Topologie en Meetkunde 2011 Lecturers: Marius Crainic and Ivan Struchiner


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1 Topologie en Meetkunde 21 Leturers: Mrius Crini nd Ivn Struhiner
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3 CHAPTER 1 The Clssifition Prolem for Compt Surfes 1. Introdution In this Chpter we will introdue nd strt deling with the lssifition prolem for ompt surfes. Giving omplete solution to this prolem is one of the min gols of the ourse. It will serve s motivtion for mny of the onepts tht will e introdue. Our pproh to the lssifition prolem will e the following: (1) We will give list of ompt onneted surfes, ll of whih will e onstruted from polygonl region in the plne y identifying its edges in pirs. (2) We will show tht ny ompt onneted surfe is homeomorphi to one in the list. (3) We will show tht ny two surfes in the list re not homeomorphi to eh other. Prts (1) nd (2) will e delt with in this hpter, while prt (3) will e done only fter we introdue the fundmentl group nd lern how to lulte it (vi the Seifert  Vn Kmpen Theorem). To e it more preise out prt (2) in the pln ove, wht we will show is tht ny tringulle ompt surfe is homeomorphi to one in the list. It turns out tht every ompt surfe is in ft tringulle, nd we hope to ome k to this t some point in the ourse. 2. Topologil Mnifolds The min ojets tht will e studied in this hpter re topologil surfes, whih re simply 2dimensionl topologil mnifolds. Definition 1.1. An ndimensionl topologil Mnifold is topologil spe (X, T ) whih stisfies the following properties: (1) X is Husdorff; (2) X dmits ountle open over {U i } i N suh tht eh U i is homeomorphi to n open set in R n. Eh open U i together with homeomorphism ϕ i : U i V i R n will e lled oordinte hrt of X. Remrk 1.2. The seond ondition in the definition ove n e restted s: X is seond ountle nd eh point of X dmits n open neighorhood whih is hrt. It the follows tht X is lolly ompt nd seond ountle, nd thus metrizle, i.e., the topology of X is indued y metri. Definition 1.3. A surfe is 2dimensionl topologil mnifold. Exmple 1.4 (The SphereS 2 ). We define the sphere S 2 to e the quotient spe otined from squre y identifying its order ording to the Figure 1. Thus, if we denote the unit intervl [0,1] y I, then S 2 = {(x,y) I I}/ where we identify (0,y) (1 y,1), nd (x,0) (1,1 x). 3
4 4 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES The sphere otined from squre glueing s indited in the piture Exerise 1.1. (2) Show tht it is surfe. Figure 1. (1) Show tht S 2 is homeomorphi to the stndrd sphere {(x,y,z) R 3 : x 2 +y 2 +z 2 = 1.} Exmple 1.5 (The Torus T 2 ). We define the torus T 2 to e the quotient spe otined from the unit squre y identifying its order ording to the Figure 2. Thus, T 2 = {(x,y) I I}/ where we identify (0,y) (1,y), nd (x,0) (x,1). Figure 2. Exerise 1.2. Let S 1 = {(x,y) R 2 : x 2 +y 2 = 1} e the stndrd irle. (1) Show tht the torus T 2 is homeomorphi to S 1 S 1. (2) Show tht it is surfe. Exmple 1.6 (The Projetive Spe P 2 ). We define the projetive spe P 2 to e the quotient spe otined from the unit squre y identifying its order ording to the Figure 3. Thus, P 2 = {(x,y) I I}/ where we identify (0,y) (1,y), nd (x,0) (1 x,1). Exerise 1.3. Let D R 2 denote the (losed) disk of rdius 1. (1) Show tht P 2 is homeomorphi to the quotient spe otined from D y identifying its order S 1 vi the ntipodl mp (Figure 4) A : S 1 S 1, A(x,y) = ( x, y).
5 2. TOPOLOGICAL MANIFOLDS 5 ~ Figure 3. (2) Show tht P 2 is homeomorphi to the quotient spe otined from the stndrd sphere y identifying point p with its ntipodl p. (3) Show tht P 2 is homeomorphi to the spe of lines through the origin in R 3. (4) Show tht it is surfe. The relevnt informtion to reover the lines Lines in the spe determined y their intersetions with the sphere Projet down (homeomorphilly) onto the dis The dis on whih we still hve to glue the ntipodl points on its irle oundry Figure 4. There is very si opertion whih llows us to onstrut new mnifold out of two given mnifolds. Definition 1.7. Given two topologil mnifolds M nd N of the sme dimension, define their onneted sum, denoted M#N s follows: remove from M nd N two smll lls B 1 nd B 2 nd glue M B 1 nd N B 2 long the sphere B 1 = B 2. For surfes, it mens tht we remove two smll disks nd we glue the remininig spes long the oundy irles (Figure 5). We n desrie this opertion with more detils: Remove n Open Disk: We remove from M nd N n open suset D 1 nd D 2 eh of whih is homeomorphi to n open disk in R 2. Glue long the Boundry: We fix homeomorphism ϕ : D 1 D 2 nd we tke the quotient spe M#N = (M D 1 ) (N D 2 )/ where x y if nd only if x = y or x D 1, y D 2, nd ϕ(x) = y.
6 6 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES mke holes glue Figure 5. Conneted Sum Exmple 1.8. Theonneted sumof two tori is thedouletorus T 2. Repeting theopertion of onneted sum, oneotins ll tori with ritrry numerof holes (see Figure 6 for theg = 2): T g = T#...#T. }{{} g times Figure 6. Doule Torus. Similrly, one onsiders the onneted sum of h opies of P 2 : P h = P 2 #...#P 2. }{{} h times Exerise 1.4. Show tht the onneted sum M#S 2 of ny surfe M with the sphere S 2 is homeomorphi to M itself. We ould, in prinipl, onsider more surfes y onsidering other exmples of onneted sums (for exmple of torus with projetive spe), ut s we will soon see, we hve lredy otined omplete list of ll ompt onneted surfes: Theorem 1.9. Any ompt onneted surfe is homeomorphi to one of the following: (1) A sphere S 2, (2) A onneted sum of Tori (plurl of Torus) T g, with g N, or (3) A onneted sum of projetive spes P h, with h N.
