### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Intro to Topology MATH 432

UO

GPA 3.8

### View Full Document

## 12

## 0

## Popular in Course

## Popular in Mathematics (M)

This 18 page Class Notes was uploaded by Henderson Lind II on Tuesday September 8, 2015. The Class Notes belongs to MATH 432 at University of Oregon taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/187188/math-432-university-of-oregon in Mathematics (M) at University of Oregon.

## Popular in Mathematics (M)

## Reviews for Intro to Topology

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/08/15

Classifying Covering Spaces Richard Koch February 277 2006 1 Previous Work We have been using covering spaces to compute fundamental groups To nd n39X7 we must guess a universal covering space 7139 X a X The theory does not help us guess But if we succeed 1 n39X7 0 is in one to one correspondence with the set 7r 1m0 of points in X over 0 to n39X7 0 is isomorphic as a group to the group P of deck transformations of X 03 if i0 if a point over 0 then for each i1 over 0 there is a unique deck transformation mapping 0 to 1 hence if we are able to guess these particular deck transformations we know the complete group P and so the group n39X7 mo Example 1 Let 7rz 527m R a 51 Under this map R is the universal cover of 51 So 7139Sl7 1 is set theoretically the collection of all points over 17 which is Z The map z a z n is clearly a deck transformation This transformation takes 0 to 71 Since these maps carry 0 to every possible image7 they form the complete set of all deck transformations Notice that R mim min B sends z to m m n z m n7 so composition in P corresponds to addition in Z Thus 7rSl7 1 Z as a group Example 2 Write down the corresponding calculations for the Klein bottle IC 2 Going Backward We are going to turn this theory on its head Suppose that we have managed to compute 7rX some other way We will then use this group to classify all possible covering spaces of X 35 a 2 i aim 55 5 x m DEE as 55 32x DEE 2 E 5 x m 153 DEE r 95 g aim um asib a 35 a 2235 8 x 5 ii 525 Essie a as a E ai w 5523 355 Si 3 2235ng x s 3 s 33st S53 lt93 ggs iasi a xx gt 3 3yi fisagima g s mxggismaa u 3233 53517 QE mwqu umas mo b mgmgm mgg g 31 E53521 E efsaaimggmga g gasf zmgmgoifg 5325323 35 i gggzig aa mEEEEasm EEE quot53235 FEE Esp aim35650xms rgr8lt sxm nw tmsg a The Lim39ng Theorem To End All Lifting Theorems a wwwm mew Wer memmmemm WW Mung mm mm 1 My 5 e Mme mu mm mng e W angspamoX mu yexMAmmMW mmwmwmm mm mm ma mm 16 Emma 1 39 T J My 75 A m MW my my come mm 5 m and mm mm 5 mm have mm gemsqu hummmhm 11y Mep hy z y We mxmemmxmenww mumspammeth 7 x niuwdmfamp4mbemeadolmmcadpm 51w KW quotT 21A Wm shpmmmmwwpzmxmemmmmme nustwe ma afxswa rd ha i m epusaithensmp wnmyxsm vcm ed 3 Hammad Immmu mts epus theassmpumtbzyxshu v mm mums mmwmmmmm sow mm M 7 use and M n mm nemgmmmwgmm mwummw quotWWW Wmme am NM 5m WMMWWW M mummmmgm gw gymgmmsmmsmm mbemmmmp 1 x1 Xsaldmgoxo m a By mm Dim mmsoiwxwms pummxmeswgbm nnda ms mp 3 Shawn haw We Wu um um sqmze Aswammse mm mm mmnmmmmemmMesqmwegwmm mm am 4 Covering Spaces of X and Subgroups of 0 medxswsiwwtbzvamngspMEXDmenspmdswsubgmpui 0 mm 5 mm2 up w mwy mam a 1 1x 1 4w gnqspam My man Mom 1 Mum 1 am mm mama to A a WP moo mmam mmmm mmmmaym hgsmmsmx mm M70 mama s ms mduethesune mmm mm m 7 ma 7 x t mmmwwWw m m mm and V Wem We t wmsmaMm7x Smewoymdwovamt emmtohrx hauehanwmpmmx m h leaxbesud hmmwm mmmmwm leainuppmgoxow mmwhmms hmm DPyHum Y39DTtm chgiPmSanm uih umdmnedmm mmmcmgmmgmx Wmka in E N X morn 3 mama W Z m