Covering Spaces 33 1 De64257nition of Covering Let topological spaces a continuous map Assume that is surjective and each point of possesses a neighborhood such that the preimage of is a disjoint union of open sets and maps each homeomorphically on ID: 10058
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ChapterVIICoveringSpacesandCalculationofFundamentalGroups33.CoveringSpaces331.De nitionofCoveringLetX,Btopologicalspaces,p:X!Bacontinuousmap.AssumethatpissurjectiveandeachpointofBpossessesaneighborhoodUsuchthatthepreimagep 1(U)ofUisadisjointunionofopensetsV andpmapseachV homeomorphicallyontoU.Thenp:X!Bisacovering(ofB),thespaceBisthebaseofthiscovering,XisthecoveringspaceforBandthetotalspaceofthecovering.NeighborhoodslikeUaresaidtobetriviallycovered.Themappisacoveringmaporcoveringprojection.33.A.LetBbeatopologicalspaceandFbeadiscretespace.ProvethattheprojectionprB:BF!Bisacovering.33.1.IfU0UBandtheneighborhoodUistriviallycovered,thentheneighborhoodU0isalsotriviallycovered.Thefollowingstatementshowsthatinacertainsenseanycoveringlo-callyisorganizedasthecoveringof33.A.33.B.Acontinuoussurjectivemapp:X!Bisacoveringi foreachpointaofBthepreimagep 1(a)isdiscreteandthereexistaneighborhoodUofa 231 232VII.CoversingSpaces andahomeomorphismh:p 1(U)!Up 1(a)suchthatpjp 1(U)=prUh.Here,asusual,prU:Up 1(a)!U.However,thecoveringsof33.Aarenotinteresting.Theyaresaidtobetrivial.Hereisthe rstreallyinterestingexample.33.C.ProvethatR!S1:x7!e2ixisacovering. Todistinguishthemostinterestingexamples,acoveringwithacon-nectedtotalspaceiscalledacoveringinanarrowsense.Ofcourse,thecoveringof33.Cisacoveringinanarrowsense.332.MoreExamples33.D.R2!S1R:(x;y)7!(e2ix;y)isacovering.33.E.Provethatifp:X!Bandp0:X0!B0arecoverings,thensoispp0:XX0!BB0.Ifp:X!Bandp0:X0!B0aretwocoverings,thenpp0:XX0!BB0istheproductofthecoveringspandp0.The rstexampleoftheproductofcoveringsispresentedin33.D.33.F.C!Cr0:z7!ezisacovering.33.2.Riddle.Inwhatsensethecoveringsof33.Dand33.Farethesame?De neanappropriateequivalencerelationforcoverings.33.G.R2!S1S1:(x;y)7!(e2ix;e2iy)isacovering.33.H.Foranypositiveintegern,themapS1!S1:z7!znisacovering.33.3.ProvethatforeachpositiveintegernthemapCr0!Cr0:z7!znisacovering.33.I.Foranypositiveintegerspandq,themapS1S1!S1S1:(z;w)7!(zp;wq)isacovering.33.J.ThenaturalprojectionSn!RPnisacovering. 33.CoveringSpaces233 33.K.Is(0;3)!S1:x7!e2ixacovering?(Cf.33.14.)33.L.IstheprojectionR2!R:(x;y)7!xacovering?Indeed,whyisnotanopeninterval(a;b)Ratriviallycoveredneighborhood:itspreimage(a;b)Ristheunionofopenintervals(a;b)fyg,whicharehomeomorphicallyprojectedonto(a;b)bytheprojection(x;y)7!x?33.4.FindcoveringsoftheMobiusstripbyacylinder.33.5.FindnontrivialcoveringsofMobiusstripbyitself.33.6.FindacoveringoftheKleinbottlebyatorus.Cf.Problem21.14.33.7.FindcoveringsoftheKleinbottlebytheplaneR2andthecylinderS1R,andanontrivialcoveringoftheKleinbottlebyitself.33.8.DescribeexplicitlythepartitionofR2intopreimagesofpointsunderthiscovering.33.9*.Findacoveringofaspherewithanynumberofcrosscapsbyaspherewithhandles.333.LocalHomeomorphismsversusCoverings33.10.Anycoveringisanopenmap.1Amapf:X!YisalocalhomeomorphismifeachpointofXhasaneighbor-hoodUsuchthattheimagef(U)isopeninYandthesubmapab(f):U!f(U)isahomeomorphism.33.11.Anycoveringisalocalhomeomorphism.33.12.Findalocalhomeomorphismwhichisnotacovering.33.13.Provethattherestrictionofalocalhomeomorphismtoanopensetisalocalhomeomorphism.33.14.ForwhichsubsetsofRistherestrictionofthemapofProblem33.Cacovering?33.15.FindanontrivialcoveringX!BwithXhomeomorphictoBandprovethatitsatis esthede nitionofacovering.334.NumberofSheetsLetp:X!Bbeacovering.Thecardinality(i.e.,thenumberofpoints)ofthepreimagep 1(a)ofapointa2Bisthemultiplicityofthecoveringataorthenumberofsheetsofthecoveringovera.33.M.Ifthebaseofacoveringisconnected,thenthemultiplicityofthecoveringatapointdoesnotdependonthepoint. 1Weremindthatamapisopeniftheimageofanyopensetisopen. 234VII.CoversingSpaces Inthecaseofcoveringwithconnectedbase,themultiplicityiscalledthenumberofsheetsofthecovering.Ifthenumberofsheetsisn,thenthecoveringisn-sheeted,andwetalkaboutann-foldcovering.Ofcourse,unlessthecoveringistrivial,itisimpossibletodistinguishthesheetsofit,butthisdoesnotpreventusfromspeakingaboutthenumberofsheets.Ontheotherhand,weadoptthefollowingagreement.Byde nition,thepreimagep 1(U)ofanytriviallycoveredneighborhoodUBsplitsintoopensubsets:p 1(U)=[V ,suchthattherestrictionpjV :V !Uisahomeomorphism.EachofthesubsetsV isasheetoverU.33.16.WhatarethenumbersofsheetsforthecoveringsfromSection332?Inproblems33.17{33.19wedidnotassumethatyouwouldrigorouslyjustifyyouranswers.Thiswillbedonebelow,seeproblems39.3{39.6.33.17.WhatnumberscanyourealizeasthenumberofsheetsofacoveringoftheMobiusstripbythecylinderS1I?33.18.WhatnumberscanyourealizeasthenumberofsheetsofacoveringoftheMobiusstripbyitself?33.19.WhatnumberscanyourealizeasthenumberofsheetsofacoveringoftheKleinbottlebyatorus?33.20.WhatnumberscanyourealizeasthenumberofsheetsofacoveringoftheKleinbottlebyitself?33.21.Constructad-foldcoveringofaspherewithphandlesbyaspherewith1+d(p 1)handles.33.22.Letp:X!Yandq:Y!Zbecoverings.Provethatifqhas nitelymanysheets,thenqp:x!Yisacovering.33.23*.Isthehypothesisof nitenessofthenumberofsheetsinProblem33.22necessary?33.24.Letp:X!BbeacoveringwithcompactbaseB.1)ProvethatifXiscompact,thenthecoveringis nite-sheeted.2)IfBisHausdor andthecoveringis nite-sheeted,thenXiscompact.33.25.LetXbeatopologicalspacepresentableastheunionoftwoopencon-nectedsetsUandV.ProvethatiftheintersectionU\Visdisconnected,thenXhasaconnectedin nite-sheetedcovering.335.UniversalCoveringsAcoveringp:X!BisuniversalifXissimplyconnected.Theappear-anceoftheworduniversalinthiscontextisexplainedbelowinSection39.33.N.Whichcoveringsoftheproblemsstatedaboveinthissectionareuniversal? 