7 3. THE BASIC BUILDING BLOCKS: POLYGONAL REGIONS 7 3. The Bsi Building Bloks: Polygonl Regions In this setion we will show how to onstrut surfes out of polygonl regions of the plne, y identifying its edges in pirs. Intuitively, polygonl region is suset of the plne whih looks like in Figure 7. Let us explin how to mke this preise. P 1 P P P P2 Polygonl region Figure 7. Polygonl Region. Fix irle in R 2, pik n+1 points on it nd order them in ounterlokwise diretion {p 0,...,p n }. For eh 0 < i n onsider the line pssing through p i 1 nd p i. It divides R 2 into two hlfplnes. Let H i e the hlfplne whih ontins ll the other points p j. Let P e the set P = H 1 H 2 H n. Definition An nsided polygonl region of the plne is ny suset of R 2 otined y the reipe ove. Assoited to polygonl region will will use the following nottion: Verties: The points p i will e lled verties of P. The set of ll verties of P will e denoted y V(P). Edges: The line segment joining p i 1 nd p i will e denoted y e i, nd will e lled n edge of P. The set of ll edges of P will e denoted y E(P) Border: The union of ll edges of P will e denoted y P nd will e lled the order of P. Interior: The omplement of P in P will e denoted y Int(P) nd will e lled the interior of P. It will lso e importnt to introdue orienttions on the edges of polygon, nd to speify wht mp etween edges is (this is how we will e le the mke preise the notion of glueing one edge to nother ).
8 8 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES Definition (1) Let L R 2 e line segment. An orienttion of L is hoie of ordering of its end points. Suh n orienttion will e represented y n rrow, nd we will sy tht L is line from to (Figure 8). (2) If L is line from to, nd L is line segment from to d, then positive liner mp from L to L is the homeomorphism h : L L whih ssoites to x = (1 t)+t L the point h(x) = (1 t)+td x the liner trnsformtion h h(x) d Figure 8. Positive Liner Mps. 4. Glueing the Edges of Polygonl Region Sine we will e onsidering (disjoint unions of) polygonl regions with severl identifitions on the orders, we must find onvenient wy of keeping trk of suh glueing proedures. For this, we will introdue the onept of lels: Definition A leling of polygonl region P is mp E(P) Λ from the set of edges of P to set Λ, whose elements will e lled lels. Given polygonl region long with: (1) leling of its edges, nd (2) n orienttion on edge, we onsider the spe X = P/ where If p Int(P), then p is equivlent only to itself, i,e,m p p; If e i nd e j re edges with the sme lel, we let h : e i e j e positive liner mp nd we set x e i h(x) e j. In this se we sy tht X ws otined from P y glueing its edges together ording to the orienttion nd the leling. We remrk tht we lso llow X to e otined from finite disjoint unit of polygonl regions with identifitions on the edges. Thus X my e either onneted or disonneted. As n illustrtion of spes otined in this wy, onsider the following exmples: Exmple The disk n e otined from tringle with two lels ( nd ) nd orienttions on the edges s shown in the Figure 9 elow. Exmple As we hve seen in Figure 1 the sphere n e otined from squre with to lels nd orienttions on the edges.
9 5. OPERATIONS ON LABELING SCHEMES 9 X = homeomorphism Figure 9. The Disk Exmple In Figure 10 we illustrte the ft tht sine we llow X to e otined y glueing the edges of more thn one polygonl regions, it follows tht X is not neessrily onneted. d e d d f e d Figure 10. X n e onneted or disonneted. Finlly, in order to keep trk of the orienttions of the edges long with the lels, we will now introdue the notion of leling sheme. Let e k is n edge of P with lel ik. If e k is oriented from p k 1 to p k, then we put en exponent +1 on ik. If e k is oriented from p k to p k 1, then we put n exponent 1 on ik. Then P, its lels, nd the orienttions on its edges is totlly speified up to homeomorphism whih respets the quotient spe X y the symol w = ǫ 1 i1 ǫ 2 i2 ǫn i n, ǫ i = ±1. Definition The symol w = ǫ 1 i1 ǫ 2 i2 ǫn i n will e lled leling sheme for P with respet to its lels nd orienttions. InFigure11resomeexmplesofhowtogokndforthfrom(disjointunionof)polygonl regions with lels nd orienttions to leling shemes. 5. Opertions on Leling Shemes It is importnt to note tht we re interested in the quotient spe X otined from polygonl region y gluing its edges nd not on the leling sheme itself. With this in mind, we will now introdue some opertions we n perform on the leling sheme (or equivlently on the polygonl region) whih will leve the resulting quotient spe unhnged. I) Cutting: The opertion of utting is desried t the level of leling shemes s follows. Suppose tht w = ǫ 1 i1 ǫp i p ǫ p+1 i p+1 ǫn i n is leling sheme nd let e
10 10 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES Leling Sheme 1 w = 1 Reonstruting the Polygonl Region d Leling Sheme Reonstruting the Polygonl Region 1 1 w = w = d Figure 11. Leling Shemes. lel whih does not pper elsewhere in the sheme. Then we my reple w y pir of leling shemes w 1 = ǫ 1 i1 ǫp i p, nd w 2 = 1 ǫ p+1 i p+1 ǫn i n. For geometri interprettion see the Figure 12. II) Glueing: The reverse opertion of utting is known s glueing. In terms of the leling sheme it n e desried s follows: If w 1 = ǫ 1 i1 ǫp i p, nd 1 ǫ p+1 i p+1 ǫn i n re leling shemes, nd the lel only ppers where it is indited ove, then we my reple w 1 nd w 2 y the leling sheme w = ǫ 1 i1 ǫp i p ǫ p+1 i p+1 ǫn i n. CUTTING GLUEING P Q1 Q2 Figure 12. Cutting & Glueing Before we go on with the desription of the opertions, let us tke smll rek to write down more formlly the result of utting nd glueing:
11 5. OPERATIONS ON LABELING SCHEMES 11 Proposition Suppose tht X is otined y glueing the edges of n polygonl regions with leling sheme w 1 = y 0 y 1,w 2,...,w n. Let e lel tht does not pper in the sheme. If oth y 0 nd y 1 hve length t lest 2, then X n lso e otined y n+1 polygonl regions with leling sheme y 0, 1 y 1,w 2,...,w n. Exerise 1.5. The purpose of this exerise is to prove the proposition ove. Denote y P 1,...,P n the originl n polygonl regions nd y Q 0,Q 1,P 2,...P n the n+1 polygonl regions otined y utting P 1. Denote lso y X = ( n i=1 P i)/ the spe otined y glueing the edges efore utting, nd y Y = (Q n 0 Q1 i=2 P i)/ the spe otined fter performing the utting opertion. Consider the ovious mp n n Φ : Q 0 Q1 P i P i. Show tht: (1) Φ indues well defined mp ϕ : Y X, i.e., if q q, then Φ(q) Φ(q ). (2) ϕ is ontinuous (use the definition of the quotient topology). (3) ϕ is injetive, i.e., if Φ(q) Φ(q ), then q q. (4) ϕ is surjetive. (5) X nd Y re oth ompt nd Husdorff. (6) ϕ is homeomorphism. i=2 With only the opertions of utting nd gluing we n now esily understnd how to onstrut the onneted sums of tori (nd projetive spes) out of polygonl regions with identifition on the orders: Exmple 1.18 (The Doule Torus T 2 ). Let P e the 8sided polygonl region with leling sheme w = 1 1 d 1 d 1. In order to see tht X = P/ is homeomorphi to doule torus, we will pply the utting nd glueing opertions desried ove (see Figure 13). Thus we first ut P into two 5sided polygonl regions Q 1 nd Q 2, with leling shemes w 1 = 1 1 e nd w 2 = e 1 d 1 d 1 respetively. Now, it is ler tht fter identifying the verties of Q 1 orrespond to the endpoints of e, we otin the usul representtion of the torus s quotient of the unit squre, ut with n open disk removed. The edge e then eomes the order of the open disk. The sme is oviously true lso for Q 2. Thus, if we now pply the glueing opertion, wht we otin is the quotient of two opies of the torus, oth with n open disk removed y identifying the order of the disk. This is preisely the onstrution of the onneted sum. Exerise 1.6. ShowthtT g isotinedfrom4gsidedpolygonl regionwithlelingsheme i=1 w = ( ) ( g g 1 g 1 g ). Exmple A similr rgument s the one presented ove shows tht for h > 1, P h n e otined from 2hsided polygonl region with leling sheme w = ( 1 1 ) ( h h ). We now ontinue to desrie the rest of the opertions tht my e performed on the leling sheme. We suggest tht you onvine yourself tht eh of these opertions leve the quotient spe unhnged.
