Lemma 1.1 Let K be a simplicial complex. Then K can be partitioned into pairwise disjoint subcomplexes K₁, K₂, . . . , K_r whose polyhedra are the connected components of the polyhedron |K| of K.

Proof Let X₁, X₂, . . . , X_r be the connected components of the polyhedron of K, and, for each j, let K_j be the collection of all simplices σ of K for which σ ⊂ X_j . If a simplex belongs to K_j for all j then so do all its faces.

Therefore K₁, K₂, . . . , K_r are subcomplexes of K. These subcomplexes are pairwise disjoint since the connected components X₁, X₂, . . . , X_r of |K| are

pairwise disjoint. Moreover, if σ ∈ K then σ ⊂ X_j for some j, since σ is a connected subset of |K|, and any connected subset of a topological space is contained in some connected component. But then σ ∈ K_j . It follows that

               K = K₁ ∪ K₂ ∪ · · · ∪ K_r and |K| = |K₁| ∪ |K₂| ∪ · · · ∪ |K_r|, as required.

 The direct sum A₁⊕A₂⊕· · ·⊕A_r of additive Abelian groups A₁, A₂, . . . , A_r is defined to be the additive group consisting of all r-tuples (a₁, a₂, . . . , a_r) with a_i ∈ A_i for i = 1, 2, . . . , r, where

                (a₁, a₂, . . . , a_r) + (b₁, b₂, . . . , b_r) ≡ (a₁ + b₁, a₂ + b₂, . . . , a_r + b_r).

Lemma 1.2 Let K be a simplicial complex. Suppose that K = K₁ ∪ K₂ ∪· · · ∪ K_r, where K₁, K₂, . . . K_r are pairwise disjoint. Then

H_q(K) ∼= H_q(K₁) ⊕ H_q(K₂) ⊕ · · · ⊕ H_q(K_r) for all integers q.

Proof We may restrict our attention to the case when 0 ≤ q ≤ dim K,  since H_q(K) = {0} if q < 0 or q > dim K. Now any q-chain c of K can be expressed uniquely as a sum of the form c = c₁ + c₂ + · · · + c_r, where c_j is a q-chain of K_j for j = 1, 2, . . . , r. It follows that

                                    C_q(K) ≅ C_q(K₁) ⊕ C_q(K₂) ⊕ · · · ⊕ C_q(K_r).

Now let z be a q-cycle of K (i.e., z ∈ C_q(K) satisfies ∂_q(z) = 0). We can express z uniquely in the form z = z₁ + z₂ + · · · + z_r, where z_j is a q-chain of K_j for j = 1, 2, . . . , r. Now

                       0 = ∂_q(z) = ∂_q(z₁) + ∂_q(z₂) + · · · + ∂_q(z_r),

and ∂q(zj ) is a (q−1)-chain of K_j for j = 1, 2, . . . , r. It follows that ∂q(zj ) = 0 for j = 1, 2, . . . , r. Hence each z_j is a q-cycle of K_j , and thus

                         Z_q(K) ≅ Z_q(K₁) ⊕ Z_q(K₂) ⊕ · · · ⊕ Z_q(K_r).

Now let b be a q-boundary of K. Then b = ∂_q+1(c) for some (q + 1)-  chain c of K. Moreover c = c₁ + c₂ + · · · c_r, where c_j ∈ C_q+1(K_j ). Thus b = b₁ + b₂ + · · · br, where b_j ∈ B_q(K_j ) is given by b_j = ∂_q+1c_j for j = 1, 2, . . . , r.  We deduce that

                        B_q(K) ≅ B_q(K₁) ⊕ B_q(K₂) ⊕ · · · ⊕ B_q(K_r).

It follows from these observations that there is a well-defined isomorphism

               ν: H_q(K₁) ⊕ H_q(K₂) ⊕ · · · ⊕ H_q(K_r) → H_q(K)

which maps ([z₁], [z₂], . . . , [z_r]) to [z₁ + z₂ + · · · + zr], where [z_j] denotes the

homology class of a q-cycle z_j of K_j for j = 1, 2, . . . , r.

