Hatcher Exercise 2.1.6 We compute the simplicial homology of the complex described in the text. First, we fix notation to refer to all the faces of the complex: v^i_j refers to the j‘th vertex of the i‘th 2-simplex. Next, we perform the indicated identifications in order to form equivalence classes of faces. We have n+1 two-simplices: t_0 = \set{[v^0_0 v^0_1 v^0_2]}, \dots, t_n = \set{[v^n_0 v^n_1 v^n_2]} We have n+1 edges: % And finally, all the vertices end up identified to a single vertex: v = \setbuilder{[v^i_j]}{0 \leq j \leq n, 0 \leq i \leq 2} Now, we compute homology groups. We have \Delta_0 = \la v \ra \cong \mathbb{Z}, \Delta_1 = \la e_0, \dots, e_n \ra, \Delta_2 = \la t_0, \dots, t_n \ra, the latter two isomorphic to \mathbb{Z}^{n+1}. For \delta_0, we have \im \delta_0 = \set{0} and \ker \delta_0 = \Delta_0. For \delta_1, we have \delta_1(e_i) = v - v = 0, for all i (since there is only one vertex). As such, \im \delta_1 = \set{0} and \ker \delta_1 = \Delta_1. For \delta_2, we have: % So, \delta_2 : \mathbb{Z}^{n+1} \rightarrow \mathbb{Z}^{n+1} can be described by: % Where the matrix on the right is the Smith Normal Form of \delta_2. \delta_3 and higher are the zero map. We have H_0 = \la v \ra / \set{0} \cong \mathbb{Z}. To find H_1, we first conclude that \beta_1 = (n+1) - (n+1) = 0, so H_1 is a torsion group. The torsion coefficients are given by the non-trivial entries of \mathsf{SNF}(\delta_2). We conclude that H_1 \cong \mathbb{Z}_{2^n}. It’s clear that \delta_2 is injective, so H_2 \cong 0.