12 12 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES Cut The doule torus otined from n otogone y identifying its sides s shown y the lels: g l u e the sides Glueing Figure 13. Conneted Sum of Two Tori. Cut 2 2 Glue P # P Figure 14. Conneted Sum of Two Projetive Spes. Definition Let w 1,w 2,...,w n e leling sheme nd let y e string of lels tht ppers in the leling sheme (it my pper in more tht one ple of the sheme). We will sy tht y is removle string if
13 5. OPERATIONS ON LABELING SCHEMES 13 (1) ll lels of y re distint, i.e., y = ǫ 1 i1 ǫ k ik with ip iq for ll i p i q, nd (2) the lels of y do not pper elsewhere (outside of y) in the leling sheme, i.e., for ll 1 p k, if ip ppers in the leling sheme, then it elongs to the string y. III) Unfolding Edges: If y is removle string of leling sheme, then we my reple y y lel tht does not pper elsewhere in the sheme. Geometrilly, this n e interpreted s repling sequene of edges ( folded line segment ), y single edge ( line segment) (See Figure 15). IV) Folding Edges: The reverse opertion to unfolding edges is tht of folding edges desried y: reple ll ppernes of single lel y removle string of lels. 1 3 UNFOLDING 2 2 FOLDING 3 1 Figure 15. Fold/Unfold V) Reversing Orienttions: We my hnge the the sign of the exponent of ll ourrenes of single lel in the leling sheme. In order to understnd why the quotient spe is left unhnged, rell tht we re identifying the points on two oriented edges with the sme lel y mens of positive liner mp. Note tht if the orienttion on oth edges re reversed, the identifition remins unhnged. VI) Cyli Permuttion: It is ler tht if insted of writing the leling sheme of polygonl region y strting with the lel on the edge e 1, we deide to strt with the lel on different edge nd then ontinue in the sme ounterlokwise diretion, then the quotient spe X is unhnged. We my think of this s performing rottion on the polygonl region. The effet on the leling sheme is to tke yli permuttion of its lels ǫ 1 i1 ǫn i n ǫ 2 i2 ǫn i n ǫ 1 i1. VII) Flip: We my reple leling sheme y its forml inverse: w = ǫ 1 i1 ǫn i n ǫn i n ǫ 1 i 1. Geometrilly this orresponds to flipping the polygonl region s in Figure 16 (nd then performing yli permuttion if neessry). Remrk The opertions of permuttion nd flipping should e thought of s instnes of the sme phenomen. If my pply ny Euliden trnsformtion of the plne(i.e., trnsltions, refletions nd rottions) to our originl polygonl region, the resulting ojet will e gin polygonl region whose quotient spe is homoemorphi to the originl one. Finlly, for ompleteness nd lso for further referene, we desrie two opertions tht re otined y omposing the opertions of utting/glueing with tht of folding/unfolding:
14 14 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES FLIP (UN)FLIP Figure 16. Flip/Unflip VIII) Cnel: We my reple leling sheme of the form y 0 1 y 1 y y 0 y 1 provided tht does not pperelsewhere in theleling sheme, nd othy 0 nd y 1 hve length t lest 2. Geometrilly, this opertion is represented y the sequene of digrm in Figure 17. IX) Unnel: Under the sme onditions s ove, we my reverse the opertion of neling y repling sheme y 0 y 1 y the leling sheme y 0 1 y 1, s indited in Figure 17. It should ler tht the opertions ove leve the quotient spe X unhnged. Thus, it is nturl to pose the following definition: Definition Two leling shemes re equivlent if one n e otined from the other y pplying the opertions (I)  (IX) desried ove. Exerise 1.7. Show tht this defines n equivlene reltion on the set of ll leling shemes. Exmple We hve seen tht the Klein ottle is the quotient of the unit squre y the identifition whose leling sheme is 1. Let us prove tht the Klein ottle is homeomorphi to the onneted sum of two projetive spes: 1 & 1 1 (utting) & 1 1 (permuting nd flipping) (glueing) (neling) (permuting). 6. Geometri Surfes In this setion we will onsider surfes whih re otined from polygonl region y identifying it edges in pirs. We wil then show tht every suh surfe is homeomorphi to one in the list given in Theorem 1.9. Definition A ompt nd onneted topologil surfe X is lled geometri surfe if it n e otined from polygonl region y glueing its edges in pirs.
