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Proof: If this were not so, the component of C 1 containing x would have, by Theorem 5.9.4, the single vertex x as its boundary. This is contrary to Theorem 5.9.3. Q.E.D. Theorem 5.9.6. [110, V.3.3] For any C 2 , |C 2 | = | C 2 |. It follows from Theorem 5.9.6 that 2 and 0 are the only 2-cycles. Hence, if C 2 is connected then so is C 2 , because either C 2 = 2 or else every component of C 2 contains a non-null component of C 2 . On the other hand, if C 2 is connected, it does not necessarily mean C 2 is connected. It is sometimes necessary to pass from one grating, G, to another, G , by introducing additional lines. Such a new grating G is called a re nement of the old. (It is convenient to agree that G is a re nement of itself.) A common re nement can be formed for any two gratings G1 and G2 , by taking all the lines of G1 and G2 as cross lines. k To each k-chain, C k on G corresponds to the subdivided k-chain C on G ; k k C is the sum of the k-chains into which the k-chain C are subdivided. (0-chains 0 are unaltered by subdivision: C = C 0 .) A subdivided chain has the same locus k k as its original: |C | = |C |. Theorem 5.9.7. [110, V.4.1]

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k k k k (C1 + C2 ) = C1 + C2 , k ( C k ) = (C )

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k Corollary 5.9.1. C is a k-cycle if and only if C k is a k-cycle.

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Separation Theorems. Let E be an open set of T 1 , and let k be a k-cycle on a grating G de ned in E. We say the cycle k bounds in E, denoted as k 0, if there is, on some re nement G of G, a (k + 1)-chain C k+1 such that |C k+1 | E and C k+1 = k on G . We say k is nonbounding in E if | k | E but k does not bound in E. The notion of bounding can be used to study the connectivity of a subset of T 1 : A simple closed curve (a 1-cycle) bounds if it separates a subset from the rest of the space; if two vertices (a 0-cycle) in G do not bound in E, then E is not connected. By Jordan Curve Theorem [110, V.10.2], a simple closed curve always bounds in a plane (a sphere). The following statement indicates this is not necessarily true in a torus.

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Theorem 5.9.8. Every 1-cycle on a rectangular grating in T 1 is the boundary of either none 2-chain or just two 2-chains. Proof: The proof is by induction on the number of lines drawn across the unit square that represents T 1 . On the grating consisting of a and b alone, the only 1-cycles are (a) the null sets, which bound two 2-chains (the zero chain and 2 ), and (b) the cycles a, b, and a + b, each of which does not bound (i.e, does not

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APPENDIX

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separate a region from T 1 ) because there is only one rectangle in the grating, which is and = 0. Assume the given grating, G1 , is formed from a grating G0 , for which the theorem holds true, by the addition of a line across the square; assume is parallel to a. Let 1 be the given 1-cycle on G1 . We denote by C 2 the sum of the 2-cells of G1 whose lower edges lie in the line and belong to 1 . The 1-cycle 1 + C 2 therefore contains no edge in . It follows that this 1-cycle 1 is the subdivided form of a 1-cycle 0 on G0 , because 1 + C 2 contains no horizontal edge at a vertex x of and therefore, since it is a cycle, contains both or neither of the vertical edges at x, which together make up an edge of G0 . 1 2 By hypothesis, if F0 bounds, then there is a 2-chain C0 on G0 such that 1 2 2 2 2 1 + C 2 . 0 = C0 ; and, if C1 is the subdivided form of C0 on G1 , then C1 = Hence by Theorem 5.9.2

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2 2 (C1 + C2 ) = ( 1

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