Homology group properties

Oberlin大学のHenle教授が書いたAlgebraic　Topologyを勉強しています。Currently, after reading about vector fields, plane homology, the Jordan curve theorem, Poincare's index lemma, surfaces, etc, I am trying to understand homology groups of complexes. After reading about the definitions of complex, simplex, and incidence coefficients, I encountered a passage about graph theory (as graphs are a type of complex). For the time being, I am skipping this (although the Seven Bridges of Konigsberg problem is of interest). The next passage recounts group theory (particularly abelian groups), and shows small groups (eg cyclic group of order 3), infinite groups (integers), and symmetry groups (eg dihedral group). Next is a discussion of chains and chain groups, including k-cycles and k-boundaries. An interesting postulate here is that all boundaries are cycles (have null boundary).
Next is the definition of Homology Group, which is a group Zk(K) of k-cycles such that homology is used in place of equality for the group axioms is called Hk(K), the kth homology group of K. So groups H0(K), H1(K), and H2(K) can be evaluated for a given complex. The homology groups are calculated for a cube and a torus.
After this, an interesting discussion of invariance (that homology groups are invariant over triangulation of a given surface) occurs.
Thereafter, the concept of Betti numbers is explored, and this is where I am somewhat bemused. A Betti number (discovered by Betti in 1870) indicates the rank of the group, which is the number of elements in its basis. The zeroth Betti number is one for all connected surfaces. Why is this? Considering the 0th homology group on the sphere (in the form of a cube). . .
In order to find the Betti numbers (written hk), we must first find the basis of the homology group. In order to find the basis, we must understand what a basis is and what the elements of the group are. A basis is a subset of the group such that: 1. all elements of the group depend on the basis and 2. no elements of the basis set depend on the others. In order to understand this definition, we must first understand what it means to depend on some set. I had always read that dependency was based on whether a quantity was a linear combination of elements in some set. However, the text uses the more restrictive definition: x depends on A {y1, y2, ..., yn} if x = a1y1+a2y2+...+anyn where values a1, a2, ..., an are either 0 or 1.
Back to the H0(sphere as a cube complex) there are two possible cycles of vertices. Cycles containing an even number of vertices are boundaries (example is 2 vertices). A case of how four vertices are a boundary of a single 1-chain is confusing, but a one-chain can consist of 2 disjoint 1-simplexes. Thus each segment is bound by two vertices. So any even number of vertices is homologous to any other set of even number of vertices. This holds even if they are not disjoint, as (A+B) + (B+C) = (A+C). A set of an odd number of vertices is not homologous to a set of an even number of vertices, but it is homologous to a set of an odd number of vertices. Thus, there are two classes of elements in this group, cycles of even vertices and cycles of odd vertices. The even vertex cycles are represented by epsilon, and the odd-vertex cycles are represented by theta. epsilon + epsilon = epsilon, and epsilon + theta = theta. theta + theta = epsilon and theta+epsilon = theta. Thus, this group is isomorphic to the cyclic group C2. The book explains that the zeroth Betti number h0 is 1 on a connected surface, because any two vertices are homologous
I am not sure how this argument proves that any 0-cycle is dependent on only one vertex (ie one vertex is the basis). In the case of the cube above, a single vertex is a 0-cycle of an odd number of vertices. If it is added to itself, the outcome is the null-cycle (even number of vertices), so a single vertex does generate all group members, since any even cycle is homologous to another even cycle and any odd cycle is homologous to another odd cycle.
In the general case of the zeroth Betti number for a connected surface.
1. Any null chain is a set of vertices.
2. All vertices are homologous to each other.
3. So any chain is dependent on only a single vertex (as it is formed by homologous vertices).
4. But then any homology group must consist of only 2 chains:
a. 1 vertex.
b. null chain (sum of 2 vertices).
5. H0(K) is thus isomorphic to C2, as adding a vertex to itself yields null.
A fundamental point to keep in mind is that the k-chains and k-cycles of available for considering the homology groups depend on the polygons of the complex.