15 6. GEOMETRIC SURFACES 15 y0 y1 y0 y1 CUT GLUE UNFOLD FOLD GLUE y0 y1 y0 y1 CUT Figure 17. Cnel/Unnel The reminder of this setion will e dedited to proving the following theorem: Theorem 1.25 (Clssifition of Geometri Surfes). Let X e geometri surfe. Then X is homeomorphi to one of the following: S 2, T g, or P h (for some g,h N). The ide of the proof is to onsider leling shemes whih give rise to geometri surfes (known s proper leling shemes) nd then to show tht ny proper leling sheme n e put into norml form y mens of the opertions introdued in the lst setion. Definition A leling sheme w 1,...,w m (for m polygonl regions) is lled proper leling sheme if eh lel ppers extly twie in the sheme. Remrk We note tht if we strt with proper leling sheme, then y pplying ny of the opertions introdued in the preeding setion gives rise to nother proper leling sheme. We n now restte Theorem 1.25 into more lgeri form: Theorem 1.28 (Norml Forms of Proper Leling Shemes). Let w e proper leling sheme of length greter or equl to 4 (of single polygonl region). Then w is equivlent to one of the following leling shemes: (1) 1 1,
16 16 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES (2), (3) ( ) ( g g 1 g 1 g ), or (4) ( 1 1 )( 2 2 ) ( h h ). Remrk Of ourse, in the list ove (1) is sphere, (2) is projetive spe, (3) is onneted sum of tori, nd (4) is onneted sum of projetive spes. The first step in the proof of Theorem 1.28 is to distinguish etween two lsses of proper lelings tht will then e treted seprtely: Definition Let w e proper leling sheme for single polygonl region. If every lel of w ppers one with n exponent +1 nd one with exponent 1 we sy tht w is of torus type. Otherwise, we sy tht w is of projetive type. We egin y deling with leling shemes of projetive type: Proposition Let w e leling sheme of projetive type. The w is equivlent to leling sheme of the following form: w ( 1 1 ) ( k k )w 1, where w 1 is leling sheme of torus type. The proof of this proposition will follow from the following lemm: Lemm If w is proper leling sheme of the form w = [y 0 ][y 1 ][y 2 ], where eh [y i ] is string of lels (whih my e empty), the w is equivlent to leling sheme of the form w [y 0 y 1 1 y 2]. Proof. We seprte the proof into two ses: Cse 1: [y 0 ] =. In this se w = [y 1 ][y 2 ]. If [y 1 ] is empty, then we re done. If [y 2 ] is empty, the we proeed s follows: w = [y 1 ] 1 [y 1 1 ] 1 (flipping) 1 1 [y 1 1 ] (permuting) [y1 1 ] (reversing orienttion of ). If oth [y 1 ] nd [y 2 ] re not empty, the we pply the opertions desried in Figure 18. Cse 2: [y 0 ]. Agin we exlude the most trivil se first. If oth [y 1 ] nd [y 2 ] re empty, then w = [y 0 ] nd permuttion rings w to the desired form. Assume now tht either [y 1 ]
17 6. GEOMETRIC SURFACES 17 y2 y2 CUT y1 y1 ROTATE AND FLIP y1 y2 y1 y2 GLUE FLIP y2 y1 y2 y1 REVERSE ORIENTATION OF AND RELABEL Figure 18. Cse 1 or [y 2 ] re nonempty. Then: w = [y 0 ][y 1 ][y 2 ] [y 0 ] & 1 [y 1 ][y 2 ] (utting) [y 1 0 ] 1 1 & [y 2 ] 1 [y 1 ] (flipping nd permuting) [y 1 0 ] 1 [y 2 ] 1 [y 1 ] (glueing nd neling) 1 [y 2 ] 1 [y 1 y 1 0 ] (permuting) 1 1 [y2 1 y 1y0 1 ] (se 1) [y 0 y 1 1 y 2] (flipping) [y 0 y 1 1 y 2] (permuting nd releling).
18 18 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES Exerise 1.8. Write the lgeri sequene of rguments presented in the proof of se 1, nd mke digrms to desrie the geometri sequene of rguments presented in se 2 of the proof. Proof (of Proposition 1.31). Let w e leling sheme of projetive type. Then there is t lest one lel of w whih ppers twie with the sme sign. Thus, w = [y 0 ][y 1 ][y 2 ] nd y using the lemm, we otin tht w is equivlent to [y 0 y1 1 y 2]. If [y 0 y1 1 y 2] is of torus type then we re done. Otherwise, there is lel is [y 0 y1 1 y 2] whih ppers twie with the sme sign, nd thus we my ssume tht w is equivlent to w [z 0 ][z 1 ][z 2 ]. We pply the lemm gin, this time to the leling sheme [z 0 ][z 1 ][z 2 ], to otin tht w [z 0 z 1 1 z 2]. If [z 0 z1 1 z 2] is of torus type we re done. Otherwise we ontinue this proess whih will end s soon s we hve put w into the desired form w ( 1 1 )...( k k )w 1 with w 1 leling sheme of torus type. Remrk We n onlude from Proposition 1.31 tht if w is proper leling sheme, then either: (1) w is of torus type, or (2) w is of the form ( 1 1 )...( k k )w 1 with w 1 leling sheme of torus type, or (3) w is of the form ( 1 1 )...( k k ) in whih se we re done (X is onneted sum of projetive spes). We must now exmine how to redue w to simpler form when w is of the form (1) or (2). Exerise 1.9. Show tht if w is proper leling sheme of length 4, then w must e equivlent to one of the following leling shemes:, 1 1, 1 1 Fromnowonwessumethtw hslengthgreterthen4, ndmoreover, thtitisirreduile, i.e., it does not ontin ny djent terms hving the sme lel, ut opposite signs (in whih se we ould perform the opertion of neling to redue the length of w). In this se we hve the following lemm: Lemm Suppose tht w is proper leling sheme of the form w = w 0 w 1, where w 1 is n irreduile sheme of torus type. Then w is equivlent to sheme of the form w 0 w 2, where w 2 hs the sme length s w 1, nd hs the form: where w 3 is of torus type or is empty. w 2 = 1 1 w 3, Proof. We will divide the proof of this lemm into severl steps. Step 1: We my ssume tht w is of the form w = w 0 [y 1 ][y 2 ][y 3 ] 1 [y 4 ] 1 [y 5 ], where some of the strings of lels [y i ] my e empty. To see this we proeed s follows. Let e the lel in w 1 whose ourrenes re s lose s possile (with the miniml mount of lels in etween them). If ppers first with n
19 6. GEOMETRIC SURFACES 19 exponent 1, then we revert the orienttion of oth ppernes of. Next, let e ny lel in etween nd 1. Then, sine nd 1 re the losest lels to eh other in w 1, it follows tht either 1 ppers fter 1, in whih se we re done, or 1 ppers in front of, in whih se we simply exhnge the lels of nd. Step 2: (1 st Surgery) w is equivlent to w w 0 [y 2 ][y 3 ] 1 [y 1 y 4 ] 1 [y 5 ]. We my ssume tht [y 1 ] (or else there is nothing to prove). Then, we perform the following opertions: w = w 0 [y 1 ][y 2 ][y 3 ] 1 [y 4 ] 1 [y 5 ] [y 2 ][y 3 ] 1 [y 4 ] 1 [y 5 ]w 0 & 1 [y 1 ] (permuting nd utting) [y 4 ] 1 [y 5 ]w 0 [y 2 ][y 3 ] 1 & 1 [y 1 ] (permuting) [y 4 ] 1 [y 5 ]w 0 [y 2 ][y 3 ] 1 [y 1 ] w 0 [y 2 ][y 3 ] 1 [y 1 y 4 ] 1 [y 5 ]. (glueing) (permuting nd releling) Step 3: (2 nd Surgery) w is equivlent to w w 0 [y 1 y 4 y 3 ] 1 1 [y 2 y 5 ]. First of ll, ssume tht w 0,y 1,y 4, nd y 5 re ll empty. Then w [y 2 ][y 3 ] 1 1 nd the result follows y permuting nd releling. Now ssume tht t lest one of the strings w 0,y 1,y 4, or y 5 is nonempty. Then, we n perform the following sequene of opertions: w w 0 [y 2 ][y 3 ] 1 [y 1 y 4 ] 1 [y 5 ] [y 2 ][y 3 ] 1 & 1 [y 1 y 4 ] 1 [y 5 ]w 0 (permuting nd utting) [y 3 ] 1 [y 2 ] & 1 [y 5 ]w 0 1 [y 1 y 4 ] (permuting) [y 3 ] 1 [y 2 ][y 5 ]w 0 1 [y 1 y 4 ] w 0 1 [y 1 y 4 y 3 ] 1 [y 2 y 5 ] w 0 [y 1 y 4 y 3 ] 1 1 [y 2 y 5 ]. (glueing) (permuting) (releling) Step 4: (3 nd Surgery) w is equivlent to w w [y 1 y 4 y 3 y 2 y 5 ].
