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# Introduction to Algebraic Topology 22M 201

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This 35 page Class Notes was uploaded by Virgil Wyman on Friday October 23, 2015. The Class Notes belongs to 22M 201 at University of Iowa taught by Jonathan Simon in Fall. Since its upload, it has received 24 views. For similar materials see /class/227987/22m-201-university-of-iowa in Mathematics (M) at University of Iowa.

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22M201 Fall 04 J Simon Notes on Homology Sec 72 November 17 2004 01 Don7t worry yet about degenerate cubes 1 suggest you study this section in the order rst class notes7 second this handout7 third read the text and do the homework problems 02 Just to make sure you understand how faces are de ned Suppose T is a singular 3 cube in X7 that is T 3 7 X Then T has six faces Each face of T is a singular 2 cube7 that is a map from 2 7 X The idea is easier than the notation we eventually end up using There are 3 directions in R3 In each direction7 T has a far face and a near face the text calls these the back and front faces respectively We have to specify six functions from 2 into X So for each of the six functions7 we have to decide where to send the points Lab 6 2 Keep saying it over and over again Each face is a function A given face is the function that sends points 17 b to Direction front face back face 7direction T0ab T1ab y7direction Ta0b Ta1b 27direction Tab7 0 Tab1 03 De ne the boundary The boundary of an n7cube T is a linear combination of n 7 17cubes The boundary of the boundary of T is then a linear combination of n 7 2 cubes We want that boundary to be 0 The boundary is actually a homomorphism from the group of n7cubes to the group of n 7 17 cubes7 denoted 3n QnX 7 Qn1X The claim is that the homomorphism 3W1 0 3n QnX 7 Qn2X is the zero homomorphism Consider the case where T is a 2 cube Then 32T is a linear combination of 17 cubes7 and the boundary of that is a linear combination of 07cubes7 essentially a linear combination of points of X Each point in this combination appears twice7 and we need to make sure that it appears once with a and once with a That is accomplished by the identities AiAJT Aj1AlT etc displayed in text 721 Here is an example T is a 3 cube What is AlAgTs77 Note don7t think of this as functional composition that is associative A1 of 1 A3TS would be A1 of a point in X which doesnt make sense Also A3Ts would not make sense since the domain of AgT is 2 not 1 AgT is a 27cube A1A3T is the front face of that 27cube in the rst coordinate direction So A1A3Ts AgT 05 Now what does AgT do to a pair t s It takes the pair to a point on the front face of 3 in the third direction and then applies T So A3T0 s T0s0 Sirnilarly A1B3Ts T0 s 1 HOMEWORK Prove the fourth identity in 721 that is prove that the functions BiAJT and Aj1BT are identical Hint Both of these are n 7 27cubes So start with a point 11 an2 and see where it is sent under each of the two rnaps 031 NOW we de ne the boundary homomomomhism The function 6 0X a n1X is given by its action on each basis element that is its action on each nicube V L MT ZltelgtiltltAiTgt e EaT i1 HOMEWORK Prove that boundaryboundary0 Since this will be true if an only if it is true for generators of QnX it is necessary and suf cient for you to prove that for each singular nicube T a1ltanltTgtgt o e OHM Hint The proof follows directly from the identities 721 you just need to do the careful bookkeeping 04 Provisional de nition of cycles bounds and homology groups We are going to de ne the cycles ZnX to be those combinations of nicubes that have 0 boundary That is ZnX ker 3 These nicycles are trying to record the existence of nidirnensional holes in X But if a cycle is actually the boundary of something then there7s no hole to record With this in mind we de ne a 2004 J Simon all rights reserved page 2 boundary to be any combination of nicubes that is itself the boundary of some combination of n 17cubes That is BAX image 3ndL We then de ne the homology groups of X by 05 But this give rise to too many cycles If we just proceed as above we have too many cycles For example if the licube f 01 a X is a constant path fs 0 for all s 6 01 then f is a licycle that is f E Z1X since the front and back 07faces of f are identical so their difference is 0 On the other hand f Z B1X the boundary of any 27cube is a linear combination of four licubes so each element of B1X is a combination of an even number of licubes A licube that actually is a constant function is called a degenerate licube Similarly suppose f 01 a X is a loop ie path with f0 f1 Consider the degenerate 27cube T given by Tst ft Then A1Tx T0z f B1Tx T1z f A2Tltzgt To 0 f0 B2Tx To 1 M So 62T 0 ie T E Z2X but T is not a boundary The moral is that if we include degenerate nicubes then HnX will be bigger than we want if our goal is to keep track of holes to describe the shape of X The solution is to get rid of these degenerate cubes from the beginning 06 De ne degenerate nicubes A singular cube T I a X is called degenerate if there is some index 239 such that the map T factors as 1 xi1zizi1zn a 1 xi1i1zn a X In the books words T is independent of the z coordinate De ne DnX subgroup of QnX generated by all degenerate nicubes Now de ne M BAX 39 After checking that the boundary homomorphism respects degeneracy that is the function 3 Q a le takes the subgroup Dn into Dn1 we can de ne the adjustedmod f edimproved corrected version of 3 as a map of 0X a n1X Then de ne Zn Q 0 and En Q 0 and HnX ZnXBnX 2004 J Simon all rights reserved page 3 0X 07 Homology groups of a onepoint set X In dimensions 71 2 17 all cubes are degenerate So OnX 0 ZnX 0 HnX 0 That leaves dimension n 0 There is only one Oicube7 so Q0X E Z There are no degenerate Oicubes7 so 00 QO 0 Z All licubes in X are degenerate7 so 01 Q1D1 Thus B0 6101 So H0X will equal Z0X7 once we decide which Oichains to call cycles If you go back through the de nitions7 you will realize that we never de ned the boundary of a Oichain7 so we dont yet know which of those we will say have 0 boundary That is for the next sub lsection Vll22 2004 J Simon all