34.TheoremsonPathLifting235 34.TheoremsonPathLifting341.LiftingLetp:X!Bandf:A!Bbearbitrarymaps.Amapg:A!Xsuchthatpg=fissaidtocoverforbealiftingoff.Varioustopologicalproblemscanbephrasedintermsof ndingacontinuousliftingofsomecontinuousmap.Problemsofthissortarecalledliftingproblems.Theymayinvolveadditionalrequirements.Forexample,thedesiredliftingmustcoincidewithaliftingalreadygivenonsomesubspace.34.A.TheidentitymapS1!S1doesnotadmitacontinuousliftingwithrespecttothecoveringR!S1:x7!e2ix.(Inotherwords,thereexistsnocontinuousmapg:S1!Rsuchthate2ig(x)=xforx2S1.)342.PathLifting34.BPathLiftingTheorem.Letp:X!Bbeacovering,x02X,b02Bbepointssuchthatp(x0)=b0.Thenforanypaths:I!Bstartingatb0thereexistsauniquepath~s:I!Xstartingatx0andbeingaliftingofs.(Inotherwords,thereexistsauniquepath~s:I!Xwith~s(0)=x0andp~s=s.)WecanalsoproveamoregeneralassertionthanTheorem34.B:seeProb-lems34.1{34.3.34.1.Letp:X!Bbeatrivialcovering.ThenforanycontinuousmapfofanyspaceAtoBthereexistsacontinuouslifting~f:A!X.34.2.Letp:X!Bbeatrivialcoveringandx02X,b02Bbepointssuchthatp(x0)=b0.ThenforanycontinuousmapfofaspaceAtoBmappingapointa0tob0,acontinuouslifting~f:A!Xwith~f(a0)=x0isunique.34.3.Letp:X!Bbeacovering,Aaconnectedandlocallyconnectedspace.Iff;g:A!Xaretwocontinuousmapscoincidingatsomepointandpf=pg,thenf=g.34.4.Ifwereplacex0,b0,anda0inProblem34.2bypairsofpoints,thentheliftingproblemmayhappentohavenosolution~fwith~f(a0)=x0.Formulateaconditionnecessaryandsucientforexistenceofsuchasolution.34.5.WhatgoeswrongwiththePathLiftingTheorem34.Bforthelocalhome-omorphismofProblem33.K?34.6.ConsiderthecoveringC!Cr0:z7!ez.Findliftingsofthepathsu(t)=2 tandv(t)=(1+t)e2itandtheirproductsuvandvu. 236VII.CoversingSpaces 343.HomotopyLifting34.CPathHomotopyLiftingTheorem.Letp:X!Bbeacovering,x02X,b02Bbepointssuchthatp(x0)=b0.Letu;v:I!Bbepathsstartingatb0and~u;~v:I!Xbetheliftingpathsforu;vstartingatx0.Ifthepathsuandvarehomotopic,thenthecoveringpaths~uand~varehomotopic.34.DCorollary.UndertheassumptionsofTheorem34.C,thecoveringpaths~uand~vhavethesame nalpoint(i.e.,~u(1)=~v(1)).Noticethatthepathsin34.Cand34.Dareassumedtosharetheinitialpointx0.Inthestatementof34.D,weemphasizethatthentheyalsosharethe nalpoint.34.ECorollaryof34.D.Letp:X!Bbeacoveringands:I!Bbealoop.Ifthereexistsalifting~s:I!Xofswith~s(0)=~s(1)(i.e.,thereexistsacoveringpathwhichisnotaloop),thensisnotnull-homotopic.34.F.Ifapath-connectedspaceBhasanontrivialpath-connectedcoveringspace,thenthefundamentalgroupofBisnontrivial.34.7.Provethatanycoveringp:X!BwithsimplyconnectedBandpathconnectedXisahomeomorphism.34.8.Whatcorollariescanyoudeducefrom34.FandtheexamplesofcoveringspresentedaboveinSection33?34.9.Riddle.IsitreallyimportantinthehypothesisofTheorem34.Cthatuandvarepaths?Towhatclassofmapscanyougeneralizethistheorem? 