20 20 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES We perform the following sequene of opertions: w w 0 [y 1 y 4 y 3 ] 1 1 [y 2 y 5 ] [y 1 y 4 y 3 ] 1 & 1 1 [y 2 y 5 ]w 0 (permuting nd utting) 1 [y 1 y 4 y 3 ] & 1 [y 2 y 5 ]w [y 1 y 4 y 3 ][y 2 y 5 ]w 0 1 w [y 1 y 4 y 3 y 2 y 5 ]. (permuting) (glueing) (permuting nd releling) The following grph summrizes the results tht we hve otined so fr: 1 1 or 1 1 Done! length =4 Lemm Torus Type 1.34 ( length ) ( g g 1 g 1 g ) Done! (1) w w = ( 1 1 ) ( k k )w 1 Lemm ( (2) with w 1 of torus type ) ( k k )( ) ( m m 1 m 1 m )??? (3) length =4 1 Done! w = ( 1 1 ) ( h h ) Done! Remrk The three rrows oming out of w orrespond to remrk 1.33, while the ses where the length of w is equl to 4 follow from exerise 1.9. Thus, in order to onlude the proof of Theorem 1.28 we need to desrie wht the onneted sum of tori nd projetive spes orrespond to, i.e., to redue w = ( 1 1 ) ( k k )( ) ( m m 1 m 1 m ) to its norml form. This follows from the following lemm: Lemm If w = w 0 ()( 1 1 )w 1 is proper sheme, then w w 0 ()w 1. Proof. We will mke use repetedly of Lemm 1.32 whih sttes tht [y 0 ][y 1 ][y 2 ] [y 0 y 1 1 y 2].
21 7. TRIANGULATED SURFACES 21 To prove the lemm we onsider the following sequene of opertions: w = w 0 ()( 1 1 )w 1 ()[][] 1 [w 1 w 0 ] (permuting) [][][w 1 w 0 ] (Lemm 1.32) [][][w 1 w 0 ] [][ 1 ][w 1 w 0 ] (regrouping the terms) (Lemm 1.32 nd regrouping the terms) w 0 ()w 1. The result then follows y releling the terms. (Lemm 1.32 nd permuting) We n thus onlude from the Lemm, y pplying it severl times if neessry, nd then releling, tht ( 1 1 ) ( k k )( ) ( m m 1 m 1 m ) ( 1 1 ) ( k+2m k+2m ). This finishes the proof of Theorem Exerise Throughout this setion we hve impliitly desried n lgorithm to redue ny proper leling sheme to one in the norml form of Theorem Write down this lgorithm expliitly. Exerise Use the lgorithm you developed in the exerise ove to determine whih surfe orresponds to the following leling shemes: (1) 1 1 (2) 1 (3) 1 dd 1 (4) d d 1 (5) dd (6) d d 1 7. Tringulted Surfes In this setion we will introdue the notion of tringultion on ompt Husdorff topologil spe. We will then show tht: Theorem Any tringulted ompt onneted surfe n e otined from single polygonl region y identifying its edges with respet to proper leling sheme. Theorem 1.9 will then follow from Theorem 1.28 nd the following result whih we will not prove now (ut we intend to ome k to it if time permits). Theorem Every ompt onneted surfe is tringulle (i.e., n e tringulted). We egin with the definition of tringultion: Definition Let X e ompt Husdorff topologil spe. A urved tringle in X is suspe A together with homeomorphism h : T A, where T is tringulr region in the plne. If v T is vertex, then h(v) is lled vertex of A. Similrly, if e T is n edge, then h(e) is lled n edge of A.
22 22 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES A tringultion of X is olletion A 1,...,A n of urved tringles of X whih over X, i A i = X, nd suh tht if A i A j, then either (1) A i A j = {v} is vertex, or (2) A i A j = e is n edge, nd furthermore, the mp h 1 j h i whih mps the edge h 1 i (e) of T i to the edge h 1 j (e) of T j is liner homeomorphism. Remrk By ondition (2) in the definition ove we men the tht if we pik n orienttion on h 1 i (e) then h 1 j h i indues hoie of orienttion on h 1 j (e) for whih h 1 j h i : h 1 i (e) h 1 j (e) eomes positive liner mp. For n exmple of tringultion of the sphere, see Figure 19. LABELING SCHEME TRIANGULATION Figure 19. Exerise For eh of the following spes exhiit n expliit tringultion. (1) A torus (2) ylinder (3) one (4) projetive spe (5) Möius nd, (6) Klein ottle
23 7. TRIANGULATED SURFACES 23 A tringultion {A 1,...,A n } of ompt Husdorff spe X indues leling sheme, whih will e lled the leling sheme of the tringultion, s follows: Polygonl Regions: For eh urved tringle A i, let h i : T i A i e the orresponding homeomorphism defined on the tringulr region T i. The polygonl region we will onsider is the disjoint union of the tringles T i s. Orienttion on Edges: Let e X e n edge ppering in the tringultion (i.e., it is the edge of t lest one of the urved tringles). Let v nd w e the verties t the endpoints of e. Choose n orienttion on e y delring it to go from v to w. Then if h 1 i (e) is n edge of T i, we orient it from h 1 i (v) to h 1 i (w). Lels: Let Λ = {e X : e is the edge of (t lest) one of the urved tringles} e the set of edges of the tringultion. Then if h 1 i (e) is n edge of T i, we ssoite to it the lel e Λ. Exmple Figure 19 exhiits tringultion on the sphere, nd lso the leling sheme of the tringultion. Exerise For eh of the spes in exerise 1.12, determine the leling sheme of the tringultion. Proposition If X is ompt tringulted surfe, then the leling sheme of the tringultion is proper. Proof. We need to show tht eh lel ppers extly twie in the leling sheme. The rguments needed to do this re intuitively ler. However, the esiest wy to mke them preise is y using the notion of fundmentl group. Thus, we will sketh the proof now, ut leve the detils s n exerise tht should e done fter the fundmentl group is introdued. The first step is to show tht eh lel pper t lest twie in the leling sheme. Thus, ssume tht lel ppers only one. This mens tht there is n edge e whih is the edge of only one urved tringle. Exerise 1.14 ellow shows tht this nnot hppen. The intuitive ide is tht if x e, then y removing x we will not rete hole in X, ut on the other hnd, if we remove ny point from n open set in R 2, then we do rete hole The next step is to show tht there is t most two ppernes of eh lel. Agin this will follow y removing one point trik. If lel ppers more thn twie, then there re more thn two tringles whih interset in single edge. Intuitively, this will men tht there is multiple orner whih nnot e smoothened into n open suset of R 2 (see figure 20). The preise rgument is given in exerise 1.15 ellow. Exerise 1.14 (To e done fter the definition of homotopy). Let T e tringle nd x T e point in one of the edges of T nd let U e ny neighorhood of x. Show tht ny loop in U x is homotopi to onstnt pth. Conlude tht x does not hve ny neighorhood whih is homeomorphi to n open set of R 2. Exerise 1.15 (To e done fter the Seifert vn Kmpen Theorem). Consider the spe otined y glueing together k tringles long ommon edge e, with k > 2 (Figure 20 shows the se when k = 3). Let x e e point in this ommon edge. Show tht ny neighorhood U of x ontins possily smller neighorhood V U suh tht V x is homotopy equivlent to ouquet of k 1 irles. Conlude y omputing the fundmentl group of V x tht x does not hve ny neighorhood whih is homeomorphi to n open suset of R 2. Proposition If X is ompt tringulted surfe, then X is homeomorphi to the spe otined from T i y glueing its edges ording to the leling sheme of the tringultion.