rights reserved page 4 HOMOTOPY OF MAPS AND SPACES Jr SIMON 22M2201 FALL 05 For this section refer to text Hatcher Ch 0 pages 174 and the Exercises on pages 18720 1 INTRODUCTION You already have studied the idea of topological equivalence two spaces that look different in their rigid geometry but are equivalent if we soften our idea of geometric equivalence to allow the objects to bend stretch etc In this way a circle and a square are topologically equivalent We now wish to expand our notion of equivalence to capture in a formal way the intuitive idea of squishing one set to make it look like another Here are two standard examples you can think of when pondering the de nitions Example 1 Contracting a disk to a point Let D2 be the standard unit disk in R2 Gradually shrink D2 within itself by making the radius smaller and smaller until ultimately we have just the point at the center Example 2 Contracting a solid torus to a circle Consider a solid torus homeomorphic to D2 gtlt 51 Apply the above quotsquishing to each of the disks D2 gtlt s s E 81 In this way we can continuously deform a solid torus to a circle namely the circle that is the core of the solid torus When we can continuously deform a set within itself to some subset we want to say that the two sets have essentially the same shape They won7t in general be homeomorphic but they will have many topological properties in common for example our much beloved Euler characteristic The two sets Ell have the same kinds of holes as discussed in our Course Description If we want to say that some set X has the same homotopy type as some subset A which we denote X 2 A then we also are saying A 2 X So the process of getting from a given set to something ALL RIGHTS RESERVED 2 J SIMON 22M2201 FALL 05 homotopy equivalent to it cannot be just a matter of going from big sets to smaller sets we have to allow also going from small sets to bigger ones 2 STRONG DEFORMATION RETRACTIONS Here we make precise the intuitive idea of squishing77 a set to a subset De nition 1 Suppose X is a space with subspace A A map f X gtlt I a X is called a strong deformation retraction ofX to A if 0 Va 6 X fp0 p 0 Va 6 X fp1 E A 0 Va 6 A7 Vt E I fat a If the third property above were weakened to just say fa7 t 6 A7 we would have what Hatcher calls a deformation retraction in the weak sense Some people say strong deformation retraction vs plain vanilla deformation retraction Others7 in particular Hatcher7 use deformation retraction vs weak deformation retraction I will try to avoid confusion by trying to remember always to say strong or weak as appropriate Example 3 Contracting a disk to a point the right way We need to de ne a map f D2 gtlt I a D2 with the right properties De ne fxt 1 itx Example 4 Contracting a solid torus to a circle the right way We need to de ne a map f D2 gtlt Sl gtlt I a D2 gtlt 1 with the right properties De ne fxst Example 5 Contracting a solid torus to a Mobius band Let T3 be the solid torus in Example 4 We can construct a Mo39bius band M2 C T3 so that M2 intersects each meridional disk of T3 in a diameter of that disk The diameters rotate continuously as we move around the torus Remark For an entertaining exercise7 try writing an explicit parametrization of a solid torus in R3 and then parametrize the desired Mobius band Note that the deformation in Example 3 moved each diameter within itself so if we eptend that in the obvious way to shrink T3 to its core circle that deformation will shrink M2 in itself to its center circle This shows that a Mo39bius band can be strong deformation retracted to a subset homeomorphic to 81 In Example 37 how would a person do a careful proof that the function does the job I wrote x in boldface7 to remind us that x is a vector7 and the product tx is scalar multiplication The three desired properties are just basic properties of scalar multiplication of vectors HOMOTOPY OF MAPS AND SPACES 1 3 1X x 0x 0 and t0 0 Sometimes we need to work to show that an alleged deformation retraction map is continuous In this example we can use the E 7 6 de nition of continuity in R2 or invoke generalities about cartesian products and scalar multiplication HOMEWORK 961 Hatcher page 18 Exercise 2 write explicit equations HOMEWORK 962 Hatcher page 18 Exercise 1 You might nd it easier to do this via a sequence of pictures rather than writing out equations though doing it via equations would be OK Pictures are part of the rhetoric of discourse in Topology so they are a language you canshould try talking Remark on varying speed There will be situations later where we want to change the speed with which certain deformations take place Just to start you thinking about this here is a function F that works just as well in Example 3 as our original function f 172m 1ft FX tf iftgt NlD NlH How would you convince a skeptic that F is continuous First invoke the same arguments as before to say that each piece of F is continuous then invoke the Pasting Lemma77 from point set topology Notice that you can realize this mapping F as a composition First map the interval 01 a 01 by the map h below 1720 iftgl W i f 0 1ft2 Then realize that Fx t fx ht 4 J SIMON 22M2201 FALL 05 3 STRONGLY CONTRACTIBLE SPACES A disk can be strong ldeformation retracted to a point so can any solid ball D Let us say that a space with this property is called strongly contractible This is not a standard term7 but saying contractible77 alone means something slightly more general we will de ne that soon See Hatcher page 4 HOMEWORK 963 0 Suppose X A U B where A B p7 a single point7 such that A is strongly contractible to p and B is strongly contractible to Prove A U B is strongly contractible 0 Suppose X A U B where A B P7 P is strongly contractible7 A strong deformation retracts to P7 and B strong deformation retracts to P Prove A U B is strongly contractible ln point set topology7 you studied Urysohn7s Lemma Here is one of its corollaries Theorem 1 Tietze Extension Theorem Suppose A is a closed subset of a normal space X Let f A a 01 be a continuous function Then there epists a continuous function F X a 01 such that FlA f Once you prove this7 you ie one can show that the Tietze theorem holds