[PDF] 01 Degrees

Math 312, Intro to Real Analysis: Homework  Solutions

10 8 Assume that (sn) is a nondecreasing sequence of real numbers Let σn be the average of the first n numbers in our given sequence: σn = s1 +··· +sn n We claim that the sequence (σn) is again nondecreasing

Series - UC Davis Mathematics

4 1 Convergence of series 61 Telescoping series of the form X1 n=1 (a n a n+1) are another class of series whose partial sums S n= a 1 a n+1 can be computed explicitly and then used to study their convergence

01 Degrees

10 Sn hasacontinuousfieldofnon-zerotangentvectorsifandonlyifnisodd Proof Suppose x7v(x) is a tangent vector field on Sn, assigning to a vector x2 Sn the vector

Chapter 3

22 3 Continuous Functions If c ∈ A is an accumulation point of A, then continuity of f at c is equivalent to the condition that lim xc f(x) = f(c), meaning that the limit of f as x → c exists and is equal to the value of f at c

The Definition of a Manifold and First Examples

WOMP 2012 Manifolds Jenny Wilson The Definition of a Manifold and First Examples In brief, a (real) n-dimensional manifold is a topological space Mfor which every point x2Mhas a neighbourhood

Homework  { corrections

Math 320: Analysis I Spring 2015 By the Completeness Axiom, supfs n: n > Ngand supft n: n > Ngexist and are real numbers Since supfs n: n > Ngand supft n: n > Ngare the least upper bounds for fs n: n > Ngand ft n: n > Ng, respectively, the last inequality in (1) implies s n+ t n supfs n: n > Ng+ supft n: n > Ng: (3)

Trig Cheat Sheet - Pauls Online Math Notes

Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p

Math 412 The Orbit Stabilizer Theorem

(c)Karen E Smith 2018 UM Math Dept licensed under a Creative Commons By-NC-SA 4 0 International License Math 412 The Orbit Stabilizer Theorem Fix an action of a group G on a set X For each point x of X, we have two important concepts: DEFINITION: The orbit of x 2X is the subset of X O(x) := fg xjg 2GgˆX:

Mathematical induction & Recursion

Recursive definition of * • Basis step: – empty string * • Recursive step: –I wf * and x then wx * M Hauskrecht Length of a String Example: Give a recursive definition of l(w), the length of the string w Solution: The length of a string can be recursively defined by: l(‘’) = 0; the length of the empty string

[PDF] math forte secondaire 5

[PDF] cours l'air qui nous entoure

[PDF] controle chimie 4ème l'air qui nous entoure

[PDF] quel est le pourcentage du rayonnement solaire qui traverse effectivement l'atmosphère terrestre

[PDF] l'air qui nous entoure 4ème

[PDF] comment appelle t on la couche dans laquelle nous vivons

[PDF] qu'est ce que l'atmosphère terrestre

[PDF] qu'est ce qui mesure la quantité de vapeur d'eau

[PDF] synthese d'un anesthesique la benzocaine correction

[PDF] synthèse de la benzocaine seconde

[PDF] synthèse du 4-nitrobenzoate d'éthyle

[PDF] synthèse de la benzocaine exercice seconde

[PDF] synthèse de la benzocaine sujet

[PDF] synthèse de la benzocaine terminale s

[PDF] la benzocaïne est utilisée en médecine comme anesthésique local d'usage externe

0.1 Degrees

Definition 0.1.1.The degree of continuous mapf:Sn!Snis defined as: degf:=f(1)(0.1.1) wheref:eHn(Sn) =Z!eHn(Sn) =Zis the homomorphism induced byfin homology, and

12Zdenotes the generator.

The degree has the followingproperties:

1.degidSn= 1.

Proof.This is because(idSn)=idwhich is multiplication by the integer 1.2.If fis not surjective, thendegf= 0.

Proof.Indeed, iffis not surjective, there is ay =2Imagef. Then we can factorfin the following way: S nf > Sn S nn fygh>g> SinceSn fyg=Rnis contractible,Hn(Snn fyg) = 0. Thereforef=hg= 0, so degf= 0.3.If f'g, thendegf= degg. Proof.This is becausef=g. Note that the converse is also true (by a theorem of

Hopf).4.deg(gf) = deggdegf.