35.CalculationofFundamentalGroups237 35.CalculationofFundamentalGroupsUsingUniversalCoverings351.FundamentalGroupofCircleForanintegern,denotebysntheloopinS1de nedbytheformulasn(t)=e2int.Theinitialpointofthisloopis1.Denotethehomotopyclassofs1by .Thus, 21(S1;1).35.A.Theloopsnrepresents n21(S1;1).35.B.FindthepathsinRstartingat02RandcoveringtheloopssnwithrespecttotheuniversalcoveringR!S1.35.C.ThehomomorphismZ!1(S1;1):n7! nisanisomorphism.35.C.1.Theformulan7! ndeterminesahomomorphismZ!1(S1;1).35.C.2.Provethataloops:I!S1startingat1ishomotopictosnifthepath~s:I!Rcoveringsandstartingat02Rendsatn2R(i.e.,~s(1)=n).35.C.3.Provethatiftheloopsnisnull-homotopic,thenn=0.35.1.Findtheimageofthehomotopyclassoftheloopt7!e2it2undertheisomorphismofTheorem35.C.DenotebydegtheisomorphisminversetotheisomorphismofTheorem35.C.35.2.Foranyloops:I!S1startingat12S1,theintegerdeg([s])isthe nalpointofthepathstartingat02Randcoverings.35.DCorollaryofTheorem35.C.Thefundamentalgroupof(S1)nisafreeAbeliangroupofrankn(i.e.,isomorphictoZn).35.E.OntorusS1S1 ndtwoloopswhosehomotopyclassesgeneratethefundamentalgroupofthetorus.35.FCorollaryofTheorem35.C.ThefundamentalgroupofpuncturedplaneR2r0isanin nitecyclicgroup.35.3.SolveProblems35.D{35.FwithoutreferencetoTheorems35.Cand31.H,butusingexplicitconstructionsofthecorrespondinguniversalcoverings.352.FundamentalGroupofProjectiveSpaceThefundamentalgroupoftheprojectivelineisanin nitecyclicgroup.Itiscalculatedintheprevioussubsectionsincetheprojectivelineisacircle.Thezero-dimensionalprojectivespaceisapoint,henceitsfundamental 238VII.CoversingSpaces groupistrivial.Nowwecalculatethefundamentalgroupsofprojectivespacesofallotherdimensions.Letn2,andletandl:I!RPnbealoopcoveredbyapath~l:I!SnwhichconnectstwoantipodalpointsofSn,saythepolesP+=(1;0;:::;0)andP =( 1;0;:::;0).Denotebythehomotopyclassofl.Itisanelementof1(RPn;(1:0::0)).35.G.Foranyn2group1(RPn;(1:0::0))isacyclicgroupoforder2.Itconsistsoftwoelements:and1.35.G.1Lemma.AnyloopinRPnat(1:0::0)ishomotopiceithertolorconstant.ThisdependsonwhetherthecoveringpathoftheloopconnectsthepolesP+andP ,orisaloop.35.4.Wheredidweusetheassumptionn2intheproofsofTheorem35.GandLemma35.G.1?353.FundamentalGroupofBouquetofCirclesConsiderafamilyoftopologicalspacesfX g.Ineachofthespaces,letapointx bemarked.TakethedisjointsumF X andidentifyallmarkedpoints.TheresultingquotientspaceW X isthebouquetoffX g.Henceabouquetofqcirclesisaspacewhichisaunionofqcopiesofcircle.Thecopiesmeetatasinglecommonpoint,andthisistheonlycommonpointforanytwoofthem.Thecommonpointisthecenterofthebouquet.DenotethebouquetofqcirclesbyBqanditscenterbyc.Letu1,...,uqbeloopsinBqstartingatcandparameterizingtheqcopiesofcirclecomprisingBq.Denoteby ithehomotopyclassofui.35.H.1(Bq;c)isafreegroupfreelygeneratedby 1,..., q.354.AlgebraicDigression:FreeGroupsRecallthatagroupGisafreegroupfreelygeneratedbyitselementsa1,...,aqif:eachelementx2Gisaproductofpowers(withpositiveornegativeintegerexponents)ofa1,...,aq,i.e.,x=ae1i1ae2i2:::aeninandthisexpressionisuniqueuptothefollowingtrivialambiguity:wecaninsertordeletefactorsaia 1ianda 1iaiorreplaceamibyariasiwithr+s=m.35.I.Afreegroupisdetermineduptoisomorphismbythenumberofitsfreegenerators. 