Lecture XVI - Lifting of paths and homotopies

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1 Lecture XVI - Liftig of paths ad homotopies I the last lecture we discussed the liftig problem ad proved that the lift if it exists is uiquely determied by its value at oe poit. I this lecture we shall prove the importat result that coverig projectios ejoy the path liftig ad coverig homotopy properties. This theorem is fudametal i the the theory of coverig projectios ad will be used i the ext lecture to defie a actio of the fudametal group o the fibers. Theorem 16.1 (path liftig lemma: Let p : X X be a coverig projectio ad γ : [0, 1] X be a path such that for some x 0 X ad x 0 X, The there exists a uique path γ : [0, 1] X such that γ(0 = x 0 = p( x 0. (16.1 p γ = γ, γ(0 = x 0 (16.2 Thus each path i X lifts to a uique path i X with a prescribed iitial poit i p 1 (γ(0. Proof: Let O be the ope cover of X by evely covered ope sets ad γ 1 (O be the family γ 1 (O = {γ 1 (G/G O} of ope sets coverig [0, 1]. There is a Lebesgue umber η for this cover ad we choose to be a atural umber such that 1/ < η. Cosider the partitio { 0, 1, 2,..., 1 For each j = 1, 2,...,, the piece γ([ j 1, j ] lies i a evely covered ope set i X. I particular if γ 0 deotes the restrictio of γ to [0, 1/] the the image of γ 0 lies i a ope set G 0 O. The coditios }. 63

2 (16.1-(16.2 say that there is a sheet G 0 lyig over G 0 ad cotaiig the poit x 0. Let p 0 deote the restrictio of p to the sheet G 0 ad q 1 0 be its iverse. O the sub-iterval [0, 1/], we defie γ 0 = q 0 γ 0 thereby obtaiig a iitial piece of the desired lift γ. We shall costruct the lift γ piece by piece defiig it o each subiterval of the partitio of [0, 1]. I what follows γ j deotes the restrictio of γ to the sub-iterval [ j, j+1 ]. Assume iductively that has bee defied such that γ j : [ j, j + 1 ] X p γ j = γ j γ j (j/ = γ j 1 (j/, i case j 1. γ 0 (0 = x 0 For the iductive step we set up the otatios for the edpoits of the lift γ j amely, let γ j = x j+1, γ j = x j+1, p( x j+1 = x j+1. Let G j+1 O be a evely covered eighborhood cotaiig x j+1 such that γ maps [ j+1, j+2 ] ito G j+1 ad G j+1 be the sheet lyig over G j+1 cotaiig the poit x j+1. The restrictio of p to G j+1 is a homeomorphism with iverse q j+1 say, so that q j+1 (x j+1 = x j+1. We set The γ j+1 is cotiuous, p γ j+1 = γ j+1 ad γ j+1 γ j+1 = q j+1 γ j+1 = q j+1 (x j+1 = x j+1 = γ j By gluig lemma, the pieces γ j may be glued together to yield a cotiuous fuctio γ : [0, 1] X such that p γ = γ, γ(0 = x 0. The proof is complete. The uiqueess has bee already proved i geeral. Liftig of homotopies: We ow examie what happes whe we lift homotopic paths with the lifts havig the same iitial poits. Theorem 16.2 (Coverig homotopy property: Let p : X X be a coverig projectio ad x 0 X, x 0 X be chose base poits such that p( x 0 = x 0. Let γ 1, γ 2 be two curves i X startig at x 0 ad havig the same termial poits ad F : [0, 1] [0, 1] X be a homotopy betwee γ 1 ad γ 2. There is a uique lift F : [0, 1] [0, 1] X of F such that F (0, 0 = x 0. I particular the uique lifts of γ 1 ad γ 2 startig at x 0 have the same termial poits. 64

3 Proof: The idea behid the proof is simple ad parallels the proof of the previous theorem except that the book-keepig gets a bit more ivolved. Cosider a coverig O of X by evely covered ope eighborhoods ad choose a Lebesgue umber ɛ for the coverig {F 1 (U/U O}. (16.3 Choose so large that ay square i [0, 1] [0, 1] of side 1/ is cotaied i oe of the sets F 1 (U i (16.3. Partitio [0, 1] [0, 1] usig the grid poits ad S j,k be the square with vertices ( j, k, {( j, k } /0 j, 0 k, k,, k + 1, ( j, k + 1. Let U 0,0 be a evely covered eighborhood i X such that F (S 0,0 U 0,0 ad Ũ0,0 be the sheet i Figure 13: Homotopy liftig property X lyig above U 0,0. Deotig by p 0,0 ad F 0,0 the restrictios of p ad F to Ũ0,0 ad S 0,0 respectively, defie F 0,0 = p 1 0,0 F. Thus F 0,0 : S 0,0 X is cotiuous, takes the value x 0 at the origi ad is a part of the lift F uder costructio. As i the previous theorem we shall costruct the lift F piece by piece ad we ow tur to the adjacet square S 1,0 which is mapped by F to a evely covered eighborhood U 1,0 i the cover O. I particular (referrig to the figure F (B U 1,0. Choose a sheet Ũ1,0 lyig above U 1,0 cotaiig F (B ad the restrictio p 1,0 = p Ũ1,0 65

