Lemma gluing

The next result shows that if they are coherent, homotopy equivalences can be glued together. We shall give an explicit proof to illustrate the combined 

Remark 15.26. The gluing lemma holds when X U B and C U B are CW complexes and X, B, C, and D are subcomplexes. More generally, it holds for cofibrations. 

Remark 15.27. A straightforward argument allows one to generalize Theorem 15.25 to the case of covering by finitely many spaces. Simply proceed by induction on the number of covering spaces, using the formula 

Proof of the gluing lemma. lfXnB = CnB = 0, then the statement is trivial, so assume that these are nonempty. Let T i be the nerve diagram 

Let Si colimPi — hocolimbe the Segal map defined in the proof of the projection lemma, where it was also shown that it is a homotopy equivalence. Finally, the projection mappj hocolim X 2 —cohmP2 is a homotopy equivalence by the projection lemma, and therefore we conclude that the composition 

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As an application of the gluing lemma, let us prove the so-called Quillen lemma, which turned out to be a useful tool in Combinatorial Algebraic Topology... 

Proof The case (r, q) = (3,3), (3,4), (4,3) follows immediately from the list of these polycycles given in Chapter 4. It is clear that doubly infinite and non-periodic (at least in one direction) sequences of glued copies of the elementary polycycles bi and ee, (from Figure 7.3) yield a continuum of infinite non-extensible (3,5> polycycles. By gluing the elementary (5,3)-polycycles Ci (from Figure 7.2) and C 2 (obtained from Ci by rotation through n), we obtain infinite non-extensible (5,3)-polycycles. Clearly, there is a continuum of such. In Lemma 8.2.4, we will construct a continuum of non-extensible (r, )-polycycles for non-elliptic (r, q). 

If sd(S ) is nonorientable, then it is clear from the preceding consideration that instead of a thick cylinder (a full torus) a thick Mobius strip is glued. This completes the proof of the lemma. 

Lemma 2.1.5. Suppose on a critical level Ba there exists exactly one critical saddle circle S. 1) Let sd S ) be orientable. Consider a round handle corresponding to the critical circle and glued with both feet to Then each of the feet lies... 

Thus, if a handle is glued to two distinct tori, Lemma 2.1.5 foUows for the orientable case. Let now a handle be glued to one torus (case 2). Now the tori... 

Now we proceed to the proof of Lemma 2.1.6. We begin with the orientable case. Let a round handle corresponding to a saddle circle be glued to distinct tori Ti and T2 along rings whose axes are noncontractible (by virtue of Lemma 2.1.5) circles 71 and 72, respectively. 

Lemma 2.1.8. It may always be assumed fin the study of surgeiy on Liouville tori) that on each critical level Ba there exists exactly one critical saddle circle. In other words, it may always be assumed that round handles or thick Mobius strips are glued successively and not simultaneously. 

Lemma 3.2.1. The number of ends of a connected graph glued of q tripods does not exceed g + 2, the number of ends being equal to q+ 2 if and only if the graph is branching. 

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