35.CalculationofFundamentalGroups239 Thenumberoffreegeneratorsistherankofthefreegroup.Forastandardrepresentativeoftheisomorphismclassoffreegroupsofrankq,wecantakethegroupofwordsinanalphabetofqlettersa1;:::;aqandtheirinversesa 11;:::;a 1q.Twowordsrepresentthesameelementofthegroupi theycanbeobtainedfromeachotherbyasequenceofinsertionsordeletionsoffragmentsaia 1ianda 1iai.ThisgroupisdenotedbyF(a1;:::;aq),orjustFq,whenthenotationforthegeneratorsisnottobeemphasized.35.J.EachelementofF(a1;:::;aq)hasauniqueshortestrepresentative.Thisisawordwithoutfragmentsthatcouldhavebeendeleted.Thenumberl(x)oflettersintheshortestrepresentativeofanelementx2F(a1;:::;aq)isthelengthofx.Certainly,thisnumberisnotwellde nedunlessthegeneratorsare xed.35.5.ShowthatanautomorphismofFqcanmapx2Fqtoanelementwithdi erentlength.Forwhatvalueofqdoessuchanexamplenotexist?Isitpossibletochangethelengthinthiswayarbitrarily?35.K.AgroupGisafreegroupfreelygeneratedbyitselementsa1,...,aqi everymapofthesetfa1;:::;aqgtoanygroupXextendstoauniquehomomorphismG!X.Theorem35.Kissometimestakenasade nitionofafreegroup.(De -nitionsofthissortemphasizerelationsamongdi erentgroups,ratherthantheinternalstructureofasinglegroup.Ofcourse,relationsamonggroupscantelleverythingabout\internala airs"ofeachgroup.)NowwecanreformulateTheorem35.Hasfollows:35.L.ThehomomorphismF(a1;:::;aq)!1(Bq;c)takingaito ifori=1;:::;qisanisomorphism.First,forthesakeofsimplicitywerestrictourselvestothecasewhereq=2.Thiswillallowustoavoidsuper\ruouscomplicationsinnotationandpictures.Thisisthesimplestcase,whichreallyrepresentsthegeneralsituation.Thecaseq=1istoospecial.Totakeadvantagesofthis,letuschangethenotation.PutB=B2,u=u1,v=u2, = 1,and = 2.NowTheorem35.Llooksasfollows:ThehomomorphismF(a;b)!(B;c)takingato andbto isanisomorphism.ThistheoremcanbeprovedlikeTheorems35.Cand35.G,providedtheuniversalcoveringofBisknown. 240VII.CoversingSpaces 355.UniversalCoveringforBouquetofCirclesDenotebyUandVthepointsantipodaltoconthecirclesofB.CutBatthesepoints,removingUandVandputtinginsteadeachofthemtwonewpoints.Whateverthisoperationis,itsresultisacrossK,whichistheunionoffourclosedsegmentswithacommonendpointc.ThereappearsanaturalmapP:K!BthattakesthecentercofthecrosstothecentercofBandhomeomorphicallymapstheraysofthecrossontohalf-circlesofB.SincethecirclesofBareparameterizedbyloopsuandv,thehalvesofeachofthecirclesareordered:thecorrespondinglooppasses rstoneofthehalvesandthentheotherone.DenotebyU+thepointofP 1(U)belongingtotheraymappedbyPontothesecondhalfofthecircle,andbyU theotherpointofP 1(U).WesimilarlydenotepointsofP 1(V)byV+andV . UV U+U U+U = U+U V+V TherestrictionofPtoKrfU+;U ;V+;V gmapsthissethomeomor-phicallyontoBrfU;Vg.ThereforePprovidesacoveringofBrfU;Vg.However,itfailstobeacoveringatUandV:noneofthesepointshasatriviallycoveredneighborhood.Furthermore,thepreimageofeachofthesepointsconsistsof2points(theendpointsofthecross),wherePisnotevenalocalhomeomorphism.Toeliminatethisdefect,wecanattachacopyofKateachofthe4endpointsofKandextendPinanaturalwaytotheresult.Butthen12newendpointsappearatwhichthemapisnotalocalhomeomorphism.Well,werepeatthetrickandrecoverthepropertyofbe-ingalocalhomeomorphismateachofthe12newendpoints.Thenwedothisateachofthe36newpoints,etc.Butifwerepeatthisin nitelymanytimes,allbadpointsbecomeniceones.235.M.FormalizetheconstructionofacoveringforBdescribedabove. 