mgmmmimmmyzowmmmw ex M7 791 ngoago m mm mm woo paw7mg z XbetpzzhimiuwZ Emmy Home 5mm mom mmme 7 pm hamd dmvemxbanmseiuwdmwyadw m V m m y Moo many Wm yam mpg ax um was to m mmm xing 74 mm by 7 71 Wk 5 WWW mm A Cmttwtfympp tye x uwthmWexrwmm h mm Minn moo MW MM W W W a mx MW h Wm 3115me 94m om ma mammmsepzmmymm IaXsmungdemdmgnmu m mmwmhamn mmmmmgwmmxmmmm Pam2 waeniu epzmim u aum ammmmamma apm 5 The Big One Hm mm mm mm M m Mm Anus5m 4mm W5 Mom 5 La 2 W x b m WW3 ox mu submps mam wmm 0 WWW W mum 9mm mm m Wm M mm mm wummmemgpumiu e mi Jim W By the Wavinus mm mumps moo conspndmg m x and x m m m mum w mad many 31ml Canada m the hummg Guam 3mg e and actually these groups are equal theorem one gives a map 1 X a X lifting the identity map Looking at the diagram backward we similarly discover a map 1 X a X lifting the identity map Consequently j o 1 X a X is a lift of the identity map But lifts are unique and the identity is one possible lift So 1 o 1 is the identity from X to X Similarly 1 0 j is the identity from X to It follows that 11 and j are both homeomorphisms and each is a covering space isomorphism as required 6 Constructing Covering Spaces Our next goal is to show that each subgroup of 7rX actually comes from a covering space This is easy provided we know that X has a universal cover lndeed we ll give a remarkably concrete method of constructing the covering space X corresponding to the subgroup So we have our second fundamental theorem Theorem 6 Suppose X is connected and locally pathwise connected and suppose X has a universal covering space C Then every subgroup C Q 7rXz0 comes from a covering space X of X Corollary 7 Suppose X is connected and locally pathwise connected and suppose X has a universal covering space Then there is a one to one correspondence between conjugacy classes of subgroups of 7rX and covering spaces of X Proof For convenience let C be the universal cover Let P be the group of deck trans formations of this universal cover Then we have an isomorphism from P to 7rXm0 Suppose we have a subgroup C Q 7rX me Let PG be the corresponding subgroup of the deck transformation group De ne an equivalence relation on the universal cover by calling two points equivalent if one can be mapped to the other by an element of PC Let X be the quotient space CPG The natural projection 7139 C a X maps points equivalent under P to the same point so it certainly maps points equivalent under PC to the same point Thus 7139 induces a map X CFG a X We claim this is a covering map and the fundamental group of this covering space is G This will complete the argument Eventually we will give the general argument but it is much more illuminating to examine some examples Example 1 Let X 51 so 7rSl Z Let G be the subgroup 3Z of all multiples of 3 The full deck transformation group contains all maps z a z n clearly the subgroup PG contains all deck transformations of the form x a z 3n xmmmwpmmvammwmmmmrg gm mmmmwsgcmmmmmam nganmmamam 5mm Won t4mm 7 A we mm mm m m me slug the in mm m we set the enamel x xWMQmeemgmmmg my mums ueaqmam Ofww ms 5 SaarJaw m s quotmmemmEmpcrgacnwmgx x nmsmep ay MAW mmmgwmmmmchummdx m lmEm Thsnzm szys ut usmnul Ammuan M We m 5333 Bamka LethebueSpaceX S xS swasama waska NW Amsda mesubmnp 22 x 22 clan Wm 5m um Mg m and ms mm a 3 I39hewnspmdmg 5qu 0139de ummms rs 5 the 591 ohms W a 2 law at E xa39ypmmm RxRxssqmvmtnnda agmngwpmmmme 092 Wemusjgluau el twdnyw gs wdthewpmdbu mg 55me m wmpsmm thesxsq wagome wtus mwxssxrfn d4wa mWDSzsl Mam Jam Emsdatme Baum 3 Hmmmempwaaampxa wumWomemswg rm Klan mu m mwasal