Proof.Indeed, we have that(gf)=gf.5.If fis a homotopy equivalence, thendegf=1. Proof.By definition, there exists a mapgso thatgf'idSn. The claim follows directly from 1, 3, and 4 above, sincefg=idSnimplies thatdegfdegg= degidSn=

1.6.If r:Sn!Snis a reflection across somen-dimensional subspace ofRn+1, that is,

r(x0;:::xn)7!(x0;x1;:::;xn), thendegr=1. 1 Proof.Without loss of generality we can assume the subspace isRnf0g Rn+1. The upperUand lower hemispheresLofSncan be regarded as singularn-simplices, via their standard homeomorphisms withn. Then the generator ofHn(Sn)is[UL].

The reflection maprmaps the cycleULtoLU=(UL). So

r ([UL]) = [LU] = [(UL)] =1[UL] sodegr=1.7.If a:Sn!Snis the antipodal map (x7! x), thendega= (1)n+1. Proof.Note thatais a composition ofn+1reflections, since there aren+1coordinates inx, each changing sign by an individual reflection. From 4 above we know that

composition of maps leads to multiplication of degrees.8.If f:Sn!Snis a continuous map, andSf:Sn+1!Sn+1is the suspension offthen

degSf= degf.

Proof.Recall that iff:X!Xis a continuos map and

X=X[1;1]=(X f1g;X f1g)

denotes the suspension ofX, thenSf:=fid[1;1]=, with the same equivalence as inX. Note thatSn=Sn+1.

The Suspension Theorem states that

e

Hi(X)=eHi+1(X):

We proved this fact by using the Excision Theorem. Here we give another proof by using the Mayer-Vietoris sequence for the decomposition

X=C+X[XCX;

whereC+XandCXare the upper and lower cones of the suspension joined along their bases: SinceC+XandCXare both contractible, the end groups in the above sequence are both zero. Thus, by exactness, we geteHi(X)=eHi+1(X), as desired. LetC+Sndenote the upper cone ofSn. Note that the base ofC+SnisSnf0g Sn. Our mapfinduces a mapC+f: (C+Sn;Sn)!(C+Sn;Sn)whose quotient isSf. The 2 long exact sequence of the pair(C+Sn;Sn)in homology gives the following commutative diagram: 0> eHi+1(C+Sn;Sn)'eHi+1(C+Sn=Sn)@ >eHi(Sn)>0 e

Hi+1(Sn+1)(Sf)_@

>eHi(Sn)f _Note thatC+Sn=Sn=Sn+1so the boundary map@at the top and bottom of the diagram are the same map. So by the commutativity of the diagram, sincefis defined by multiplication by some integerm, then(Sf)is multiplication by the same integerm.Example 0.1.2.Consider the reflection map:rn:Sn!Sndefined by(x0;:::;xn)7! (x0;x1;:::;xn). Sincernleavesx1;x2;:::;xnunchanged we can unsuspend one at a time to get degrn= degrn1== degr0; whereri:Si!Siby(x0;x1;:::;xi)7!(x0;x1;:::;xi). Sor0:S0!S0is given by x

07! x0. Note thatS0is two points but in reduced homology we are only looking

at one integer. Consider

0!eH0(S0)!H0(S0)!Z!0

where eH0(S0) =f(a;a)ja2Zg,H0(S0) =ZZ, and: (a;b)7!a+b. Then (r0):eH0(S0)!eH0(S0)is given by(a;a)7!(a;a) = (1)(a;a). Sodegrn=1. 9.

I ff:Sn!Snhas no fixed points thendegf= (1)n+1.Proof.Consider the above figure. Sincef(x)6=x, the segment(1t)f(x) +t(x)

fromxtof(x)does not pass through the origin inRn+1. So we can normalize to obtain a homotopy: g t(x) :=(1t)f(x) +t(x)j(1t)f(x) +t(x)j:Sn!Sn: Note that this homotopy is well defined since(1t)f(x)tx6= 0for anyx2Snand t2[0;1], becausef(x)6=xfor allx. Thengtis a homotopy fromftoa, the antipodal map.3

10.Snhas a continuous field of non-zero tangent vectors if and only ifnis odd.