Let K be a simplicial complex, and let y and z be vertices of K. We say that y and z can be joined by an edge path if there exists a sequence v₀, v₁, . . . , v_m of vertices of K with v₀ = y and v_m = z such that the line segment with endpoints v_j−1 and v_j is an edge belonging to K for j = 1, 2, . . . , m.

Lemma 1.3 The polyhedron |K| of a simplicial complex K is a connected topological space if and only if any two vertices of K can be joined by an edge path.

Proof It is easy to verify that if any two vertices of K can be joined by an edge path then |K| is path-connected and is thus connected. (Indeed any two points of |K| can be joined by a path made up of a finite number of straight line segments.)

We must show that if |K| is connected then any two vertices of K can be joined by an edge path. Choose a vertex v₀ of K. It suffices to verify that every vertex of K can be joined to v₀ by an edge path.

Let K₀ be the collection of all of the simplices of K having the property that one (and hence all) of the vertices of that simplex can be joined to v₀ by an edge path. If σ is a simplex belonging to K₀ then every vertex of σ can be joined to v₀ by an edge path, and therefore every face of σ belongs to K₀.

Thus K₀ is a subcomplex of K. Clearly the collection K₁ of all simplices of K which do not belong to K₀ is also a subcomplex of K. Thus K = K₀ ∪ K₁,  where K₀ ∩ K₁ = ∅, and hence |K| = |K₀| ∪ |K₁|, where |K₀| ∩ |K₁| = ∅.

But the polyhedra |K₀| and |K₁| of K₀ and K₁ are closed subsets of |K|. It follows from the connectedness of |K| that either |K₀| = ∅ or |K₁| = ∅. But v₀ ∈ K₀. Thus K₁ = ∅ and K₀ = K, showing that every vertex of K can be joined to v₀ by an edge path, as required.

Theorem 1.4 Let K be a simplicial complex. Suppose that the polyhedron |K| of K is connected. Then H0(K) ≅Z.

Proof Let u₁, u₂, . . . , u_r be the vertices of the simplicial complex K. Every 0-chain of K can be expressed uniquely as a formal sum of the form

                                         n₁〈u₁〉 + n₂〈u₂〉 + · · · + n_r〈u_r〉

for some integers n₁, n₂, . . . , n_r. It follows that there is a well-defined homomorphism

ε: C₀(K) → Z defined by

ε (n₁〈u₁〉 + n₂〈u₂〉 + · · · + n_r〈u_r〉) = n₁ + n₂ + · · · + n_r.

ow ε(∂₁(〈y, z〉)) = ε(〈z〉 − 〈y〉) = 0 whenever y and z are endpoints of an edge of K. It follows that ε ◦ ∂₁ = 0, and hence B0(K) ⊂ ker ε.

Let v₀, v₁, . . . , v_m be vertices of K determining an edge path. Then

Now |K| is connected, and therefore any pair of vertices of K can be joined by an edge path (Lemma 1.3). We deduce that 〈z〉− 〈y〉∈ B₀(K) for all vertices y and z of K. Thus if c ∈ ker ε, where

and hence

c ∈ B₀(K). We conclude that ker ε ⊂ B₀(K), and hence ker ε = B₀(K).

Now the homomorphism ε: C₀(K) → Z is surjective and its kernel is B₀(K). Therefore it induces an isomorphism from C₀(K)/B₀(K) to Z.

However Z₀(K) = C₀(K) (since ∂₀ = 0 by definition). Thus H₀(K) ≡C₀(K)/B₀(K) ∼= Z, as required.

On combining Theorem 1.4 with Lemmas 1.1 and 1.2 we obtain immediately the following result.

Corollary 1.5 Let K be a simplicial complex. Then

H₀(K) ≅ Z ⊕ Z ⊕ · · · ⊕ Z (r times),