ma af us Swans x x Rand 105 We map a is Z gt4 Z Recall that Z gt4 Z is the set of all 77171 with group law 7711711 o 7712712 7711 71 1 711 712 There is an exact sequence OHZLZNZLZHO where the rst map sends 771 to 77171 and the second map sends 77171 to 71 Suppose that we have a subgroup C Q Z gt4 Z The image of this group under 7 must be a subgroup of the Z on the right let us suppose that this subgroup is 3Z The kernel of 7 must be a subgroup of the left Z suppose this subgroup is 2Z The generator of 3Z comes from an element of G but we cannot assume that this element has the form 03 We can assume that it is either 03 or 1 3 since we can modify it by elements coming from 2Z For fun suppose that 1 3 E G Then 13 0 13 1 71313 3 06 6 G etc In general 0371 6 G for even 71 and 1 371 E G for 71 odd From exactness we conclude that G is exactly 2771371 l 71 is even U 2771 1371 l 71 is odd Notice that the group is generated by the two elements 20 and 1 3 The element 771 71 in the full group corresponds to the transformation z7y 1 x my 77 So the generators correspond to the transformations x37 a m 1 237 and 137 7 90 17 y 3 Multiples of the map x 37 a ix 1 37 l 3 raise or lower points by multiples of three By applying an appropriate such power we can nd an equivalent point 137 with 0 g y E 3 Multiples of the map x37 a m 1 237 slide points horizontally by multiples of 2 By applying an appropriate such power we can nd an equivalent point z37 with 0 g x g 2 Consequently every point of the plane is equivalent under PG to a point in the shaded rect angle below However it Will soon turn out that this is not the best choice of representatives for We must glue the left and right sides of this rectangle together since these points are equivalent under 9631 gt x l 2y We must also glue the heavily shaded lines on the top and bottom together as indicated since these points are equivalent under 96311 gt x 1 y3 Finally we must glue the remaining points on the top and bottom together as indicated in the second picture because they are equivalent under 16 y gt x1y3 gt x12y3 x3y3 It is a little difficult to understand the resulting space But notice that we can nd another shaded region by sliding horizontal segments toward the left like a series of dinner plates as indicated below This time the entire bottom is glued to the top by 96311 gt x 1 y3 It is clear that this space is another Klein bottle covering the original 1C six times The Klein bottle can also be covered by tori and cylinders Explain how 7 The General Theory We now give the general proof of theorem six We earlier proved that the full deck trans formation group I of the universal cover C gt X is transitive on 7r1r Whenever x E X 10 So when we glue equivalent points under F together we identify all the points over x to a single point It immediately follows that C F X De ne X C PG with the quotient topology If two points are equivalent under Fa they certainly are equivalent under the full P so there is a natural rnapping C PG gt C F Said another way there is a natural rnapping X gt X It remains to show every point of X has an evenly covered open neighborhood and that the fundamental group of 7TX is G Let x E X Choose an open neighborhood U of x which is evenly covered in C Since X is locally pathwise connected we can suppose that U is pathwise connected Write 7T1Zl U Ma in C Pick 1 E C over x this point is in one of the Ma say U1 The deck transforrnation group F acts sirnply transitively on the Ma This is a fancy way of saying that if fa is the point in Ha over x