Proof.Supposex7!v(x)is a tangent vector field onSn, assigning to a vectorx2 S nthe vectorv(x)tangent toSnatx. Regardingv(x)as a vector at the origin, tangency implies thatxandv(x)are orthogonal inRn+1. Ifv(x)6= 0for allx, we may normalize so thatjv(x)j= 1for allx. Assuming this has been done, the vectors (cost)x+ (sint)v(x)lie in the unit circle in the plane spanned byxandv(x). Letting tgo from0to, we obtain a homotopy: f t(x) = (cost)x+ (sint)v(x) from the identity map ofSnto the antipodal map. In terms of degree, this yields (1)n+1= 1, which implies thatnis odd.

Conversely, ifn= 2k1, the vector field defined by

v(x1;x2;;x2k1;x2k) = (x2;x1;;x2k;x2k1) is a nowhere vanishing tangent vector field, sincev(x)is orthogonal tox, andjv(x)j= 1 for allx2Sn.Exercises

1.Letf:Sn!Snbe a map of degree zero. Show that there exist pointsx;y2Snwith

f(x) =xandf(y) =y.

2.Letf:S2n!S2nbe a continuous map. Show that there is a pointx2S2nso that

eitherf(x) =xorf(x) =x.

3.A mapf:Sn!Snsatisfyingf(x) =f(x)for allxis called aneven map. Show that

an even map has even degree, and this degree is in fact zero whennis even. Whennis odd, show there exist even maps of any given even degree.

0.2 How to Compute Degrees?

Assumef:Sn!Snis surjective, and thatfhas the property that there exists some y2Image(Sn)so thatf1(y)is a finite number of points, sof1(y) =fx1;x2;:::;xmg. Let U ibe a neighborhood ofxiso that allUi"s get mapped to some neighborhoodVofy. So f(Uixi)Vy.fis continuous, we can choose theUi"s to be disjoint.

LetfjUi:Ui!Vbe the restriction offtoUi. Then:

4 H n(Ui;Uixi)f !Hn(V;Vy)'(excision)'(excision) H n(Sn;Snxi)Hn(Sn;Sny)'l.e.s.'l.e.s. eHn(Sn)eHn(Sn)'' Z Z Define thelocal degreeoffatxi,degfjxi, to be the effect off:Hn(Ui;Uixi)!Hn(V;V y). We then have the following result: Theorem 0.2.1.The degree offequals the sum of local degrees at points in a generic fiber, that is, degf=mX i=1degfjxi: Proof.Consider the commutative diagram, where the isomorphisms labelled by "exc" follow from excision, and "l.e.s" stands for a long exact sequence. Z =Hn(Ui;Uixi)f degfjxi> Hn(V;Vy)=Z =;exc< Z =Hn(Sn;Snxi)

Hn(Sn;Sny) =exc _Z =Hn(Sn)l.e.s.j^f degf> =;l.e.s< H n(Sn)=Z =l.e.s.^By examining the diagram above we have: k i(1) = (0;:::;0;1;0;:::;0) where the entry1is in theith place. Also,Pij(1) = 1, for alli, so j(1) = (1;1;:::;1) =mX i=1k i(1): 5

The commutativity of the lower square gives:

degf=fj(1) =f mX i=1k i(1) =mX i=1f (0;:::;0;1;0;:::;0) =mX i=1degfjxi;

where the last equality follows from the commutativity of the upper square.Example 0.2.2.Let us consider the power mapf:S1!S1,f(x) =xk,k2Z. We claim

thatdegf=k. We distinguish the following cases: Ifk= 0thenfis the constant map which has degree 0. Ifk <0we can composefwith a reflectionr:S1!S1by(x;y)!(x;y). This reflection has degree1. So since composition leads to multiplication of degrees, we can assume thatk >0. Ifk >0, then for ally2S1,f1(y)consists ofkpoints (thekroots ofy), call them x

1;x2;:::;xk, andfhas local degree 1 at each of these points. Indeed, for the above

y2Snwe can find a small open neighborhood centered aty, call this neighborhood V, so that he pre-images ofVare open neighborhoodsUicentered at eachxi, with fjUi:Ui!Va homeomorphism (which has possible degree1). In our case, these homeomorphisms are restrictions of a rotation, which is homotopic to the identity, and thus the degree offjUiis1for eachi. So the degree offis indeedk. Note that this implies that we can construct mapsSn!Sn of arbitrary degrees for anyn, simply by suspending the power mapf.

