Simplicial Structures in Topology

60 II Simplicial Complexes
orderings differ by an odd permutation). Now let us move to K. Begin by taking a
partial ordering of the set X in such a way that the set of vertices of each simplex
σ ∈ Φ is totally ordered; in this way, we obtain an ordering class – that is to say,
an orientation – for each simplex. A simplicial complex whose simplexes are all
oriented is said to be oriented.
Let K =(X,Φ) be an oriented simplicial complex. For every n ∈ Z, with n ≥ 0,
let Cn(K) be the free Abelian group defined by all linear combinations with coefficients
in Z of the oriented n-simplexes of K; in other words, if {σ i n} is the finite
set of all oriented n-simplexes of K, thenCn(K) is the set of all formal sums
∑i miσ i n , mi ∈ Z (called n-chains), together with the addition law
∑ i
piσ i n +∑ i
qiσ i n := ∑(pi + qi)σ
i
i n.
If n < 0, we set Cn(K) =0. Now, for every n ∈ Z, we define a homomorphism
∂n = ∂ K n : Cn(K) −→ Cn−1(K) as follows: if n ≤ 0, ∂n is the constant homomorphism
0; if n ≥ 1, we first define ∂n over an oriented n-simplex {x0,x1,...,xn}
(viewed as an n-chain) as
∂n({x0,x1,...,xn})=
n
∑
i=0
(−1) i {x0,...,�xi,...,xn} ;
finally, we extend this definition by linearity over an arbitrary n-chain of oriented
n-simplexes. The homomorphisms of degree −1, that we have just defined, are
called boundary homomorphisms.
(II.2.15) Lemma. For every n ∈ Z, the composition ∂n−1∂n = 0.
Proof. The result is obvious if n = 1. Let {x0,x1,...,xn} be an arbitrary oriented
n-simplex with n ≥ 2. Then
∂n−1∂n({x0,x1,...,xn})=∂n−1
n
∑
i=0
(−1) i {x0,...,�xi,...,xn}
= ∑(−1) ji
i (−1) j−1 {x0,...,�xi,..., �x j,...,xn}.
This summation is 0 because its addendum {x0,..., �x j,...,�xi,...,xn} appears twice,
once with the sign (−1) i (−1) j and once with the sign (−1) i (−1) j−1 . �
This important property of the boundary homomorphisms implies that, for every
n ∈ Z,theimageof∂n+1 is contained in the kernel of ∂n; using the notation Zn(K)=
ker∂n and Bn(K)=im∂n+1, we conclude that Bn(K) ⊂ Zn(K) for every n ∈ Z. Thus,
to each integer n ≥ 0, we can associate the quotient group
Hn(K;Z)=Zn(K)/Bn(K);
to each n < 0, we associate Hn(K;Z)=0.