2ThissoundslikeastoryaboutabattlewithHydra,butthehappyendingdemonstratesthatmodernmathematicianshaveamagicpowerofthesortthattheherosofmythsandtalescouldnotevendreamof.Indeed,wemeetaHydraKwith4heads,chopo alltheheads,but,accordingtotheoldtraditionofthegenre,3newheadsappearinplaceofeachoftheoriginalheads.Wechopthemo ,andthestoryrepeats.Wedonoteventrytopreventthismultiplicationofheads.Wejustchopthemo .Butcontrarytotherealherosoftales,weactoutsideofTimeandhencehavenotimelimitations.Thusafterin niterepetitionsoftheexercisewithanexponentiallygrowingnumberofheadswesucceed!Noheadsleft!Thisisatypicalsuccessstoryaboutanin niteconstructioninmathematics.Sometimes,asinourcase,suchaconstructioncanbereplacedbya niteone,butdealingwithin niteobjects.However,thereareimportantconstructionsinwhichanin nitefragmentisunavoidable. 35.CalculationofFundamentalGroups241 ConsiderF(a;b)asadiscretetopologicalspace.TakeKF(a;b).ItcanbethoughtofasacollectionofcopiesofKenumeratedbyelementsofF(a;b).TopologicallythisisadisjointsumofthecopiesbecauseF(a;b)isequippedwithdiscretetopology.InKF(a;b),weidentifypoints(U ;g)with(U+;ga)and(V ;g)with(V+;gb)foreachg2F(a;b).DenotetheresultingquotientspacebyX.35.N.ThecompositionoftheprojectionKF(a;b)!KandP:K!Bdeterminesacontinuousquotientmapp:X!B.35.O.p:X!Bisacovering.35.P.Xispath-connected.Foranyg2F(a;b),thereexistsapathcon-necting(c;1)with(c;g)andcoveringtheloopobtainedfromgbyreplacingawithuandbwithv.35.Q.Xissimplyconnected.356.FundamentalGroupsofFiniteTopologicalSpaces35.6.Provethatifathree-pointspaceXispath-connected,thenXissimplyconnected(cf.31.7).35.7.ConsideratopologicalspaceX=fa;b;c;dgwithtopologydeterminedbythebaseffag;fcg;fa;b;cg;fc;d;agg.ProvethatXispath-connected,butnotsimplyconnected.35.8.Calculate1(X).35.9.LetXbea nitetopologicalspacewithnontrivialfundamentalgroup.Letn0betheleastpossiblecardinalityofX.1)Findn0.2)Whatnontrivialgroupsariseasfundamentalgroupsofn0-pointspaces?35.10.1)Finda nitetopologicalspacewithnon-Abelianfundamentalgroup.2)Whatistheleastpossiblecardinalityofsuchaspace?35.11*.LetatopologicalspaceXbetheunionoftwoopenpath-connectedsetsUandV.ProvethatifU\Vhasatleastthreeconnectedcomponents,thenthefundamentalgroupofXisnon-Abelianand,moreover,admitsanepimorphismontoafreegroupofrank2.35.12*.Finda nitetopologicalspacewithfundamentalgroupisomorphictoZ2. 242VII.CoversingSpaces ProofsandComments33.ALetusshowthatthesetBitselfistriviallycovered.Indeed, prB 1(B)=X=Sy2F(By),andsincethetopologyinFisdiscrete,itfollowsthateachofthesetsByisopeninthetotalspaceofthecovering,andtherestrictionofprBtoeachofthemisahomeomorphism.33.B Weconstructahomeomorphismh:p 1(U)!Up 1(a)foranarbitrarytriviallycoveredneighborhoodUBofa.Bythede nitionofatriviallycoveredneighborhood,wehavep 1(U)=SU .Letx2p 1(U),consideranopensetsU containingxandtakextothepair(p(x);c),wherefcg=p 1(a)\U .Itisclearthatthecorrespondencex7!