Let P₁, P₂, P₃, P₄, P₅ and P6 be the vertices of a regular hexagon in the plane, listed in cyclic order, and let K be simplicial complex consisting of the triangles P₁P₂P₃, P₃P₄P₅ and P₅P₆P₁, together with all the edges and vertices of these triangles. Then

It follows that c₁ is a 1-cycle of K if and only if

                              m₂ = m₃, m₄ = m₅, m₆ = m₁

and

                             m₁ + m₇ = m₃ + m₈ = m₅ + m₉.

Moreover c₁ is a 1-boundary of K if and only if

                   m₂ = m₃ = −m₈, m₄ = m₅ = −m₉, m₆ = m₁ = −m₇.

We see from this that not every 1-cycle of K is a 1-boundary of K. Indeed

             Z₁(K) = {n₁∂₂τ₁ + n₂∂₂τ₂ + n₃∂₂τ₃ + n_z : n₁, n₂, n₃, n ∈ Z},

where z = ρ₇ + ρ₈ + ρ₉. Let θ:Z₁(K) → Z be the homomorphism defined such that

                     θ (n₁∂₂τ₁ + n₂∂₂τ₂ + n₃∂₂τ₃ + n_z) = n

for all n₁, n₂, n₃, n ∈ Z. Now

            n₁∂₂τ₁ + n₂∂₂τ₂ + n₃∂₂τ₃ + n_z ∈ B₁(K) if and only if n = 0.

It follows that B₁(K) = ker θ. Therefore the homomorphism θ induces an isomorphism from H₁(K) to Z, where H₁(K) = Z₁(K)/B₁(K). Indeed H₁(K) = {n[z] : n ∈ Z}, where z = ρ₇ + ρ₈ + ρ₉ and [z] denotes the homology class of the 1-cycle z.

It is a straightforward exercise to verify that