if we replace 1 by any cell I or R E 01 We call a space solid if the Tietze theorem holds with that space as the target De nition 2 A subspace A of a space X is called a retract ofX if there epists a map r X a A such that rlA id The map r is called a retraction ofX to A and A is called a retract of X Remark We will see later that the famous Brouwer Fixed Point Theorem is equivalent to the theorem that there does not exist a retraction of a ball D to its boundary SW1 HOMEWORK 964 A retract of a solid space is solid HOMEWORK 966 A solid space is strongly contractible 22M201 Fall 06 J Simon Homology of a Cell Complex A nite cell complex X is constructed one cell at a time7 working up in dimension Each time a cell is added7 we can analyze the effect on homology and7 by this inductive process7 calculate the homology groups of X In these notes7 we will develop some of the main properties of homology of nite cell complexes by using this inductive approach Recall that each nitely generated abelian group G can be expressed as the direct product F gtlt T7 where F is a free abelian group Z gtlt Z gtlt gtlt Z and T is nite The number of Z factors is called the rank of G The rank of G is well de ned that is7 given any two decompositions of G as F1 gtlt T1 or F2 gtlt T27 the ranks of the free abelian parts must be the same However7 there are different ways to express the torsion parts for example Z2 gtlt Z3 Z6 Theorem 1 Suppose X is a nite CW complep of maximum dimension n For each h 2 0 let ck denote the number of cells ofX of dimension k and let Bk denote the rank of Then 1 For each h 2 n 1 HkX 2 HnX is afree abelian group of rank 3 en 3 For each h HkX is generated by some set of ck or fewer elements In other words HkX is a quotient group of the free abelian group ch 4 22204ka ZLoPUk k Note that part 4 says the Euler characteristic of X is a topological invariant in fact7 a homotopy type invariant The alternating sum of the numbers of cells in each dimension is the same7 regardless of how we triangulate X or otherwise express X as a CW complex Part 3 does Lot say that Bk ck typically7 Bk lt ck But it does imply that each Bk is nite This is why the sum of Betti numbers in part 4 makes sense J Simon all rights reserved page 1 Theorem 1 is a corollary of the following Theorem 2 Suppose a space X is obtained as a quotient space by attaching a cell B to a space Y via a map f S 1 a Y S 1 is the boundary sphere of En Then the homology groups ofX and Y are related as follows 0 For all h 31 7171 71 HkX o Eccactly one of the following must happen n1X n1Y i1 and HnX 2 HnY or else n1X n1Y and HnX E HnY gtlt Z Remark Theorem 2 says exactly what HnX must be in the two cases but it does not say exactly what Hn1X must be it just speci es the rank Bibi For example7 if Y is a Mobius band7 and we attach a 2 cell B2 to Y via a homeomorphism f of their boundaries so X is R132 then H1X gZ2 51X051Y 1 H2X0 lf7 instead7 the attaching map f identi es the boundary of B2 with the center line of Y7 then we have H1X gio 51X 051Y 1 H2X 0 The basic idea of Theorem 2 is that when we add a new n cell to Y7 we either increase the rank of Hn or else decrease the rank of Hn1 More precisely7 adding an n cell to Y kills77 an element of Hn1Y if that element has in nite order in Hn1Y7 then we change the rank of Hn1Y if that element is zero7 or has nite order7 in Hn1Y7 then we dont change the rank of Hn1Y7 but we do create a new free generator for H HOMEWORK Due Friday Dec 1 Prove Thereom 1 assuming Theorem 2 Hint First prove the theorem in the case n 0 and on all values Let c07 c17 7on be the list of numbers of cells of X Do the proof by induction on this n tuple For the inductive step7 Let Y be all of X except for the nal n cell Then7 by inductive assumption7 Y satis es the theorem Now apply Theorem 2 and work out the bookkeeping J Simon all rights reserved page 2 PROOF OF THEOREM 2 Lemmas for Theorem 2 We need a topology lemma and an additional small algebra lemma Lemma 21 Suppose a space X is obtained as a quotient space by attaching a cell B to a space Y via a map f Sn 1 7 Y Then foiquot each h HkX7Y g HkBn7Sn71 Proof View B as the unit ball in R Let Sl boundary B and let 512 be the sphere of radius centered at the origin Let U be the closed spherical shell bounded by 512 and 51 Let BlZ denote the closed ball bounded by 512 Note that 1 is a strong deformation retract of U When we attach the ball B to Y via map f 81 7 Y the set U U Y becomes the mapping cylinder of f Let YlZ denote this set U Uf Y We claim that the pair X Ylg is homotopy equivalent to the pair X Y The pair of line segments 01 1 is homotopy equivalent to the pair 011 just gradually shrink 1 to its right endpoint while gradually expanding 0 to ll 01 Apply this radially to B to get a homotopy equivalence B U 2 B Sl Do the same deformations but now with the identi cations made on 51 via f to see how to deformation retract YlZ to Y while expanding B 7 int U to ll B This is just a slight modi cation of the proof that a mapping cylinder deforms to its target end 7 see Hatchep s discussion of mapping cylinders 7 the target end is the easy end Thus we have a homotopy equivalence YUf BnYUf 2 Uf ie XY12 2 XY The next step is to excise Y from X Ylg We introduced the neighborhood U in order to ensure that Y is contained in the interior of Ylg By the Excision Theorem for each h HkXY12 2 HkX 7 YY12 7 Y But X 7 Y is just the open ball B 7 81 and Y12 7 Y is the half open shell U 7 51 so X 7 Y Y12 7 Y 2 Bl2 512 2 En SW1 J Simon all rights reserved page 3 Lemma 22 Suppose o A a B is a group homomorphis Then o de nes an injection 15 G ker o a H Proof Easy unassigned exercise D We can use Lemma 22 to break a long exact sequence into a collection of short exact sequences Suppose we have groups and hornornorphisrns Aiwio where z mf kerg Then Aeiapn fega0 and A 0 a Bz mf A o are exact Proof of Theorem 2 Consider the long exact sequence for the pair WW gt Hk1XY a HkY a HkX a HkXY a By Lemma 21 for each k HkX Y E HkB SW1 which we know from earlier work recall how we calculated hornology groups of spheres is isomorphic to So in the long exact sequence we can replace each HkX Y by 0 if k 31 n and Z for k