(p(x);c)determinesahomeomorphismh:p 1(U)!Up 1(a). Byassertion33.1,Uisatriviallycoveredneighborhood,hence,p:X!Bisacovering.33.CForeachpointz2S1,thesetUz=S1rf zgisatriviallycoveredneighborhoodofz.Indeed,letz=e2ix.ThenthepreimageofUzistheunionSk2Z(x+k 1 2;x+k+1 2),andtherestrictionofthecoveringtoeachoftheaboveintervalsisahomeomorphism.33.DTheproduct(S1rf zg)Risatriviallycoveredneighborhoodofapoint(z;y)2S1R;cf.33.E.33.EVerifythattheproductoftriviallycoveredneighborhoodsofpointsb2Bandb02B0isatriviallycoveredneighborhoodofthepoint(b;b0)2BB0.33.FConsiderthediagramR2h !Cq??y??ypS1Rg !Cr0;whereg(z;x)=zex,h(x;y)=y+2ix,andq(x;y)=(e2ix;y).Theequal-ityg(q(x;y))=e2ixey=ey+2ix=p(h(x;y))impliesthatthediagramiscommutative.Clearly,gandharehomeomorphisms.Sinceqisacoveringby33.D,pisalsoacovering.33.GBy33.E,thisassertionfollowsfrom33.C.Certainly,itisnotdiculttoproveitdirectly.Theproduct(S1rf zg)(S1rf z0g)isatriviallycoveredneighborhoodofthepoint(z;z0)2S1S1.33.HLetz2S1.Thepreimage zundertheprojectionconsistsofnpoints,whichpartitionthecoveringspaceintonarcs,andtherestriction ProofsandComments243 oftheprojectiontoeachofthemdeterminesahomeomorphismofthisarcontotheneighborhoodS1rf zgofz.33.IBy33.E,thisassertionfollowsfrom33.H.33.JThepreimageofapointy2RPnisapairfx; xgSnofantipodalpoints.Theplanepassingthroughthecenterofthesphereandorthogonaltothevectorxsplitsthesphereintotwoopenhemispheres,eachofwhichishomeomorphiallyprojectedtoaneighborhood(homeomorphitoRn)ofthepointy2RPn.33.KNo,itisnot,becausethepoint12S1hasnotriviallycoveredneighborhood.33.LTheopenintervalsmentionedinthestatementarenotopensubsetsoftheplane.Furthermore,sincethepreimageofanyintervalisaconnectedset,itcannotbesplitintodisjointopensubsetsatall.33.MProvethatthede nitionofacoveringimpliesthatthesetofthepointsinthebasewithpreimageofprescribedcardinalityisopenandusethefactthatthebaseofthecoveringisconnected.33.NThosecoveringswherethecoveringspaceisR1,R2,Rnr0withn3,andSnwithn2,i.e.,asimplyconnectedspace.34.AAssumethatthereexistsaliftinggoftheidentitymapS1!S1;thisisacontinuousinjectionS1!R.Weshowthattherearenosuchinjections.Letg(S1)=[a;b].TheIntermediateValueTheoremimpliesthateachpointx2(a;b)istheimageofatleasttwopointsofthecircle.Consequently,gisnotaninjection.34.BCoverthebasebytriviallycoveredneighborhoodsandpartitionthesegment[0;1]bypoints0=a0a1:::an=1,suchthattheimages([ai;ai+1])isentirelycontainedinoneofthetriviallycoveredneigh-borhoods;s([ai;ai+1])Ui,i=0;1;:::;n 1.Sincetherestrictionofthecoveringtop 1(U0)isatrivialcoveringandf([a0;a1])U0,thereexistsaliftingofsj[a0;a1]suchthates(a0)=x0,letx1=es(a1).Similarly,thereexistsauniqueliftingesj[a1;a2]suchthates(a1)=x1;letx2=es(a2),andsoon.Thus,thereexistsaliftinges:I!X.Itsuniquenessisobvious.Ifyoudonotagree,useinduction.34.CLeth:II!Bbeahomotopybetweenthepathsuandv,thus,h( ;0)=u( ),h( ;1)=v( ),h(0;t)=b0,andh(1;t)=b12B.Weshowthatthereexistsamap~h:II!Xcoveringhandsuchthath(0;0)=x0.TheproofoftheexistenceofthecoveringhomotopyissimilartothatofthePathLiftingTheorem.WesubdividethesquareIIintosmallersquaressuchthattheh-imageofeachofthemiscontainedinacertaintriviallycoveredneighborhoodinB.Therestrictionhk;lofthehomotopyhtoeach 