there is a unique deck transforrnation rnapping 1 to fa and thus rnapping U1 to Lia Thus we can identify the plates over U in C with the elements of F Now write F as a union of right PG cosets F FGeUrgggUrggggU This decornposition then induces a similar decomposition of the plates over U into equiva lence classes Here is the picture 3 x W a W M 22 Plates in a particular coset are equivalent under Fa So when we identify points equivalent under Pg these equivalent plates are glued together Thus the various cosetscorrespond to the open sets in X which cover M This shows that U is evenly covered in X 11 Finally we must nd the fundamental group of X By our general theory this group is isomorphic to the deck transformation group of the covering C gt X We can see this deck transformation group in the above picture Indeed take one of the small stacks of plates Which Will glue together in X to form a subset V The deck transforms of C gt X must permute these stacks But the small stacks correspond to cosets of F0 and the deck transformations Which permute a particular coset are just the deck transformations in Pg So the fundamental group of X is PG as desired CD O O Q g C U O Q E 1 pan Jo 4272 v There Ba 12 Existence of a Universal Cover Richard Koch February 26 2006 1 The Theorem Theorem 1 Suppose the topological space X is connected locally pathwise connected and semi locally simply connected Then X has a uniuersal couer C Remark All nice spaces satisfy these hypotheses so the essential point is that every reasonable space has a universal cover Remark The hypothesis that X be semi locally simply connected is necessary Indeed if p E X nd an evenly covered neighborhood L of p We claim that every loop in L starting at p is homotopic to a constant in X lndeed nd q E C projecting to p and let Ma Q C be the neighborhood of q corresponding to L Q X We want to prove that 7rIlp a 7rXp is the zero map but 7139 Lia a L is a homeomorphism so it suf ces to prove that llmg a 7rIlp a 7rX p is the zero map This map can be written as the composition 711le q a 7rC q a 7rX p and this map is zero because C is simply connected 2 Motivating the Proof If we only know X how can we get our hands on points of its univeral cover C Suppose we had a point c E C over a point z E X We may as well assume that C has a base point co over a base point mo 6 X Choose a path y I a C starting at co and ending at 0 Then 7139 o 39y will be a path in X starting at 0 and ending at m But conuersely if we had 7139 o y in X we could uniquely lift it to C and thus nd c If we had two paths y and r in C from co to 0 these paths would be homotopic with xed endpoints because C is simply connected and therefore 7139 o 39y and 7139 o 739 would be homotopic in X But conuersely if7r o 39y and 7139 o 739 were homotopic in X then 39y and 739 would end at the same point ofC by lifting the homotopy h I x I gt X to C PuMmga ui uswga a wedsmva uzamppmmm0mzxsmavevdmmmedw hmmwm4bssoiplkszwaegmnwamaudnz mamm 1 we Wmam me mm mX mmsmz mu by7 7 w m 7 m a usawz W w W mmw mx mama gm La c 7zaxmo 7w7m1k muu77 m caxmaw a A prology and Namemmmmpdwunc xmmmgmepmgshmrpmA mmummmxgmmmuupmsmmenmmMm mmmmmmm xnmmmqmmapu mxmmnmm atmam 5z 4mmt waivcmaai p lh ghbmhmduizmx ME mmwmmauMaupmawmwwmmmmum mmm mam thu39wemnsuum4 M57 5 mammmmmaanugm mnelvmu mammnmp km h uwiwx saudamdu mmmmmnelvmuSn mmmmmeban mu McmbeMnaduM39 oiapPWmepzms Inpxacmewxsnmnanmwmmtuwb pmhcm mgmeomwgm sf men lluw n5 Desmqu Lay z x MMstsz ngtubtmzopcrzm o wod o uuadoymx MnltuxygtmsmsttoWoCmmcmbtkmtddby yamWinWynWau M WltQmusth