24 24 1. THE CLASSIFICATION PROBLEM FOR COMPACT SURFACES e Figure 20. Exerise Consider the mp h : T i X otined y putting together ll of the mps h i : T i A i X. Consider the spe X otined y identifying two points p nd q of T i if nd only if h(p) = h(q). Show tht X is homeomorphi to X. Proof. Let us denote Y the quotient spe otined from T i y identifying it edges with respet to the leling sheme of the tringultion. It is n immedite onsequene of the exerise ove, tht h ftors through ontinuous mp f : Y X, i.e., i T i h Y Moreover, sine h is surjetive, it follows tht f is lso surjetive. Thus, in order to prove the proposition, it suffies to show tht f is injetive (euse Y is ompt nd X is Husdorff). Let us denote y [p] Y the equivlene lss with respet to the leling sheme of the tringultion of point p in T i. Assume tht f([p]) = f([q]), for some p q. Then, y definition, it follows tht h(p) = x = h(q). Thus, either x elongs to some edge e of the tringultion on X, in whih se it is ler tht [p] = [q], or x is vertex. In this se, in order to show tht [p] = [q] (so tht f is injetive) we must verify tht the identifition of p with q is fored s onsequene of the identifition of the edges of the tringles T i s (see lso exerise 1.20). Suppose tht A i nd A j interset t vertex v. Wht we need to show is tht we n find sequene A i = A i1,a i2,...a im = A j, suh tht A ik intersets A ik+1 on ommon edge whih ontins v s its endpoint (s illustrted in Figure 21). This is the ontent of the following exerise. Exerise Given v, define two urved tringles A i nd A j with vertex v to e equivlent if we n find sequene A ik s ove. Use the remove the one point trik to show tht if there is more tht one equivlene lss of urved tringles with vertex v, then v does not hve ny neighorhood in X whih is homeomorphi to n open set of R 2, nd thus X is not surfe. We now re redy to finish the proof of Theorem Wht we will show is tht we my glue the tringles T i together in order to otin the desired polygonl region. In ft, strt y f X.
25 7. TRIANGULATED SURFACES 25 Aj = Aim Aim1 Ai2 Ai = Ai1 Figure 21. hoosing one of the tringles, sy T 1. If T i is nother tringle whih hs lel on one of its edges whih is equl to lel of T 1, then (fter possily flipping T i ) we my glue oth tringles together. The effet of this is to redue the originl numer of tringles y two, t the expense of dding one polygonl region (whih in this se hs 4 sides) whih we denote y P 1. Next, we look t the edges of P 1. If one of the tringles T j, with j 1,i, hs lel equl to one of the lels of P 1, then, fter flipping T j if neessry, we my glue it to P 1 otining in this wy new polygonl region P 2. We ontinue this proess s long s we hve two polygonl regions ontining edges tht hve ommon lel. At some point we will reh sitution where either we otin the polygonl region P n+1 tht we were looking for, or we otin more then one disonneted polygonl region in whih none of the lels ppering in one of them pper lso in the other region. However, it is esy to see tht this nnot hppen, for in this se the quotient spe X will neessrily e disonneted. Exerise Determine the spe otined from the following leling shemes: (1),de,ef,df. (2),,def,dfe 1. Exerise Show tht the projetive spe P 2 n e otined from two Moius nds y glueing them long there oundry. Exerise Let X e the spe otined from sphere y identifying its north nd south poles (X is not surfe). Find tringultion on X suh tht the leling sheme of the tringultion determines sphere (i.e., the surfe otined y glueing the edges of the tringles with respet to the leling sheme of the tringultion is homeomorphi to sphere). Conlude tht two nonhomeomorphi ompt Husdorff spes n hve tringultions whih indue the sme leling sheme. (We remrk tht this exerise gives n exmple of tringulted spe for whih the mp f from the proof of Proposition 1.43 is not injetive.)
