Algebraic Topology, Homotopy and Homology by R. Switzer

By R. Switzer

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Example text

Similarly, the edge set G1L,M is the set of those (L+1)×(M +1) arrays of entries of G1 obtained as e0 (yl , zm ), 0 ≤ l ≤ L, 0 ≤ m ≤ M , where (y0 , . . , yL , z0 , . . , zM ) is in ΣL,M (π). The definition of the maps i and t are obvious. 5, and we omit it. 9. Let π be an s/u-bijective pair for (X, ϕ) and suppose that G is a presentation of π. Then for every L, M ≥ 0, (ΣL,M (π), σ) ∼ = (ΣGL,M , σ). Although it is not needed now, it will be convenient for us to have other descriptions of these systems.

Let p be in GK . By definition i ◦ t∗ = t(q)=p i(q), while t∗ ◦ i(p) = t(q)=i(p) q. We claim that i : {q | t(q) = p} → {q | t(q ) = i(p)} is a bijection. Since we suppose K ≥ 1, if q is such that t(q) = p, then t(i(q)) = i(t(q)) = i(p). Moreover, the map sending q 1 · · · q k to q 1 · · · q K pK is the inverse of i and this establishes the claim. The conclusion follows at once from this. The second part is proved in the same way and the last two statements are easy applications of the first two. For a fixed graph G, its higher block presentations, GK , K ≥ 1, all have the same Ds and Du invariants , stated precisely as follows.

Let (y0 , . . , yL ) be in YL (πs ). As πu is onto, we may find z in Z such that πu (z) = πs (y0 ). Then (y0 , . . , yL , z) is in ΣL,0 (π) and its image under ρL, is (y0 , . . , yL ). Next, we check that ρL, is u-resolving. Suppose that (y0 , . . , yL , z0 ) and (y0 , . . , yL , z0 ) are unstably equivalent and have the same image under ρL, . The first fact implies, in particular, that z0 and z0 are unstably equivalent. The second fact just means that (y0 , . . , yL ) = (y0 , . . , yL ). Since the points are in ΣL,0 , we also have πu (z0 ) = πs (y0 ) = πs (y0 ) = πu (z0 ).

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