Proposition 1.1 Let K be the simplicial complex consisting of all the proper faces of an (n + 1)-dimensional simplex σ, where n > 0. Then

                       H₀(K) ≅ Z, H_n(K) ≅ Z, H_q(K) = 0 when q ≠0, n.

Proof Let M be the simplicial complex consisting of the (n+1)-dimensional simplex σ, together with all its faces. Then K is a subcomplex of M, and C_q(K) = C_q(M) when q ≤ n.

It follows from Proposition 1.4in(Simplicial Homology Groups) that H₀(M) ≅ Z and H_q(M) = 0 when q > 0. (Here 0 denotes the zero group.) Now Z_q(K) = Z_q(M) when q ≤ n,  and B_q(K) = B_q(M) when q < n. It follows that H_q(K) = H_q(M) when q < n. Thus H₀(K) ≅Z and H_q(K) = 0 when 0 < q < n. Also H_q(K) = 0 when q > n, since the simplicial complex K is of dimension n. Thus, to determine the homology of the complex K, it only remains to find H_n(K).

Let the (n+1)-dimensional simplex σ have vertices v₀, v₁, . . . , v_n+1. Then

                 C_n+1(M) = {n〈v₀, v₁, . . . , v_n+1〉 : n ∈ Z}.

and therefore Bn(M) = {nz : n ∈ Z}, where

                 z = ∂_n+1 (〈v₀, v₁, . . . , v_n+1〉).

Now H_n(M) = 0 . It follows that Z_n(M) = B_n(M). But Z_n(K) = Z_m(M), since C_n(K) = C_n(M) and the definition of the boundary homomorphism on C_n(K) is consistent with the definition of the boundary homomorphism on C_n(M). Also B_n(K) = 0, because the simplicial complex K is of dimension n, and therefore has no non-zero n-boundaries. It follows that

                          H_n(K)≅Z_n(K) = Z_n(M) = B_n(M) ≅ Z.

Indeed H_n(K) = {n[z] : n ∈ Z}, where [z] denotes the homology class of the n-cycle z of K defined above.

Remark Note that the n-cycle z is an n-cycle of the simplicial complex K,  since it is a linear combination, with integer coefficients, of oriented nsimplices of K. The n-cycle z is an n-boundary of the large simplicial complex M. However it is not an n-boundary of K. Indeed the n-dimensional simplicial complex K has no non-zero (n + 1)-chains, therefore has no nonzero n-boundaries. Therefore z represents a non-zero homology class [z] of Hn(K). This homology class generates the homology group Hn(K).