - A_{i-1} + B_{i-1} - B_i \end{align*}$$

It's clear that $H_0 \cong \mathbb{Z}$. Let's represent the rest of the boundary maps as matrices, to compute the homology:

$$\delta_1 : \mathbb{Z}^{n+2} \rightarrow \mathbb{Z}^2 : \left[ \begin{array}{c|ccccccc} & p & b & f_1 & f_2 & f_3 & \dots & f_n \\ \hline v & 0 & 0 & 1 & 1 & 1 & \dots & 1 \\ w & 0 & 0 & -1 & -1 & -1 & \dots & -1 \end{array} \right] \xrightarrow{SNF} \left[ \begin{array}{c|ccccccc} & p & b & f_1 & f_2 & f_3 & \dots & f_n \\ \hline v & 1 & 0 & 0 & 0 & 0 & \dots & 0 \\ w & 0 & 0 & 0 & 0 & 0 & \dots & 0 \end{array} \right]$$ $$\delta_2 : \mathbb{Z}^{2n} \rightarrow \mathbb{Z}^{n+2} : \left[\begin{array}{c|cccccccccc} & A_1 & A_2 & A_3 & \dots & A_n & B_1 & B_2 & B_3 & \dots & B_n \\ \hline p & 1 & 1 & 1 & \dots & 1 & 0 & 0 & 0 & \dots & 0 \\ b & 0 & 0 & 0 & \dots & 0 & 1 & 1 & 1 & \dots & 1 \\ f_1 & 1 & -1 & 0 & \dots & 0 & 1 & -1 & 0 & \dots & 0 \\ f_2 & 0 & 1 & -1 & \dots & 0 & 0 & 1 & -1 & \dots & 0 \\ f_3 & 0 & 0 & 1 & \dots & 0 & 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots \\ f_{n-1} & 0 & 0 & 0 & \dots & -1 & 0 & 0 & 0 & \dots & -1 \\ f_n & -1 & 0 & 0 & \dots & 1 & -1 & 0 & 0 & \dots & 1 \end{array} \right] \xrightarrow{SNF} \left[\begin{array}{c|cccccccccc} & A_1 & A_2 & A_3 & \dots & A_n & B_1 & B_2 & \dots & \dots & B_n \\ \hline p & 1 & 0 & 0 & \dots & 0 & 0 & 0 & \dots & \dots & 0 \\ b & 0 & 1 & 0 & \dots & 0 & 0 & 0 & \dots & \dots & 0 \\ f_1 & 0 & 0 & 1 & \dots & 0 & 0 & 0 & \dots & \dots & 0 \\ \vdots & \vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots \\ f_{n-2} & 0 & 0 & 0 & \dots & 1 & 0 & 0 & \dots & \dots & 0 \\ f_{n-1} & 0 & 0 & 0 & \dots & 0 & n & 0 & \dots & \dots & 0 \\ f_n & 0 & 0 & 0 & \dots & 0 & 0 & 0 & \dots & \dots & 0 \end{array} \right]$$ $$\delta_3 : \mathbb{Z}^n \rightarrow \mathbb{Z}^{2n} : \left[ \begin{array}{c|ccccc} & \mathcal{T}_1 & \mathcal{T}_2 & \mathcal{T}_3 & \dots & \mathcal{T}_n \\ \hline A_1 & 1 & -1 & 0 & \dots & 0 \\ A_2 & 0 & 1 & -1 & \dots & 0 \\ A_3 & 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ A_{n-1} & 0 & 0 & 0 & \dots & -1 \\ A_n & -1 & 0 & 0 & \dots & 1 \\ B_1 & -1 & 1 & 0 & \dots & 0 \\ B_2 & 0 & -1 & 1 & \dots & 0 \\ B_3 & 0 & 0 & -1 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ B_{n-1} & 0 & 0 & 0 & \dots & 1 \\ B_n & 1 & 0 & 0 & \dots & -1 \\ \end{array} \right] \xrightarrow{SNF} \left[ \begin{array}{c|ccccc} & \mathcal{T}_1 & \mathcal{T}_2 & \dots & \mathcal{T}_{n-1} & \mathcal{T}_n \\ \hline A_1 & 1 & 0 & \dots & 0 & 0 \\ A_2 & 0 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ A_{n-1} & 0 & 0 & \dots & 1 & 0 \\ A_n & 0 & 0 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ B_n & 0 & 0 & \dots & 0 & 0 \\ \end{array} \right]$$

Where the matrices on the right are the corresponding Smith normal forms. Then, we compute the Betti numbers:

$$\begin{align*} & \beta_1 = ((n+2) - 1) - (n+1) = 0 \\ & \beta_2 = (2n - (n+1)) - (n-1) = 0 \\ & \beta_3 = (n - (n-1)) - 0 = 1 \end{align*}$$

So, besides $H_0$, $H_3$ is the only homology group with non-trivial free rank, and $H_1$ has a non-trivial torsion coefficient coming from $\mathsf{SNF}(\delta_2)$. We conclude:

$$H_0 \cong \mathbb{Z} \qquad H_1 \cong \mathbb{Z}_n \qquad H_2 \cong 0 \qquad H_3 \cong \mathbb{Z}$$