0.3 CW Complexes

Start with a discrete setX0, whose points are called0-cells. Inductively, form then-skeleton X nfromXn1by attachingn-cellsen=Int(Dn)via maps@Dn=Sn1 n!Xn1, i.e., X n=Xn1qDn with the identificationx'n(x)for allx2@Dn. As a set,Xn=Xn1qen, whereenis the homeomorphic image ofDnn@Dnunder the quotient map. We can either stop this inductive process at a finite stage, settingX=Xlfor somel, or continue indefinitely, in which case we setX=S nXn. Such a spaceXis called aCW (cell-) complex. One can think ofXas a disjoint union of cells of various dimensions, or asqn;Dn, wheremeans that we are attaching the cells via their respective attaching maps. Each cellenhas a characteristic mapndefined by the composition: D n,!Xn1qDn!Xn,!X: Note thatnjInt(Dnis homeo toen, while the restriction ofnlambdato@Dnis the attaching map'n. 6 A CW complex is endowed the weak topology:AXis open()A\Xnis open for anyn. Ann-cell will be denoted byen=Int(Dn). One can think ofXas a disjoint union of cells of various dimensions, or asqn;Dn, wheremeans that we are attaching the cells via their respective attaching maps. A CW complexXisfiniteif it has finitely many cells. A CW complex is offinite type if it has finitely many cells in each dimension. Note that a CW complex of finite type may have cells in infinitely many dimensions. IfX=[nXnandXm=Xnfor allm > nfor some n, thenX=Xnand we say that the skeleton stabilizes. The smallestnfor whichX=Xn is called the dimension ofX. Example 0.3.1.On then-sphereSnwe have a CW structure with one 0-cell(e0)and one n-cell(en). The attaching map for then-cell is':Sn1=@Dn!point. There is only one such map, the collapsing map. Think of taking the diskDnand collapsing the entire boundary to a single point, givingSn. Example 0.3.2.A different CW structure onSncan be constructed so that there are two cells in each dimension from0ton. Start withX0=S0=fe01;e02g. ThenX1=S1where the two1-cellsD11,D12are attached to the 0-cells by homeomorphisms on their boundary. Similarly, two 2-cells can be attached toX1=S1by homeomorphism on their boundary givingX2=S2. Keep working in this manner adding two cells in each new dimension. Note that if we identify each pair of cells in the same dimension by the antipodal map, we get a CW structure onRPnwith one cell in each dimension from0ton. Example 0.3.3.Thecomplex projective spaceCPn=Cn+1=Cis identified with the col- lection of complex lines through the origin. It is also the orbit space of theC-action on C n+1n f0ggiven by (z0;:::;zn)7!(z0;:::;zn): Let[z0:::::zn]be the equivalence class of(z0;;zn)under this action. Define :D2n!CPn by (z0;:::;zn)7!2 4 z1:::::zn:v uut1nX i=1jzij23 5 Thentakes@D2ninto the set of points withzn= 0, i.e., intoCPn1. Let':= j@D2n. It is easy to check thatfactors throughCPn1['D2nand, moreover, the resulting map

CPn1['D2n!CPn

is a homeomorphism. So it follows inductively thatCPnhas a CW structure with one cell in each even dimension0;2;:::;2n, where the attaching maps are the maps labelled by'. 7 Example 0.3.4.A covering space of a CW complex has a canonical structure as a CW complex. Letf:X!Ybe a covering map so thatYis a CW complex with characteristic maps:Dn!Y. AsDnis simply-connected, eachlifts to a mapen:Dn!X, which are unique upon specification of the image of any point. The collection of all such liftings of allndefine a cell structure onX. 8quotesdbs_dbs15.pdfusesText_21