n Now consider thecases k gt n k lt n 7 1 and the intermediate situations Caseskgtnorkltn71 We have H Y NEO a Cases kn n 1 We have Hn1X Y 0 Hn1X Y so the homology groups in dimensions n and n 7 1 are connected by the nite exact sequence 0 a HAY a HnXY 71Y 71X a 0 We also know HX Y E Z so we actually have mamm mm n m pamm m We have been omitting the names of the various hornornorphisrns but now we need to refer to the function g The image of j is some subgroup pZ of Z There are two separate cases Either p 0 or else p 31 0 and these are the two different cases for Theorem 2 J Simon all rights reserved page 4 Subcase p 0 By Lemma 22 see the discussion following that lemma our sequence breaks into two short exact sequences memmemm m and 0 a Z 0 Hn1Y 71X a 0 This says that HnX E HAY and by the rank nullity theorem77 for nitely generated abelian groups n1Y n1X 17 which is the rst alternative in Theorem 2 Subcase p 31 0 Again by Lemma 227 the sequence breaks into short exact sequences memmemm wem and 0 a ZpZ a Hn1Y n71X 0 The subgroup 19 is isomorphic to Z so the rst short exact sequence splits7 to give HnX HAY gtlt Z Meanwhile7 the group ZpZ is nite7 so the rank nullity theorem says n1X n1Y This completes the proof of Theorem 2 D J Simon all rights reserved page 5 22M201 Fall 06 J Simon Brief note on Quotient Spaces We often construct spaces as quotient spaces or identi cation spaces For example 1 On the disk D2 de ne an equivalence relation by saying z N y for all Ly E boundary SZ ie using a standard but sometimes ambiguous notation for the boundary of something z N y ltgt Ly E 6D2 The quotient space is or at least appears to be homeomorphic to 82 2 Let X I gtlt I C R2 De ne an equivalence relation on X by For each t E I t 1 N t 0 Then the quotient space X 1 gtlt 01 3 Let X I gtlt I C R2 De ne an equivalence relation on X as follows For each t E I t 1 N t 0 for each s E I 0 s N 15 Then the quotient space X N is homeomorphic to a torus That is X Sl gtlt 81 4 Let X I gtlt I C R2 De ne an equivalence relation on X as follows For each t E I t 1 N t 0 for each s E I 0 s N 11 7 5 Then the quotient space X N is homeomorphic to a Klein bottle In the rst three examples we already have an a ppioii understanding of the spaces being constructed we know what a 2 sphere is before we try to represent it as a quotient space In the fourth example we may have only a vague intuitive vision of what this surface is supposed to be and we make it into an unambiguous reality by using the quotient space de nition A quotient space is not just a set of equivalence classes it is a set together with a topology How do we know that the quotient spaces de ned in examples 1 3 really are homeomorphic to the familiar spaces we have stated The following lemma is often what tells us we are ok 1 Lemma 1 Suppose X is a topological space w an equivalence relation on X and Q X N the quotient space ie the set of equivalence classes with the quotient topology Suppose Y is a space and f X a Y is a continuous surjective map such that the equivalence classes in X are precisely the point inverses under f that is for each Ly E X z y ltgt u IfX is compact and Y is Hausdoi then X N is homeomoiphic to Y HOMEWORK PROBLEM 1 Prove Lemma 1 Hint Have you ever seen a theorem that started off If f is a map from a compact space to a Hausdorff space such that 77 HOMEWORK PROBLEM 2 Let X E R2 l S x2 y2 S 1 which is homeomorphic to 1 gtlt I De ne an equivalence relation on X by identifying all the points zy with 2 y2 to each other In other words there is one big equivalence class consisting of all the points of the inner circle of X and all the other equivalence classes consist of single points Prove X N 2 D2 Note For this problem so early in the course you are asked to do the proof carefully write equations and use Lemma 1 appropriately Gluing spaces together Here is one very important kind of quotient space Gluing two spaces together via a map Suppose W X H Y the disjoint union of spaces X and Y Suppose A Q X and f A a Y is a continuous function De ne an equivalence relation on W by de ning a partition of W into equivalence classes The partition has 3 kinds of sets If x E X 7 A then is an equivalence class If y E Y 7 fA then M is an equivalence class The third kind of equivalence class consists of a point y E fX together with all the points of A that map to y in other words U f 1y We de ne XUfYXUY all rights reserved page 2 HOMEWORK PROBLEM 3 We de ne below two spaces7 Q1 and Q2 Show that Q1 and Q2 are horneornorphic First7 let Di7y71 6R31x2y2 S1 D3 7y771 E R3 l x2y2 1 Let S be the boundary Circle of Di and let T be the boundary Circle of DE De ne f S a T by ay 1 Ly7 71 and et Q1 Di Uf DE De ne g S a T by fx7y71 7y7z7 71 and et Q2 Di U9 D3 Remark The question of which attaching rnaps77 give the same spaces is almost always interesting all rights reserved page 3 HOMOTOPY PROPERTIES OF CELL COMPLEXES Jr SIMON 22M2201 FALL 05 For this handout refer to text Hatcher Ch 0 and the Appendix 1 MAKING WISHES We have seen how the idea of strong deformation retraction77 can be generalized to the notion of two spaces being homotopy equivalent ie of the same homotopy type In particular De nition 1 A space X is contractible ifX has the homotopy type of a point Notice in the above de nition we are not saying that X deforrnation retracts to any particular point of X We would welcome a theorem saying that if a space X is contractible then X does deforrnation retract to one of its points Perhaps we could even hope for a theorem saying that X can be strong deforrnation retracted to any point we wish Wish 1 IfX is contractible then 3p 6 X such that X strong deformation retracts to X Perhaps we might even hope for Vp E X X s d r s to X Example 1 Let X1 be the in nite fan of arcs consisting of all the line segments in R2 from the origin to the points Li n 123 Let X be the closure ofX1 that is adjoin to X1 the segment from the origin to the point 10 The we can strong deformation retract X to the origin but we cannot deform it to the point p 0 without moving p It is possible to build a more complicated space see Hatcher Exercises that is contractible but does not deforrnation retract