4 maps Ũ1,0 homeomorphically oto U 1,0. Now we defie the ext piece of the lift F 1,0 as F 1,0 = p 1 1,0 F which is cotiuous o the square S 1,0 ad p F 1,0 = F S1,0 I order to glue together the pieces F 0,0 ad F 1,0 we must esure that they agree all alog the commo edge BC of the adjacet squares S 0,0 ad S 1,0. Their restrictios alog BC where t = 0 ad 0 s 1/ agree at B amely F 0,0 (0, 1 = F 1,0 (0, 1 ad are both lifts of the map which implies, by uiqueess of lifts, s F (s, 1, 0 s 1 F 0,0 (s, 1 = F 1,0 (s, 1, 0 s 1, as desired. It is ow clear how the costructio ought to proceed ad we get a lift F : [0, 1] [0, 1] X of F. We ow have to check that F is ideed a homotopy of paths with fixed edpoits. Well, so that the coected set p F (s, 0 = F (s, 0 = x 0, for all s [0, 1] { F (s, 0/0 s 1} is cotaied i the discrete set p 1 (x 0 ad so must reduce to a sigleto. Likewise F (s, 1 is costat as s varies over [0, 1]. Also p F (0, t = F (0, t = γ 1 (t ad p F (1, t = F (1, t = γ 2 (t showig that F is the desired homotopy betwee the lifts of γ 1 ad γ 2 startig at x 0. Theorem 16.3: Give a coverig projectio p : X X, for ay x0 X ad x 0 X the iduced group homomorphism p : π 1 ( X, x 0 π 1 (X, x 0 is ijective. Proof: Let γ be a loop i X based at x 0 that represets a elemet of ker p. This meas the loop γ = p γ is homotopic to the costat loop i X based at x 0. But the costat loop ε x0 at x 0 lifts as the costat loop ε x 0 at x 0 X. By the coverig homotopy theorem we coclude that γ ad the costat loop ε x0 are homotopic. That is to say [ γ] is the trivial elemet i π 1 ( X, x 0. Remark: The above theorem eables us to idetify π 1 ( X, x 0 as a subgroup of π 1 (X, x 0. We shall ow discuss aother importat cosequece of the path liftig property. 66

5 Theorem 16.4: Give a coverig projectio p : X X where X ad X are path-coected, for ay poits x 1, x 2 X the fibers p 1 (x 1 ad p 1 (x 2 have the same cardiality. Proof: We shall costruct ijective maps from p 1 (x 1 ito p 1 (x 2 ad vice versa. Fix a path γ i X joiig x 1 ad x 2. Pick x 1 p 1 (x 1 ad let γ be the lift of γ startig at x 1 ad defie a map T : p 1 (x 1 p 1 (x 2 by the prescriptio T : x 1 γ(1. Likewise let S : p 1 (x 2 p 1 (x 1 be the map i the reverse directio costructed usig the path γ 1. Sice the iverse path γ 1 is the uique lift of γ 1 startig at γ(1, we see that S( γ(1 = γ(0 = x 1, whereby we coclude S T is the idetity map o p 1 (x 1. By symmetry T S is the idetity map o p 1 (x 2 as desired. Exercises 1. Use the geeral results of this sectio to give a efficiet ad trasparet proof that π 1 (S 1, 1 = Z. First show that for ay loop γ based at 1, the map π 1 (S 1, 1 Z give by [γ] γ(1 is well defied by theorem 16.1, is a group homomorphism usig uiqueess of lifts. Show that surjectivity follows from uiqueess of lifts ad ijectivity follows from theorem Let X be a topological spaces ad a, b X. A simple chai coectig a ad b is a fiite sequece U 1, U 2,..., U of ope sets such that a U 1, b U ad for 1 i < j, U i U j implies j = i + 1. Show that if X is a coected metric space ad U is a ope coverig of X the ay Figure 14: Chai coectedess two poits a, b X ca be coected by a simple chai. This property is referred to as chai coectedess. Is Q chai coected? 67

6 3. Use the above exercise to show that if X is a chai-coected space ad p : X X is a coverig projectio the for ay pair of poits x, y X the fibers p 1 (x ad p 1 (y have the same cardiality. The poit here is that X eed ot be path coected ad the idea of usig a path joiig x ad y as was doe i the proof of theorem 14.4 is o loger available. 4. A toral kot is a group homomorphism κ : S 1 S 1 S 1 give by z (z m, z where m, N. Regardig the toral kot as a loop o the torus determie its lifts with respect to the coverig projectio R R S 1 S For the group homomorphism κ of the previous exercise describe the iduced map κ. 68