Simplicial Structures in Topology

II.2 Abstract Simplicial Complexes 63
0 1 2 0
3 5 6 3
4 7 8 4
0 1 2 0
Fig. II.10 A triangulation
of the torus with oriented
simplexes
groups of boundaries and cycles). We notice that C2 ∼ = Z 18 , C1 ∼ = Z 27 , C0 ∼ = Z 9 .We
represent the boundary homomorphisms in the next diagram
0
��
C2
∂2 ��
C1
∂1 ��
C0
Clearly, each vertex (and hence, each 0-chain) is a cycle; hence, Z0 = C0. The
elements {0}, {1}−{0}, ...{8}−{0} form a basis of Z0. Any two vertices can
be connected by a sequence of 1-simplexes and so the 0-cycles {1}−{0}, ...,
{8}−{0} are 0-boundaries. Since the boundary of a generic 1-simplex {i, j} can
be written as
∂1 ({i, j})={ j}−{i} = { j}−{0}−({i}−{0}),
��
0.
we have that B0 ⊂ Z0 is generated by {1}−{0},...,{8}−{0} and thus,
H0(T 2 ;Z) ∼ = Z .
The homology class of any vertex is a generator of this group.
Next, we compute H1(T 2 ;Z). The two 1-chains
z 1 1 = {0,3} + {3,4} + {4,0} and z 2 1 = {0,1} + {1,2} + {2,0}
are cycles and generate (in Z1) a free Abelian group of rank 2 which we denote by
S ∼ = Z ⊕ Z. Let z ∈ Z1 be a 1-cycle z = ∑i kiσ i 1 ,inwhichσi 1 are the 1-simplexes
and ki ∈ Z. By adding suitable multiples of 2-simplexes, it is possible to find a
1-boundary b such that the 1-cycle z − b does not contain the terms, which correspond
to the diagonal 1-simplexes {0,5}, {1,6}, ..., {7,2}, {8,0}. Similarly,
adding suitable pairs of adjacent 2-simplexes (those forming squares with a common
diagonal) it is possible to find a 1-boundary b ′ such that the cycle z−b−b ′ contains
only the terms corresponding to the 1-simplexes {0,3}, {3,4}, {4,0},{0,1}, {1,2},
and {2,0} (we leave the details to the reader, as an exercise). Because z−b−b ′ is a
1-cycle, it follows that z − b − b ′ ∈ S. This argument shows that B1 + S = Z1. Letus
now suppose that B1 ∩ S �= 0; then there exists a linear combination of 2-simplexes
have a common
∑ j h jσ j
2 such that ∑ j h j∂σ j
2 ∈ S. If two 2-simplexes σ i 2
and σ j
2