to any of its points Note on an English language subtlety in use of the word any If we say This space contracts to any of its points that means Vp E X X contracts to x If we say This space does not contract to any of its points that means Ed 6 X such that X contracts to x There is ALL RIGHTS RESERVED 2 J SIMON 22M2201 FALL 05 no quick one word way to say This space contracts to some points but not to others Here is another example of a theorem that is too good to be true Wish 2 Suppose A is a contractible subset of a space X Then the quotient space ofX modulo A is of the same homotopy type as X that is X 2 XA To see that this is too much to hope for we rst prove Lemma 01 If XY are spaces with X path connected and X 2 Y then Y is path connected The next example is a space X that is not path connected with a contractible subset A such that XA is path connected Thus X cannot be homotopy equivalent to XS Example 2 Let X be the closure of the graph of the function y sin1z That is X96y R2l967 0 ysin196U0yl 713131 Since wishes are free here are more Wish 3 IfA Q X and the inclusion map i A a X is a homotopy equivalence then X strong deformation retracts to X Wish 4 IfX A U B where A B and A B are all contractible then X is contractible Wish 5 Suppose XY are spaces A Q X f A a Y a continuous function and Wf X Uf Y that is Wf is the space obtained by attaching X to Y yia f Suppose we have another map g A7 gt Y that is homotopic to f and let W9 be the space obtained by attaching X to Y yia g Then Wf 2 W9 Wish 6 Every space X is Hausdor eyen normal So theorems like Urysohn s Lemma and the Tietze Eptension Theorem can be used Wish 7 Each point z E X has an open neighborhood Uz such that U strong deformation retracts to x Note this would imply that X is locally path connected Wish 8 Each closed subset A Q X has an open neighborhood UA such that U strong deformation retracts to A In fact ifX A U B where A and B are closed then there are neighborhoods UA VB such that U sdr s to A V sdr s to B and U V sdr s to A B HOMOTOPY PROPERTIES OF CELL COMPLEXES 1 3 Remark on the last wish One of our goals is to prove theorems that let us understand the homology groups and fundamental group of a space by seeing the space as being built from simple pieces eg cells Typically7 the theorems require that the pieces be open sets But7 just as typically7 we do the actual building with compact sets such as cells The last wish77 would extend the open set versions to the compact set constructions 2 WISHES CAN COME TRUE In the above list of wishes7 assume that each space being discussed is a CW complex7 and each subset discussed is a sub complex Then all your wishes come true Here is just one sample of the kinds of calculations that become possible when we restrict our attention to spaces that are nice Example 3 Let X 52 V 81 Let A be an arc Let W be obtained by attaching A to X at the endpoints of A Then all seven check that topologically di erent spaces are homotopy equivalent 22M201 Fall 06 J Simon Mayer Vietoris Sequence If a space X is the union of two open subsets A7 B or at least the interiors of A and B cover X7 then there is a long exact sequence relating the homology groups of X to the homology groups of A7 B and A B As with the ecoeision theorem7 the sequence still is valid if we are dealing with cell complexes7 or some other situation7 where A7 B are deformation retracts of open sets whose intersection deformation retracts to A B iHnm m BLgtHnA ea HnBigtHnA u mi n1A m B a To de ne the homomorphism Ar and later to prove the sequence exact7 we need a lemma based on the idea of small cubes77 see text for details Lemma 01 Assuming X intA U int B Suppose w is an n ehain in X Then there epists n ehains 046 in X and an nZ ehain c in X such that 0 04 is supported in A o B is supported in B o wa 6w as chains When we say a chain is supported77 in some set7 we mean that the chain is a linear combination of singular cubes whose images lie in that set The functions I is based on the homomorphisms ii induced by inclusions of A B into A and B respectively lel Alelzl 7 324M which we can think of as ZlAmB V lZlA7 lZlB The function Jr is based on the homomorphisms j induced by inclusions of A and B into A U B7 but notice the minus sign L ali ml A Mal B l l which we can think of as lalm MB M a 7 lAuB J Simon all rights reserved page 1 We now de ne the function Ar Suppose z is an n cycle in X By Lemma 017 we can nd chains 047670 such that z 04 B 30 7 where 04 is in A ie 04 is supported in A and B is in B For notational convenience7 replace 6 with 767 which7 of course7 also is supported in B Since Z is a cycle7 we know 32 07 so we have 2 Oz 7 B 30 7 and 0626a7666606a766 This says 0 304 36 as chains7 o 304 is supported in A7 but also it is supported in B7 so 304 is supported in A B7 0 and 304 is a cycle in A B We de ne AleX laalAmB DRAW SOME PICTURESH to help Visualize the function Ar For exarnple7 if X is the n sphere expressed as the union of Aupper hernisphere7 and Blower hernisphere7 then all of X is an n cycle7 and this cycle is mapped by Al to the n 1 dirnensional homology class represented by the equator n 1 sphere Lemma 02 Unassigned Show the homomorphisms m particular Ar are well de ned We next consider how to prove that the sequence is exact We work out one part below7 another part is assigned HW7 and the third part is unassigned HW Proof that the sequence is exact at HnA U B We have LAMA ea HnBigtHnA u mi mm m B a First7 lalm MB M la 7 lm V Mlsag But 04 is a cycle7 so its boundary 0 J Simon all rights reserved page 2 The converse takes a little more work Suppose z is a cycle in X and Ar 0 E Hn1A B This means that when we write 2 a 7 B 30 7 there is an n chain w in A B such that 304 3w7 ie 6a 7 w 0 Since it is a chain in A B7 it is7 in particular7 supported in A Thus a 7 w is a cycle in A Sirnilarly7 B 7 w is a cycle in B But then we have J laiwlm 67 3 aiw 67wlAUB7 which says 2 7 30 6 image Jr We are almost done We want 2 6 image Jr But since 2 and z 7 30 differ just by a boundary7 they de ne the same hornology class7 ie z z 7 30 6 image Jr J Simon all rights reserved page 3 22M201 Fall 05 J Simon De ning Cell Complex77 and Why We would