26 CHAPTER 6 Atthing ells 1. Cells Definition 6.1. Let X e Husdorff topologil spe. An open nell in X is suspe e X together with homeomorphism he: Dn e X. It is lled n nell if he extends to ontinuous mp h e : D n X. We ll h e the defining mp of the nell, nd we lso sy tht e is n nell with defining mp h e,orthte is the imge of the nell. The ell oundry of e is defined y: ell (e) :=e e, where e is the losure of e in X. Thehrteristimpχ e of e is defined s the restrition χ e := h e S n 1 : S n 1 X. Remrk 6.2. Sine Dn is dense in D n (hene eh point x D n n e written s the limit of sequene of points in the interior) nd h e must e ontinuous (hene preserves the limits), given he, theextensionh e will e unique. In onlusion, nnell in the spe X is just suspe e X together woth ontinuous mp h e : D n X whih, when restrited to Dn,ishomeomorphismetween D n nd e. Here re some simple exmples (more will ome lter). Exmple 6.3. Note tht, for n =0,0ellinX is the sme thing s point of X. An interesting 1ell is e = S 1 {(1, 0)} whih is 1ell in S 1 with defining mp h e (t) =(os(πt),sin(πt)). Exmple 6.4. Vrious ells inside the sphere re shown in Figure 1. Exmple 6.5. In generl, n open nell my fil to e n nell. This is lredy ler when n =1ndX = R. A suspe e R is n open nell (with some defining mp) if nd only if e =(, ) isnopenintervl(with nd  relnumersorplus/minusinfinity). Indeed, ny suh open ell will e onneted suspe of R hene it must e n intervl. From the disussions inthe previous hpter, it must e n open intervl. On the other hnd, if e =(, ) isoundedintervl(hene, re finite), then e (together with some defining mp) is 1ell. However, unounded intervls nnot e mde into 1ells. This will lso follow from the next proposition (whih imply tht the losure e of e in X must e ompt), ut let s hek it diretly here for the open 1ell e =(1, ) togetherwiththe defining mp: he: ( 1, 1) R,t 2 t +1 83
27 84 6. ATTACHING CELLS 0ell 2ell (the sphere minus the north pole) 0ell 1ell 2ells six 0ells eight 1ells (the qurters of the irles) four 2ells (the qurter spheres) Figure 1. This nnot hve ontinuous extension to [ 1, 1] euse x n = 1+ 1 n onverges to 1, ut he (x n )=2n does not hve finite limit. Exerise 6.1. Let X e the onepoint omptifition of the spe otined from D 2 y removing two points on its oundry. Desrie X in R 3 nd show tht it is the losure of 2ell. Here re the min properties of nells. Proposition 6.6. If e X is n nell with defining mp h e : D n X, then h e (D n )=e, h e (S n 1 )= ell (e). Moreover, s mp from D n into e, h e is topologil quotient mp. Proof. Sine h e is ontinuous, we hve h e (B) h e (B) for ll B D n (prove this!). Choosing B = Dn,weotinhe (D n ) e. On the other hnd, e = h e ( Dn ) he (D n ), nd h e (D n )isompt(stheimgeofomptyontinuousmp),henelosedinx (sine X is Husdorff). This implies e h e (D n ). Sine the opposite inlusion hs een proven, we get h e (D n )=e. Wenowprove h e (S n 1 )=e e. We first show the inverse inlusion: for y e e, ythefirstprt,y = h e (x) forsomex D n nd, sine e = h e ( Dn ), x nnot e in D n ;henex S n 1.Wenowprovethediretinlusion. So let y = h e (x) withx S n 1,ndwewnttoprovethty / e. Assume the ontrry, i.e. y = h e (x )withx Dn.Sinex nd x re distint, we find U, V D n opens (in D n )suhtht x U, x V,U V =. We my ssume tht V Dn. Sine he is homeomorphism, h e (V )isopenine, henelsoin e. Sine h e : D n e is ontinuous, h 1 e (h e (V )) will e open in D n ;utitontinsx, hene
28 2. ATTACHING ONE ncell 85 we find ɛ>0suhtht D n B(x, ɛ) h 1 e (h e(v )) (where the ll is with respet to the usul Euliden metri). Sine x U nd U is open in D n,wemyhooseɛ so smll so tht D n B(x, ɛ) U. Pik up n element z Dn B(x, ɛ). By the inlusion ove, we find z V suh tht h e (z) = h e (z ). But then oth z nd z re in Dn hene we must hve z = z.henez V.Butz U hene we otin ontrdition with U V =. Finlly,h e is topologil quotient mp s ontinuous surjetion from ompt spe to Husdorff spe. 2. Atthing one nell Definition 6.7. Let X e Husdorff spe nd A X losed. We sy tht X is otined from A y tthing n nell if there exists n nell e X (with some defining mp h e )suhtht X = A e, A e =. Exmple 6.8. It is ler tht the nll D n is otined from D n = S n 1 y tthing n nell (the defining mp eing the identity mp). From exmple 6.3, we see tht S 1 n e otined from point (whih is 0ell!) y tthing 1ell. In generl, S n n e otined from point y tthing n nell (see lso exmple 6.13 elow). To tret exmples treted in the previous letures suh s the torus, the Moeius nd et, the following lemm is very useful. Lemm 6.9. Let X = D n or X =[0, 1] n,ndlety e quotient of X otined y gluing (ertin) points on the oundry (X). Assume tht Y is Husdorff, Let π : X Y e the quotient mp, nd let B = π( (X)) (i.e. the spe otined from (X) y the originl gluing). Then Y is otined from B y tthing n nell. Proof. Sine D n nd X =[0, 1] n re homeomorphi y homeomorphism whih preserves their oundry, we my ssume tht X = D n. We then hoose s defining mp for the nell the quotient mp h : D n Y, hene the nell will e e := h( Dn ). Note tht B = π(s n 1 ). Clerly, Y = B e. Next, sine no element on the oundry of D n is equivlent (identified) with n interior element, we hve B e =. Next,sinenotwointeriorpointsofD n re equivlent, the restrition of h to Dn is injetive, hene h D n : Dn e is ontinuous ijetion. We still hve to prove tht this mp is homeomorphism, nd for this it is enough to show tht it sends losed sets to losed sets. So, let F e losed in Dn ;write F = Dn K with K D n losed. Then K is ompt (s losed inside ompt), hene h(k) will e ompt. Sine Y ws ssumed to e Husdorff, we dedue tht h(k) is losed in Y.So,toshowthth(F )islosedine, itsuffiestoshowtht h( Dn K) =e h(k).
29 86 6. ATTACHING CELLS The diret inlusion is ler. For the onverse, note tht if y elongs to the right hnd side we hve y = h(x) ndy = h(x )withx Dn nd x K, utsineinteriorpointsrenoteing identified with ny other points, we must hve x = x,heney elongs to the left hnd side. Remrk The se n = 2 is prtiulrly importnt: mny spes X n e otined from D 2 y identifying ertin prts of D 2 = S 1,ndtheidentifitionneshownonthepiture y leling y letters the prts tht re to e identified. In the quotient B = S 1 / = π(s 1 ), eh letter will pper only one (euse we identified ll the prts leled y the sme letter). When going one time round the irle, we will meet vrious lels tht will give us word whose letters re lels. Reding this word in the spe B desries the hrteristi mp. Exmple Consider the torus s quotient of X =[0, 1] [0, 1] (see Setion 5) of the first hpter, or of D 2. We n pply the previous lemm. The resulting spe B is shown in the piture (Figure 2) nd it onsists of two irles on the torus touhing eh other in one point ( nd on the piture). This spe n e drwn (is homeomorphi to) the spe onsisting of two irles in the plne touhing in one point only ( ouquet of two irles). In onlusion, T 2 n e otined from ouquet of two spheres y tthing 2ell. q p Figure 2. Note tht, ording to the onventions from the previous remrk, the hrteristi mp χ : S 1 S 1 S 1 n e desried symolilly s: nd whih is further pitured in Figure 3. χ = 1 1 q the desription of the hrteristi mp 1 1 Figure 3.
