整数ホモロジーの続き

本を読みながら証明出来た事がありました。
1. Firstly, all chains on the torus are cycles under the definitions of integral homology (i.e. of chain, boundary, incidence coefficient).
Consider:
a. 2-chains- all 2-chains on the torus are of the form nA, where n is the lone face in the normal (plane) form. Looking at the diagram, the two 1-simplexes touching A are a and b. Both of these are pointed once in the same direction as the face and once in the opposite direction of the face. Thus, their incidence coefficients are both 0. So the sum of the 1-chain bounding A is 0. Thus, the boundary of any sum of n faces is zero. So there are no 1 boundaries, and every two chain must be a cycle.
b. 1-chains- every 1-chain on the torus is of the form ma + nb. Consider the individual simplexes a and b. Each is bound by the same point P. As P is at both the head and tail of each, d(a) = d(b) = 0. By the additivity of boundaries, the boundary of any 1-chain is 0, so it is a cycle
c. 0-chains- by definition, every 0-chain is a cycle.

2. Every boundary is by definition a cycle, for any complex of a topological space.
a. 2-boundary - the only 2 boundary is the zero 2-chain, which is a cycle, as d(0) = 0
b. 1-boundary - The boundary of any 1 boundary of a single polygon can be written as d(d(C)), where C is a polygon. The boundary of a one boundary of a 2-chain can be written as a sum of expressions like that above. So proving d(d(C)) is 0 is sufficient to prove that any one-boundary is a cycle (i.e. has a boundary of 0). On a two-simplex (e.g. square), every point will occur on two distinct edges. If a point occurs at the beginning of one edge and the end of another, then both edges must be oriented in the same way relative to the direction of the polygon (picture a circular arrow counterclockwise in a square, and 2 adjoining edges both directed clockwise). So the incidence coefficients of those edges are both +1 or -1. However, the incidence coefficients of the points on the opposing edges (which are considered when we take the boundaries of the sums of the edges) will be of opposite sign. Thus, the 0 boundary of the one-boundary will be a sum of the contributions of each point, which is 0. Thus, the 1-boundary around any polygon has 0 boundary, making any 1-boundary a cycle. In the case where a point P is on the head of both edges or the tail of both edges, those edges are oriented in opposite direction relative to the polygon. Thus, the coefficients of the edges in d(C) will be of opposite sign, while the coefficients of the two point occurrences in d(d(C))
c. Every 0-boundary is a zero-chain, and every 0-chain is a cycle

3. Homology groups of a torus
H0 isomorphic to Z. This is because any zero chain is of the form nP, where n is an integer. As the only zero boundary is the zero boundary no two terms nP and mP are distinct if n!=m.
H1 isomorphic to Z^2. A 1-chain is of the form na+mb. There is no one boundary other than the zero boundary, so any two chains na+mb, n'a+m'b are distinct under homology if n!=n' or m!-m'. The set of all ordered pairs (n,m) is isomorphic to Z^2
H2 isomorphic to Z. As mentioned above, any two-chain is of the form nA, where n is an element of Z. As the only two boundary is zero, all nA elements are distinct when n distinct. So the homology group is isomorphic to Z (integers).

4. Homology groups of a Klein bottle
H0 isomorphic to Z - same as torus
H1 isomorphic to C2 (+) Z, where (+) is the direct sum, indicating that the group consists of elements obtained from summing an element of C2 and an element of Z, where the only member the two groups have in common is zero. This is because the boundary of the face of a Klein bottle equals 2b (or -2b, depending how you direct the edges and face). Any one-chain is of the form na+mb. Consider cases where n=0. If m is even, mb is a boundary, so is homologous to null. If m is odd, mb-b is a boundary, so mb is homologous to b. Thus, the two elements in the subgroup of H1 for mb are null and b, so the subgroup is isomorphic to the cyclic group C2. If m is 0, the number set of elements in the subgroup for na is isomorphic to Z. Thus, the group H1 equals A(+)B, which is isomorphic to C2(+)Z.
H2 isomorphic to Z. Same as torus.
Thus, the torus and Klein bottle differ in their first homology group. It makes sense that two topologically different spaces should show a difference in their homology groups, but no difference appears in homology mod 2.