Multivariable Advanced Calculus

352 THE JORDAN CURVE THEOREMA solid line indicates the corresponding 1 cell is in C. It is there because a changetook place either from top to bottom or from left to right. Note that in both of thosesituations the vertex right in the middle of the crossed lines will occur in ∂C and so C isnot a 1 cycle. There are 8 similar pictures you can draw and in each case this happens.The vertex in the center gets added in an odd number of times. You can also noticethat if you start with the contradictory 2 cell and move counter clockwise, crossing 1cells as you go and starting with B, you must end up at A as a result of crossing 1 cellsof C and this requires crossing either one or three of these 1 cells of C.ABThus that center vertex is a boundary point of C and so C is not a 1 cycle afterall. Similar considerations would hold if the contradictory 2 cell were labeled BA. Thusthere can be no contradiction in the two labeling schemes. They label the 2 cells in Geither A or B in an unambiguous manner.The labeling algorithm encounters every 1 cell of C (in fact of G) and gives a labelto every 2 cell of G. Define the two 2 chains as A and B where A consists of thoselabeled as A and B those labeled as B. The 1 cells which cause a change to take placein the labeling are exactly those in C and each is contained in one 2 cell from A and one2 cell from B. Therefore, each of these 1 cells of C appears in ∂A and ∂B which showsC ⊆ ∂A and C ⊆ ∂B. On the other hand, if l is a 1 cell in ∂A, then it can only occurin a single 2 cell of A and so the 2 cell adjacent to that one along l must be in B and sol is one of the 1 cells of C by definition. As to uniqueness, in moving from left to right,you must assign adjacent 2 cells joined at a 1 cell of C to different 2 chains or else the1 cell would not appear when you take the boundary of either A or B since it would beadded in twice. Thus there are exactly two 2 chains with the desired property. The next lemma is interesting because it gives the existence of a continuous curvejoining two points.yLemma 13.0.18 Let C be a bounded 1 chain such that ∂C = x + y. Then both x, yare contained in a continuous curve which is a subset of |C|.Proof: There are an odd number of 1 cells of C which have x at one end. Otherwise∂C ≠ x + y. Begin at x and move along an edge leading away from x. Continue tillthere is no new edge to travel along. You must be at y since otherwise, you would havefound another boundary point of C. This point would be in either one or three one cellsof C. It can’t be x because x is contained in either one or three one cells of C. Thus,there is always a way to leave x if the process returns to it. IT follows that there is acontinuous curve in |C| joining x to y. x