like to have a clean77 de nition of cell complecc to go along with our intuitive idea of what it means to cut a space into cells that intersect each other in a controlled way We found in our class discussions that the act of making this intuitive idea precise is more complicatedsubtle than one might rst expect This is a theme that reappears in mathematics In the class handouts7 you can see various examples of de nitions The goal always is to describe a class of spaces that is 0 general enough to include many familiarimportant spaces7 and 0 narrow enough to allow us to prove theorems that say IfX A U B where AB7 and A B are nice and we understand something about some algebraic invariant of A7 B7 and A B and how A B sits within each of A7 B then we can deduce something about that algebraic invariant for X People take two general approaches to de ning some kind of cell complex At one extreme7 one might restrict the cells77 to be very special7 triangles or other simplexes simplices77 for the language purists but why dont we say complices see the de nition of triangulation of a surface in the Massey text At the other extreme7 we can allow the individual cells to be attached to one another in a very general way see the de nition of CW complex in the Hatcher text note Massey also gives a de nition of CW complex7 but it is stated in a spread out way that may be hard to decipher In our course7 I want us to try to use the idea of cell complex in an informal way7 but have the machinery of CW complex as developed in Hatcher as something we can fall back on ifas needed One of the main theorems of our course is the theorem saying that Euler characteristic of a cell complex is a topological invariant Speci cally7 Theorem 1 Suppose X is a space realized as a nite dimensional cell complep of dimension N with a nite number of cells in each dimension For each i let BAX denote the rank of the ith homology group Then the alternating sum of the numbers of cells in each dimension equals the alternating sum of the 61 1 2 22Mz201 Fall 05 J Simon That is if 71 denotes the number of cells of dimension j N 00 24 71739 ZEN BAX 70 i0 I used different letters 27 to index the above sums to emphasize that the respective numbers nk and Bk are not claimed to be the same 7 though they are related We will see that BAX 3 7 Thus in fact the sum of Betti numbers is not really in nite it stops at N or sooner For example when we represent the 2 sphere 52 as the surface of a cube with 710 8711 12712 6 we have later in the semester H0SZ E Z H1SZ 0 H2SZ E Z and zero thereafter so 5017 5107 521 and zero thereafter On the other hand if we consider the solid cube then we have the same numbers of cells in dimensions 0 1 and 2 and acquire one new cell in dimension 3 Meanwhile the solid cube has 60 1 and all other Betti numbers 0 All this will come much later I am hoping that by introducing some words and notation now they will seem normal to you later so you7ll be able to focus on the ideas and not get tied up inby the syntax 22M201 Fall 04 J Simon Brief note on Quotient Spaces We often construct spaces as quotient spaces or identi cation spaces For example 1 On the disk D27 de ne an equivalence relation by saying z N y for all Ly E boundary SZ7 ie using a standard7 but sometimes ambiguous7 notation for the boundary of something7 z N y ltgt Ly E 6D2 The quotient space is or at least appears to be homeomorphic to 82 2 Let X I gtlt I C R2 De ne an equivalence relation Rel on X by For each t 6 I7 t71 t7 0 Then the quotient space XRel E 1 gtlt 01 3 Let X I gtlt I C R2 De ne an equivalence relation Rel on X by For each t 6 I7 t71 t70 w for each s 6 I7 05 N 15 Then the quotient space XRel is homeomorphic to a torus That is7 XRel E 1 gtlt 1 4 Let X I gtlt I C R2 De ne an equivalence relation Rel on X by For each t 6 I7 t71 t70 for each s 6 I7 05 1717 5 Then the quotient space XRel is homeomorphic to a Klein bottle In the rst three examples7 we already have an a ppioii understanding of the spaces being constructed we know what a 2 sphere is before we try to represent it as a quotient space In the fourth example7 we may have only a vague intuitive vision of what this surface is supposed to be7 and we make it into an unambiguous reality by using the quotient space de nition A quotient space is not just a set of equivalence classes7 it is a set together with a topology How do we know that the quotient spaces de ned in examples 1 3 really are homeomorphic to the familiar spaces we have stated The following lemma is what tells us we usually are ok 1 Lemma 01 Suppose X is a topological space Rel an equivalence relation on X and XRel the quotient space ie the set of equivalence classes with the quotient topology Suppose Y is a space and f X a Y is a continuous supjective map such that the equivalence classes in X are precisely the point inverses undeiquot f that is for each Ly E X z y ltgt u IfX is compact and Y is Hausdoi then XRel is homeomoiphic to Y 2005 all rights reserved page 2 HOMOTOPY PROPERTIES OF CELL COMPLEXES Jr SIMON 22M2201 FALL 06 For this handout7 refer to text Hatcher7 Ch 0 and the Appendix 1 MAKING WISHES We have seen how the idea of strong deformation retraction77 can be generalized to the notion of two spaces being homotopy equivalent7 ie of the same homotopy type In particular7 De nition 1 A space X is contractible ifX has the homotopy type of a point Notice in the above de nition7 we are not saying that X deforrnation retracts to any particular point of X We would welcome a theorem saying that if a space X is contractible7 then X does deforrnation retract to one of its points Perhaps we could even hope for a theorem saying that X can be strong deforrnation retracted to any point we wish Wish 1 IfX is contractible then 3p 6 X such that X strong deformation retracts to X Perhaps we might even hope for Vp E X X s d r s to X Example 1 Let X1 be the in nite fan of arcs consisting of all the line segments in R2 from the origin to the points 17 7 n 172737 Let X be the closure ole that is adjoin to X1 the segment from the origin to the point 17 0 The we can strong deformation retract X to the origin but we cannot deform it to the point p a 0 without