Compute the simplicial homology groups of the \(\Delta\)-complex obtained from \(n+1\) 2 simplices \(\Delta_0^2, \cdots, \Delta_n^2\) by identifying all three edges of \(\Delta_0^2\) to a single edge, and for \(i > 0\) identifying the edges \([\nu_0, \nu_1]\) and \([\nu_1, \nu_2]\) of \(\Delta_i^2\) to a single edge and the edge \([\nu_0, \nu_2]\) to the edge \([\nu_0, \nu_1]\) of \(\Delta_{i-1}^2\).
Solution
Let \(\sigma_i\) denote the image of \(\Delta_i^2\) and let \(e_i\) denote the common image of the two edges \([\nu_0,\nu_1]\) and \([\nu_1,\nu_2]\) of \(\Delta_i^2\) for \(i>0\). For \(i=0\), let \(e_0\) be the common image of all three edges.
All vertices are identified, so \[ C_0\cong\mathbb Z,\qquad C_1\cong\mathbb Z^{n+1},\qquad C_2\cong\mathbb Z^{n+1}. \]
1. The map \(\partial_1\) is zero.
Every edge begins and ends at the unique vertex of the quotient.
2. With the ordered bases \(\sigma_0,\dots,\sigma_n\) and \(e_0,\dots,e_n\),\[ \partial_2(\sigma_0)=e_0, \qquad \partial_2(\sigma_i)=2e_i-e_{i-1}\quad(i>0). \]
For \(\sigma_0\) the three oriented faces are all identified with \(e_0\), preserving the ordering of vertices, hence \[ \partial_2(\sigma_0)=e_0-e_0+e_0=e_0. \] For \(i>0\), the first and third faces are both \(e_i\), while the middle face is \(e_{i-1}\), so \[ \partial_2(\sigma_i)=e_i-e_{i-1}+e_i=2e_i-e_{i-1}. \]
3. The map \(\partial_2\) is injective.
Its matrix is triangular after ordering the bases as above, with diagonal entries \[ 1,2,2,\dots,2. \] Its determinant is therefore \(2^n\neq0\), so it is injective over \(\mathbb Z\).
4. The cokernel of \(\partial_2\) is cyclic of order \(2^n\).
The relations in the cokernel are \[ e_0=0, \qquad 2e_i=e_{i-1}\quad(1\le i\le n). \] Hence \[ e_{n-k}=2^k e_n, \] and in particular \[ 2^n e_n=e_0=0. \] Conversely, all \(e_i\) are generated by \(e_n\), and the determinant computation in 3 shows that the cokernel has order \(2^n\). Thus \[ \operatorname{coker}\partial_2\cong\mathbb Z/2^n\mathbb Z. \]
5. Therefore\[ \boxed{ H_0(X)\cong\mathbb Z,\qquad H_1(X)\cong\mathbb Z/2^n\mathbb Z,\qquad H_k(X)=0\ (k\ge2). } \]
Since \(\partial_1=0\), one has \[ H_1=C_1/\operatorname{im}\partial_2\cong\operatorname{coker}\partial_2. \] Injectivity of \(\partial_2\) gives \(H_2=0\), and there are no simplices in higher dimensions. The quotient is connected, so \(H_0\cong\mathbb Z\).