244VII.CoversingSpaces ofthe\little"squaresIk;liscoveredbythecorrespondingmapehk;l.Inordertoobtainahomotopycoveringh,wemustonlyensurethatthesemapscoincideontheintersectionsofthesesquares.By34.3,itsucestorequirethatthesemapscoincideatleastatonepoint.Letusmakethe rststep:leth(I0;0)Ub0andleteh0;0:I0;0!Xbeacoveringmapsuchthateh0;0(a0;c0)=x0.Nowweputb1=h(a1;c0)andx1=eh(a1;c0).Thereisamapeh1;0:I1;0!XcoveringhjI1;0suchthateh1;0(a1;c0)=x1.Proceedinginthisway,weobtainamapehde nedontheentiresquare.Itremainstoverifythatehisahomotopyofpaths.Considerthecoveringpatheu:t7!eh(0;t).Sincepeuisaconstantpath,thepatheumustalsobeconstant,whenceeh(0;t)=x0.Similarly,eh(1;t)=x1isamarkedpointofthecoveringspace.Therefore,ehisahomotopyofpaths.Inconclusion,weobservethattheuniquenessofthishomotopyfollows,oncemore,fromLemma34.3.34.DFormallyspeaking,thisisindeedacorollary,butactuallywealreadyprovedthiswhenprovingTheorem34.C.34.EAconstantpathiscoveredbyaconstantpath.By34.D,eachnull-homotopicloopiscoveredbyaloop.35.AConsiderthepaths~sn:I!R:t7!nt,~sn 1:I!R:t7!(n 1)t,and~s1:I!R:t7!n 1+tcoveringthepathssn,sn 1,ands1,respectively.Sincetheproductesn 1es1isde nedandhasthesamestartingandendingpointsasthepathesn,wehaveesnesn 1es1,whencesnsn 1s1.Therefore,[sn]=[sn 1] .Reasoningbyinduction,weobtaintherequiredequality[sn]= n.35.BSeetheproofofassertion35.A:thisisthepathde nedbytheformulaesn(t)=nt.35.CBy35.C.1,themapinquestionisindeedawell-de nedhomo-morphism.By35.C.2,itisanepimorphism,andby35.C.3itisamonomor-phism.Therefore,itisanisomorphism.35.C.1Ifn7! nandk7! k,thenn+k7! n+k= n k.35.C.2SinceRissimplyconnected,thepathsesandesnarehomotopic,therefore,thepathssandsnarealsohomotopic,whence[s]=[sn]= n.35.C.3Ifn=0,thenthepathesnendsatthepointn,hence,itisnotaloop.Consequently,theloopsnisnotnull-homotopic. ProofsandComments245 35.DThisfollowsfromtheabovecomputationofthefundamentalgroupofthecircleandassertion31.H:1(S1:::S1| {z }nfactors;(1;1;:::;1))=1(S1;1):::1(S1;1)| {z }nfactors=Zn:35.ELetS1S1=f(z;w):jzj=1;jwj=1gCC.Thegeneratorsof1(S1S1;(1;1))aretheloopss1:t7!(e2it;1)ands2:t7!(1;e2it).35.FSinceR2r0=S1R,wehave1(R2r0;(1;0))=1(S1;1)1(R;1)=Z.35.G.1LetubealoopinRPn,andlet~ubethecoveringuthepathinSn.Forn2,thesphereSnissimplyconnected,andifeuisaloop,theneuandhencealsouarenull-homotopic.Nowifeuisnotaloop,then,oncemoresinceSnissimplyconnected,wehaveeuel,whenceul.35.GBy35.G.1,thefundamentalgroupconsistsoftwoelements,there-fore,itisacyclicgroupofordertwo.35.HSee355.35.MSeetheparagraphfollowingthepresentassertion.35.NThisobviouslyfollowsfromthede nitionofP.35.OThisobviouslyfollowsfromthede nitionofp.35.PUseinduction.35.QUsethefactthattheimageofanyloop,asacompactset,inter-sectsonlya nitenumberofthesegmentsconstitutingthecoveringspaceX,anduseinductiononthenumberofsuchsegments.