moving p It is possible to build a more complicated space see Hatcher Exercises that is contractible but does not deforrnation retract to any of its points Note on an English language subtlety in use of the word any If we say7 This space contracts to any of its points 7 that means Vp 6 X7 X contracts to x If we say7 This space does not contract to any of its points 7 that means Ed 6 X such that X contracts to x There is no quick one word way to say7 This space contracts to some points but not to others 1 HOMOTOPY PROPERTIES OF CELFHM OMPDEMWI FALL 06 Here is another example of a theorem that is too good to be true Wish 2 Suppose A is a contractible subset of a space X Then the quotient space ofX modulo A is of the same homotopy type as X that is X 2 XA To see that this is too much to hope for we rst prove Lemma 01 If XY are spaces with X path connected and X 2 Y then Y is path connected The next example is a space X that is not path connected with a contractible subset A such that XA is path connected Thus X cannot be homotopy equivalent to XS Example 2 Let X be the closure of the graph of the function y sin1z That is Xy R2l7 07 ysin1U0yl 713131 Since wishes are free here are more Wish 3 IfA Q X and the inclusion map i A a X is a homotopy equivalence then X strong deformation retracts to X Wish 4 IfX A U B where A B and A B are all contractible then X is contractible Wish 5 Suppose XY are spaces A Q X f A a Y a continuous function and Wf X Uf Y that is Wf is the space obtained by attaching X to Y yia f Suppose we have another map g A7 gt Y that is homotopic to f and let lVg be the space obtained by attaching X to Y yia g Then Wf 2 W9 Wish 6 Every space X is Hausdor eyen normal So theorems like Urysohn s Lemma and the Tietze Eptension Theorem can be used Wish 7 Each point z E X has an open neighborhood Uz such that U strong deformation retracts to x Note this would imply that X is locally path connected Wish 8 Each closed subset A Q X has an open neighborhood UA such that U strong deformation retracts to A In fact ifX A U B where A and B are closed then there are neighborhoods UA VB such that U sdr s to A V sdr s to B and U V sdr s to A B Remark on the last wish One of our goals is to prove theorems that let us understand the homology groups and fundamental group of a space by seeing the space as being built from simple pieces eg J Simon all rights reserved page 2 J SIMON WEOPXWPERTIES OF CELL COMPLEXES cells Typically7 the theorems require that the pieces be open sets But7 just as typically7 we do the actual building with compact sets such as cells The last wish would extend the open set versions to the cornpact set constructions 2 WISHES CAN COME TRUE In the above list of wishes7 assume that each space being discussed is a CW cornplex7 and each subset discussed is a sub cornplex Then all your wishes come true Here is just one sample of the kinds of calculations that become possible when we restrict our attention to spaces that are nice Example 3 Let X 52 V 81 Let A be an arc Let W be obtained by attaching A to X at the endpoints of A Then all seven check that topologically di epent spaces are homotopy equivalent J Simon all rights reserved page 3 22M201 Fall 05 J Simon GROUP PRESENTATIONS AND TlETZE TRANSFORMATIONS continued Here is a step by step proof using Tietze transformation that the pre sentations lta b l 12 b3gt and lt9 l 96W yzygt present isomorphic groups Start with ltxy l syx yxygt and introduce new generating symbols as nicknames77 for existing elements A H 721 l 96W yzygt g 2170 l 96W Wych 2cm 1170le l yzywyawnbwgt 3 A 3 VV Add new relation that is a consequence of the ones already there Since a2 xyzxyx and b3 xyxysy um our group now is seem to be isomorphic to lt7y707blyyy7ay7byyazb3gt 4 But now the relation syx yzy can be deduced from the other three that is in any group with elements aby the three equations a xys b zy and a2 b3 together imply that syx yxy So in the presence of the other three relations the rst one is redundant and can be deleted without changing the group More formally in the free group of rank 4 generated by aby the normal subgroup mu4964114 7 flu967 Flu7 025 5 049611967 b lxy 7 azb 3gtgt 6 The rst two relators which remind us how we introduced a77 and b as words in z and y can also be used to express z and y in terms of a and b We just solve the two equations for unknowns z and y The second equation says y x lb Substitute that into the rst equation to get a 1zx 1bx 1 ie a lbx 1 which implies z b la Now substitute that back into the equation y x lb to get y b la 1b a lbz Since the equations giving z and y in terms of a and b are consequences of the other equations our group is now isomorphic to ltabzy l a lsyx b lxy azb s z b la y a 1b2gt But now you can check that the last two equations imply the rst two even in the free group on 1 bxy symbols Just replace each x and each y by b la and 1 le respectively to show a lzys 1 and b lsy 1 Thus we have ltabzy l azb s z b la y a 1b2gt Finally we eliminate each generator my and the single relation that de nes it as some word in the other generators This leaves just ltabzy l azbis gt Another way to prove WWW 9620 lt01 l 02 53gt would be to exhibit isomorphisms De ne b ltzylxyx yzygt a ltabla2 b3gt by Ms b la and y a lbz De ne 1 ltabla2 b3gt a ltzylzyx yygt by 7a syx and 7b xy Then show that the functions b and 7 de ne homomorphisms and that they are inverses of each other D end of handout 3 on group presentations FUNDAMENTAL GROUP OF 51 J1 SIMON 22M2201 FALL 08 The purpose of these notes is to summarize part of the text Hatcher proof and then to nish the proof in a slightly more elementary way than the text Hatcher proves a certain homomorphism is bijective by proving one big lemma In this handout we will see separate proofs for 1 177 and onto After you understant them as separate arguments then you may be better equipped to appreciate the conceptual unity in Hatcher7s approach The uni ed approach is the general homotopy lifting theorem for covering spaces which we will see later 1 THE FUNCTION lt1gt FROM Z TO 7T1Slz0 Let p R1 a 1 be the wrapping function pt cos27rt sin27rt For n E Z let En 01 a R1 be the function 1749 