Simplicial Structures in Topology
Simplicial Structures in Topology
Simplicial Structures in Topology
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II.4 <strong>Simplicial</strong> Homology 75<br />
splits (or is split) if there exists a homomorphism hn−1 : Gn−1 → Gn such that<br />
fnhn−1 = 1Gn−1 (or if there exists a homomorphism kn : Gn → Gn+1 such that<br />
). Prove that if the short exact sequence<br />
kn fn+1 = 1Gn+1<br />
splits, then<br />
2. Prove Theorem (II.3.3).<br />
II.4 <strong>Simplicial</strong> Homology<br />
Gn+1 ��<br />
fn+1 ��<br />
Gn<br />
Gn ∼ = Gn+1 ⊕ Gn−1.<br />
fn ��<br />
��<br />
Gn−1<br />
In this section, we give some results which allow us to study more <strong>in</strong> depth the<br />
homology of a simplicial complex. We beg<strong>in</strong> with some important remarks on the<br />
homology of a simplicial complex K. The groups Cn(K) of the n-cha<strong>in</strong>s are free,<br />
with rank equal to the (f<strong>in</strong>ite) number of n-simplexes of K; hence, also the subgroups<br />
Zn(K) and Bn(K) of Cn(K) are free, with a f<strong>in</strong>ite number of generators. F<strong>in</strong>ally,<br />
the homology groups Hn(K) are Abelian and f<strong>in</strong>itely generated; therefore, by the<br />
decomposition theorem for f<strong>in</strong>itely generated Abelian groups, they are isomorphic<br />
to direct sums<br />
Z β (n) ⊕ Z n(1) ⊕ ...⊕ Z n(k)<br />
where Z n(i) is cyclic of order n(i). The number β(n) – equal to the rank of the<br />
Abelian group Hn(K) –isthenth-Betti number of the complex K.<br />
Let p be the dimension of the simplicial complex K; for each 0 ≤ n ≤ p,lets(n)<br />
be the number of n-simplexes of K (remember that K is f<strong>in</strong>ite). Hence, the rank<br />
of the free Abelian group Cn(K) is s(n). We <strong>in</strong>dicate with z(n) and b(n) the ranks of<br />
the groups Zn(K) and Bn(K), respectively, where n = 0,...,p. S<strong>in</strong>ce the boundary<br />
homomorphism ∂n : Cn(K) → Cn−1(K) is a surjection on Bn−1(K),byNöther’s Homomorphism<br />
Theorem, for each n ≥ 1<br />
(1) s(n) − z(n)=b(n − 1);<br />
if n = 0, we have s(0)=z(0) because C0(K)=Z0(K) and B−1(K)=0; on the other<br />
hand, Hn(K,Z)=Zn(K)/Bn(K) and so<br />
(2) β(n)=z(n) − b(n)<br />
for n ≥ 0. Subtract<strong>in</strong>g (2) from (1) (when n ≥ 1) it follows that<br />
(3) s(n) − β (n)=b(n) − b(n − 1).