Let K be a simplicial complex. A q-chain z is said to be a q-cycle if ∂_qz = 0. Aq-chain b is said to be a q-boundary if b = ∂_q+1c΄for some (q+1)-chain c΄ . Thegroup of q-cycles of K is denoted by Z_q(K), and the group of q-boundaries of K is denoted by B_q(K). Thus Z_q(K) is the kernel of the boundary homomorphism ∂_q: C_q(K) → C_q−1(K), and B_q(K) is the image of the bound ary homomorphism ∂_q+1: C_q+1(K) → C_q(K). However ∂_q ◦ ∂_q+1 = 0, by  (∂_q−1 ◦ ∂_q = 0 for all integers q.). Therefore B_q(K) ⊂ Z_q(K). But these groups are subgroups of the Abelian group C_q(K). We can therefore form the quotient group H_q(K),  where H_q(K) = Z_q(K)/B_q(K). The group H_q(K) is referred to as the q^th homology group of the simplicial complex K. Note that H_q(K) = 0 if q < 0   or q > dim K (since Z_q(K) = 0 and B_q(K) = 0 in these cases). It can be shown that the homology groups of a simplicial complex are topological invariants of the polyhedron of that complex.

The element [z] ∈ H_q(K) of the homology group H_q(K) determined by z ∈ Z_q(K) is referred to as the homology class of the q-cycle z. Note that  [z₁ + z₂] = [z₁] + [z₂] for all z₁, z₂ ∈ Z_q(K), and [z₁] = [z₂] if and only if z₁ − z₂ = ∂_q+1c for some (q + 1)-chain c.

Proposition 1.1 Let K be a simplicial complex. Suppose that there exists a vertex w of K with the following property:

• if vertices v₀, v₁, . . . , v_q span a simplex of K then so do w, v₀, v₁, . . . , v_q.

Then H₀(K) ≅ Z, and H_q(K) is the zero group for all q > 0.

Proof Using Lemma 6.2, we see that there is a well-defined homomorphism D_q: C_q(K) → C_q+1(K) characterized by the property that

                     D_q(〈v₀, v₁, . . . , v_q〉) = 〈w, v₀, v₁, . . . , v_q〉

whenever v₀, v₁, . . . , v_q span a simplex of K. Now ∂₁(D₀(v)) = v − w for all vertices v of K. It follows that

for all ∑^s_r=1 n_r〈v_r〉 ∈ C₀(K). But Z₀(K) = C₀(K) (since ∂₀ = 0 by definition),  and thus H₀(K) = C₀(K)/B₀(K). It follows that there is a well-defined surjective homomorphism from H₀(K) to Z induced by the homomorphism from C0(K) to Z that sends ∑^s_r=1 n_r〈v_r〉∈ C0(K) to ∑^s_r=1 n_r. Moreover this induced homomorphism is an isomorphism from H₀(K) to Z.

Now let q > 0. Then

whenever v₀, v₁, . . . , v_q span a simplex of K. Thus

                   ∂_q+1(D_q(c)) + D_q−1(∂_q(c)) = c

for all c ∈ C_q(K). In particular z = ∂_q+1(D_q(z)) for all z ∈ Z_q(K), and hence Z_q(K) = B_q(K). It follows that H_q(K) is the zero group for all q > 0, as required.