tmu 47 m 2 1 1 a 2 c 4 axemvulam u PM umquot mun paused J wwwmmu emm m m mv mgmimupmnm WWW WWWWWWHQ Wm WWW WE wmmvmwiunvmmspwpaw Eysymnmmxtsxmswpzmeum 7 SW mung l0 ltu 1gt unltu agt lt1Aw1gt m 2y mmp wn h p haxshanwmwwy a whmammmnaymu mathemoi39y mmmwmmwmmmm mmymm azuhesaznepmm magmaaltmygtmsnaammveomwmawwmwmm mmmugmmnmmm nmawxsmmmmwawmmmg w memwmm mamwm gmmwmwwm mam mpm mmmu Hame a xepx a sm mmtufltuwygtmd ma xsm ussat katwno uwoo ummz Mm caxgmmumuummm humane u pmmmswmmmu 4 c a X is 2 Covering Space TbAmplmmeptmim mxmthm ws mswsiwwmtc X54m7 mspmawduuszssxmphwmma mmmmmoxmmgmm mm s x Amusesamrlnunv imply mwgop mamthdu an m 3505 by Mmhss m u m mum Mahatma Du mm 5 mm 492mm ms 5 mebsmex 5 1043le PINEeme quotwe um 51mm smwlno1v amply Ammada i 5a Pmdus mama mhsa View New 14 mm mm 59mm quot2 Pm m w m 2 m hamvpy BlurWe mg and la m be weummamm Mme p 0 up Wm mampm awwwmmmmmmumm MWwwmgmwmwmummmwquotWWW wammmmgm Let B be 3 traced backward Then a 3 N y a 3 and 0 N W BB and so WMWBN WNW But 3 3 is homotopic to a constant and so the right side of this similarity is homotopic to w Also 043 is a loop at z and by hypothesis such loops are homotopic to constants in X So the left side of the above similarity is homotopic to w Thus m N w contradicting the choice of the m From here on everything is easy We must prove lt Ll 39yi gtH L a homeomorphism It is onto because L is pathwise connected so any u 6 L is the end of a path 04 starting at z and thus the image of an element y Oz in lt um gt It is oneto one for if y Oz and w 3 map to the same point then 04 and B begin and end at the same point of LI Since the loop 04 3 is homotopic to a constant in X 04 and B are homotopic in X with xed endpoints and so 39yi Oz and w B are homotopic paths and represent the same point of The map lt Ll y gtH L is continuous because 7139 C a X is continuous It is an open map and thus has continuous inverse because every open set in C is a union of lt 117 gt with L pathwise connected why and the image of such a lt 117 gt is LI 5 C is Pathwise Connected The nal step of the proof is fun because it is hokus pokus You ll need to sit in a closet and see if you buy it We must prove that C is connected and that every loop beginning and ending at co is homotopic to a constant We ll prove C pathwise connected Let c E C Then 0 is an equivalence class of paths in X let 39y I a X be a representative of c For each u E I consider the path W from 0 to 39yu obtained by following the rst part of 39y to We want this path to be parameterized by 01 so write mt wt 0 t 1 Then each 39yu represents a point of C We can then think of the entire collection of ms as a path in C this path assigns to u the point ofC represented by the path yu Please think about this slight of hand until you understand it it is the central point of the arguments which follow We call this new path 7y because it de nes a lift of 39y to C So is the point of C represented by my and thus by the portion of 39y from 0 to Notice that yo is the path which is constantly me this path represents the base point co 6 C Thus 7y is a path in C from co to the element 0 represented by 39y So C is pathwise connected 6 c is Simply Connected Nwweptmecsxmplthmaded Letbelmpm sumngamdmdmgnm 12mm mumpmpzmmxmgmmnw mmMueMwem madamem mana hapwhyimyww za athaweu a uspnhm TMSPMhma s pmmufc asuwmweob am pl hm Asbdme us mhmmmgmumawwmm m m4mm9wmhgmam pmaaph In mm w m 7 mm WeMWMtCaXxs4 mmmmm x 5 Emma mam ohmmamWepzmam 4 Wmquot Mege awmsmmxm Wm puny mm mm man be me mu am

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.