n5 Note that 47 maps the interval 01 to the interval 071 We de ne Lon 01 a 1 by ion po n Let x0 1 0 E 81 We de ne the function lt1gt Z a 7T1Slx0 by lt1gtn Our goal is to show that lt1gt is a group isomorphism 2 SHow lt1gt IS AN ISOMORPHISM In this section and again later we will use the following lemma Lemma 01 Suppose 52 01 a R1 is a path with 520 0 and 521 n Then p o 52 ion 6 7T1Slz0 The same conclusion holds under the weaker assumption that 520 and 521 6 Z with 521 7 520 n Proof First consider the case where 520 0 Since R1 is convex the path 52 is path homotopic to the any other path from 0 520 to 521 in particular 52 is path homotopic to Q where n 521 But then composing with p gives the desired homotopy in 81 For the case where 520 h 31 0 consider the path 3 given by 65 525 7 h Then 6 satis es the rst case in our proof and p o 52 p o D 31 Simon all rights reserved 1 2 J SIMON 22Mz201 FALL 08 Proposition 01 ltIgt is a homomorphism Proof We need to check that ltlgtm n m ltlgtn in 7T1Slz0 First ltlgtm n p o ltmn by de nition of ltlgt Note we are composing p with a constant speed path in R1 from 0 to m n On the other hand by de nition of the group operation in 7T1Slz0 m ltlgtn p o amxp o If we let Q denote the straight line constant speed path in R1 from n to n m then p o 17 p o 47 identically so lt12 o wmgtltp o a lt12 o wmgtltp o m p o mam By Lemma 01 p0 mam P0mnl D Proposition 02 ltIgt is surjective We need to show that each loop in 1 is in one of the loop classes w We will do this by showing that each loop f in 1 actually is of the form po f for some path f 01 a R1 By Lemma 01 any path f from 0 to some n E Z is homotopic to Q and that will give f w Thus the following lemma shows that ltIgt is surjective Lemma 02 Let f 01 a 1 be a loop in 1 with f0 f1 0 Then there epists a path f 01 a R1 such that p o f f Moreover there epists eccactly one such function f with f0 0 Remark Since p o 0 f1 E Z Remark This part of the proof is the quotbig deal The relationship between 1 and R1 more precisely our map p R1 a 81 is an eccample of a covering space We plan to show in our particular case that loops in 1 lift to paths in R1 and that homotopies in 1 lift to homotopies in R1 The ability to lift loops and homotopies is the reason quotcovering spaces are important We will see spaces with elegant and powerful relationship between their fundamental groups and in the special case of a simply connected covering space such as R1 covering 51 we can E the fundamental group of the quotbase space here 51 as a group of translations of the covering space here Z group of integer translations of R1 Additional Remark We are using the word quotcovering in two unrelated ways one use is the one you have been using before to describe a bunch of sets whose union contains some particular set the other is this new idea of quotcovering map and quotcovering space Proof The map p R1 a 1 is not 1 1 but it is locally 1 1 In particular we can nd an open cover U1U2 UL of S 1 such that FUNDAMENTAL GROUP OF S1 3 for each 239 p 1UZ is a set of pairwise disjoint open intervals 711 712 such that for each j7 plUZj is a homeomorphism of 71 onto Ui Note the number L of sets Ul is not important the point is that there is some open cover of 1 consisting of connected open sets each of which lifts in the sense of pre image under p to a bunch of homeomorphic copies of itself In the case of p R1 a 817 we can take L 2 Since f is continuous7 the sets f 1U17f 1U27 7fquotlUL form an open cover of 01 So7 by the Lebesgue number lemma7 we can partition 01 into intervals 07517 5152 7 5K71 such that each sub interval 515 is contained in some fquotlUl7 that is fltl5q7 Srl g Ui39 We now construct the map finductively Start by de ning 0 The set f0751 is contained in some Ui One of the components and only one of p lUZ7 call it 7107 contains the point The map p has a local inverse map p l UZ a 710 use that inverse to de ne 7 on 07 51 That is7 07 51 p 1 0 f7 where here p l de ned homeomorphism of UZ a 710 is the well The next step is almost identical to the rst The point f51 has been de ned The interval 517 52 is small enough to lie in some Uj Let 771 be the component of p 1UJ that contains the point sl Use the fact that plUJl has an inverse to extend fa little further7 to now include 517 52 in its domain We continue inductively7 until all the sub intervals comprising 01 are included in the domain of Note that once we declare f0 07 there are no more free choices in constructing In addition to showing that jfvexists7 the above proof shows that fis unique If we had started with some other intger k7 then the resulting fwould be a translation of the one with f0 0 D Proposition 03 ltIgt is 1 Proof We wish to show that m ltIgt0 gt m 0 This is accomplished by the following lemma Lemma 03 Special case of homotopy li ting Suppose F 12 a 1 is a homotopy of loops at 0 between a loop fs Fs7 0 and the constant loop e10s zo So Vt 4 J SIMON 22M2201 FALL 08 130t F1t 230 Then there exists a map 1312 gt R1 with F00 0 such thatpoF F Proof We start by de ning the map I3 on the boundary of I 2 On the bottom it is f on the top and sides it is the constant map 3 t gt 0 We are going to construct I3 inductively by building up the domain in stages Let s call the boundary of I 2 Do as the initial partial domain for F As in the previous lemma we can subdivide I 2 into rectangles small enough that F takes each rectangle into one of the sets U C S 1 Number the rectangles sequentially starting from the lower left completing each row before starting the next As in the previous lemma the first Nrectangle intersects D0 in a connected set Thus HOMEWORK 1 F D0 has an unique extension to this first rectangle Let D1 denote DO U the first rectangle Then the second rectangle intersects D1 in a connected set so there is SAME HOMEWORK 1 an unique extension of I3 to D1U the second rectangle Check that we can ennumerate all the rectangles so that each rectangle meets the union of D0 and the Previous rectangles in a connected set In this way we build the map F D HOMEWORK 2 Show how Lemma 03 implies that I is 1 1 FIGURE 1 The first few sub rectangles of I 2

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