Compute the simplicial homology groups of the triangular parachute obtained from \(\Delta^2\) by identifying its three vertices to a single point.
Solution
Let \(v\) be the common image of the three vertices. Let \[ a=[v_0,v_1], \qquad b=[v_0,v_2], \qquad c=[v_1,v_2] \] denote the three distinct \(1\)-simplices, and let \[ \sigma=[v_0,v_1,v_2] \] be the unique \(2\)-simplex.
1. The simplicial chain groups are\[ C_2\cong\mathbb Z, \qquad C_1\cong\mathbb Z^3, \qquad C_0\cong\mathbb Z. \]
There is one \(2\)-simplex, three \(1\)-simplices, and one vertex in the quotient \(\Delta\)-complex.
2. The first boundary map is zero:\[ \partial_1=0. \]
Every edge begins and ends at the same quotient vertex \(v\), so \[ \partial_1(a)=\partial_1(b)=\partial_1(c)=v-v=0. \]
3. The second boundary map is\[ \partial_2(\sigma)=a-b+c. \]
The simplicial boundary formula gives \[ \partial[v_0,v_1,v_2] =[v_1,v_2]-[v_0,v_2]+[v_0,v_1] =c-b+a. \]
4. The map \(\partial_2:\mathbb Z\to\mathbb Z^3\) is injective and its image is a primitive rank-one subgroup.
The vector of coefficients is \[ (1,-1,1), \] which is nonzero and whose coordinates have greatest common divisor \(1\). Thus the map is injective and the quotient by its image is torsion-free of rank two.
5. Therefore\[ \boxed{H_0\cong\mathbb Z,\qquad H_1\cong\mathbb Z^2,\qquad H_k=0\ (k\ge2).} \]
Since \(\partial_1=0\), \[ H_0=C_0\cong\mathbb Z. \] Also \[ H_1=\ker\partial_1/\operatorname{im}\partial_2 \cong \mathbb Z^3/\langle(1,-1,1)\rangle \cong\mathbb Z^2. \] Injectivity of \(\partial_2\) gives \(H_2=0\), and there are no higher simplices.