30 2. ATTACHING ONE ncell 87 Exmple AsimilrdisussionppliestotheMoeiusnd(setion4inthefirst hpetr). The resulting spe B, showninfigure4) ismdeofthethreesegments, nd on the Moeius nd (hene B is the oundry of the Moeius nd plus the segment ). The spe B lone (s topologil spe itself) is homeomorphi to the spe onsitsting of the irle S 1 together with segment joining the north nd the south pole (see the piture). In onlusion, the Moeius nd n e otined from this spe y tthing 2ell., nd together form spe B homeomorphi to: n n s s Figure 4. The hrteristi mp n e desried in symols s: χ = (on the piture, put the diretions for nd so tht this formul is orret!). This, s mp defined on S 1,isfurtherpituredinFigure5. B the diretion of the pth followed y the hrteristi mp Figure 5. Exerise 6.2. Show tht S 2 n e otined y tthing 2ell to the intervl [0, 1]. Desrie t lest two different wys of relizing this (nd explin the hrteristi mps). (Hint: see Figure 12 nd Figure 13 in Chpter 1).
31 88 6. ATTACHING CELLS Exerise 6.3. Do the sme for P 2 (see Setion 7 of the first hpter). Exerise 6.4. Do the Klein ottle. Exerise 6.5. Show tht P 2 n e otined y tthing 2ell to the the Moeius nd. Exmple Consider the sphere S n. We lredy know tht S n = D n / D n,i.e. S n n e otined from D n y gluing ll the points of D n = S n 1 together into single point (See Setion 3 nd Figure 11 in the first hpter nd Exmple 3.24.iii in the previous hpter). Hene we n pply the previous lemm. The resulting spe B lerly onsists of one point only, hene S n n e otined from one point y djoining n nell (the tthing mp is, of ourse, just the onstnt mp). Exmple Consider the projetive spe P n. We rell tht P n is equl to the spe D n / otined y identifying (gluing) the ntipodl points on D n = S n 1 (see e.g. Exerise 1.27 in the first hpter). Hene, we n pply the previous lemm. The spe B resulting from the lemm will e S n 1 /  thespeotinedfroms n 1 y gluing its ntipodl points. This is just nother desription of the projetive spe nd we see tht B is equl to P n 1. In onlusion, P n n e otined from P n 1 y djoining n nell. Exerise 6.6. In the previous exmple, is equl to relly mens tht is homeomorphi to. Vi the sequene of is equl to tht is used in the exmple, it ppers tht P n 1 is suspe of P n.writeoutthe equlities (i.e.homeomorphisms)thtweusedtoonludethtthewy we see P n 1 s suspe of P n is vi the nonil inlusion P n 1 P n whih ssoites to line l inside R n the line inside R n+1 = R n R given y l {0} = {(x, 0) : x l} R n The hrteristi mp Note tht, using Proposition 6.6, we see tht the hrteristi mp of e n e viewed s mp χ e : S n 1 A. The role of the hrteristi mp is tht it desries the wy tht A nd e intert inside X, or, equivlently, tht it desries the reltionship etween the inlusion i : A X nd the defining mp h : D n X. Thisreltionshipish S n 1 = i χ e whih should e interpreted s ommuttive digrm: S n 1inlusion χ e A i But the rel reson tht the hrteristi mp χ e is importnt omes from the ft tht X n e reonstruted from the suspe A nd the hrteristi mp χ e.theimofthissetionistoprovethe following: Theorem For ny Husdorff spe A nd ny ontinuous mp χ : S n 1 A, thereisspe X whih is unique up to homeomorphism, with the property tht X is otined from A y djoining n nell with hrteristi mp equl to χ. We first estilish the following universl property from whih we will e le to dedue unqueness y formlrgument. D n X h e
32 3. THE CHARACTERISTIC MAP 89 Proposition (the universl property) Assume tht X is otined from A y djoining n nell e with defining mp h e : D n X. LetY e nother topologil spe. Then mp f : X Y is ontinuous if nd only if f A : A Y, f h e : D n Y re ontinuous. Moreover, the orrespondene f (f A,f h e ) defines 11 orrespondene etween ontinuous mps f : X Y pirs (f A,f e ) with f A : A Y nd f e : D n Y ontinuous stisfying f A χ e = f e S n 1. S n 1 χ e D n h e i f A X e f f A Y Proof. If f is ontinuous, it is ler tht f A nd f h e re ontinuous. For the onverse, ssume tht f A nd f h e re ontinuous. To show tht f is ontinuous, we will show tht, for B Y losed, f 1 (B) islosedinx. Notetht f 1 (B) A =(f A ) 1 (B),f 1 (B) e = h e ((f h e ) 1 (B)). From the first equlity nd the ontinuity of f A,wededuethtf 1 (B) A is losed in A, hene(sine A is losed in X), lso in X. On the other hnd, (f h e ) 1 (B) islosedind n,heneompt. Then h e ((f h e ) 1 (B)) is ompt inside the Husdorff X, heneitislosedinx. Sinef 1 (B) istheunion of two losed suspes of X (nmely f 1 (B) A nd f 1 (B) e), it is itself losed in X. Thisonludes the proof of the equivlene. For the seond prt we remrk tht, given f A nd f e,theonditions determine f uniquely euse X = A Im(h e ): f A = f A,f e = f h e f(x) =f A (x) ifx A, f(x) =f e (v) ifx = h e (v) e. Moreover, under the ssumption f A χ e = f e S n 1, thepreviousformulsdefinef unmiguously: if x is oth in A nd of type h e (v) withv D n,thenf A (x) =f e (v). Indeed, sine h e (v) A, v must e in S n 1,henewenwritex = χ e (v) ndwenusethessumption. Theontinuityofthempsinvolved follows from the first prt. The ft tht X only depends on A nd χ e is even stronger indited y the following: Corollry Let A e Husdorff spe nd let χ : S n 1 A e ontinuous mp. For i {1, 2}, ssumethtx i is spe whih is otined from A y djoining n nell e i, nd let χ i : S n 1 A e the hrteristi mp of e i. If χ 1 = χ 2,thenX 1 nd X 2 re homeomorphi. Proof. Let χ = χ 1 = χ 2,ndleth i e the mps defining the nell e i.wepplythelstprtofthe previous proposition to X = X 1, Y = X 2 nd to the pir (i 2,h 2 ), where i 2 : A X 2 is the inlusion. We find tht there is one nd only one ontinuous mp f 1,2 : X 1 X 2 ontinuous suh tht f 1,2 i 1 = i 2,f 2,1 h 1 = h 2. These onditions men tht: for A, f 1,2 () =, whilefory h 1 (D n ), writing y = h 1 (x), f 1,2 (y) = h 2 (x). Sine X 1 = A h e1 (D n ), these formuls do define f 1,2 uniquely, ut wht the universl property is telling us is tht f 1,2 is well defined nd it is ontinuous. Of ourse, this ould hve een heked diretly
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