Simplicial Structures in Topology

80 II Simplicial Complexes
Φi = {σ ∈ Φ|(∃x ∈ Xi|x ∈ σ)} , i = 1,...,k
are disjoint; moreover, the pairs Ki =(Xi,Φi), i = 1,...,k, called connected components
of K, are simplicial subcomplexes of K. Hence, the relation of connectedness
subdivides the complex K into a union of disjoint simplicial subcomplexes of K.
From this point of view, a complex K is connected if and only if it has a unique
connected component.
(II.4.4) Lemma. A simplicial complex K is connected if and only if |K| is connected.
Proof. Suppose K to be connected and let p and q be any two points of |K|. Join
p to a vertex x of its carrier s(p) by means of the segment with end points p and x;
this segment is contained in |s(p)| and is therefore a segment of |K|; similarly, join
q to a vertex y of its carrier s(q). However, the vertices x and y are also vertices of K
and since K is connected, there is a path of 1-simplexes of K which links x to y. In
this manner, we obtain a path of |K| that links p to q; hence, |K| is path-connected
and so, |K| is connected (see Theorem (I.1.21)).
Conversely, suppose |K| to be connected and let Ki be a connected component
of K; sinceKi and K � Ki are subcomplexes of K, wehavethat|Ki| is open and
closed in |K|; since|K| is connected, |Ki| = |K|, thatistosay,Ki = K and so, K is
connected. �
The reader is encouraged to review the results on connectedness and pathconnectedness
in Sect. I.1; note that these two concepts are equivalent for polyhedra.
(II.4.5) Lemma. The following properties regarding a simplicial complex K =
(X,Φ) are equivalent:
1. K is connected;
2. H0(K;Z) � Z;
3. the kernel of the augmentation homomorphism
ε : C0(K) → Z ,
coincides with the group B0(K).
n
∑
i=1
Proof. 1 ⇒ 3: We first notice that the inclusion
is always true: indeed,
�
ε
�
k
∂1 ∑ gi{x
i=0
i 0 ,xi1 }
B0(K) ⊂ ker ε
��
=
gi{xi} ↦→
k k
∑ gi − ∑
i=0 i=0
n
∑ gi
i=1
gi = 0.
Let x be a fixed vertex of K. The connectedness of K means that, for every vertex
y of K, the 0-cycles {x} and {y} are homological and so, for every 0-chain
c0 = ∑ k i=0 gi{xi}, there exists a 1-chain c1 such that