The degree of a map from a sphere to itself

Cellular Homology

2.4 The degree of a map from a sphere to itself

To exhibit maps of every degree we start with the case n= 1, the circle.

For d ∈ Z define f1,d : S¹ → S¹ by e^2πit → e^2πidt, a map with “winding number”d.

Proposition 2.4.2.The map f1,d has degreed.

Proof. This is an exercise in applying the definitions given in Sect. 2.2. Apply f1,d to the singular 1-simplex (1−t)p0+tp1→e^2πit. Whenever g : Sⁿ → Sⁿ is a map, its suspension Σg : Sⁿ⁺¹ → Sⁿ⁺¹ is illustrated in Fig. 2.1; a formula is:

(Σg)(x, t) =

⎧⎪

⎨

⎪⎩ (√

1−t²g(√¹

1−t²x), t) if−1< t <1 (0, . . . ,0,1) ift= 1 (0, . . . ,0,−1) ift=−1

In the exercises, the reader is asked to check thatΣg is continuous, and that g₁ g₂ implies Σg₁ Σg₂. Define Σ^rg =Σ(Σ^r−1g) by induction. Σ^rg is the r-fold suspension of g. It is convenient to define Σ⁰g to beg. For each d ∈ Z, and each n ≥ 2, define f_n,d : Sⁿ → Sⁿ to be Σⁿ⁻¹f_1,d. The map f_n,0 maps Sⁿ onto a proper subset ofSⁿ, half of a great (n−1)-sphere. By consideringf2,d:=Σf1,d:S² →S², and then generalizing, we may think of fn,d as wrappingSⁿ |d|times around itself, in a “positive” sense ifd >0 and a “negative” sense ifd <0. This is made precise in 2.4.4.

g

Sg

on every level g

Fig. 2.1.

Proposition 2.4.3.For n≥ 1, f_n,1 is the identity map of Sⁿ;f_n,−1 is the map (x₁, x₂, x₃, . . . , x_n+1) → (x₁,−x₂, x₃, . . . , x_n+1); f_n,0 is homotopic to a constant map.

2.4 The degree of a map from a sphere to itself 45 Proof. The statements about fn,1 andfn,−1 are obvious. For the statement aboutfn,0, one proves by induction onnthatPn does not lie in the image of fn,0, wherePn= (−1,0, . . . ,0)∈Sⁿ⊂Rⁿ⁺¹. This is clear forn= 1; assume it forn; iffn+1,0(x, t) =Pn+1, thent= 0, andfn,0(x) =Pn, a contradiction. By 1.3.7,Sⁿ− {Pn}is contractible. So, by 1.3.4,fn,0is homotopic (inSⁿ− {Pn},

hence inSⁿ) to a constant map.

Proposition 2.4.4.deg(fn,d) =d.

Proof. The casen= 1 is 2.4.2, so we may assumen≥2. Leteⁿ_±=Sⁿ∩Rⁿ⁺¹_± . ThenSⁿ =eⁿ₊∪eⁿ₋ andSⁿ⁻¹ =eⁿ₊∩eⁿ₋. The spaces eⁿ_± are homeomorphic to Bⁿ, hence they are contractible. The Mayer-Vietoris sequence in singular homology gives (omitting the coefficient ringZ):

H_n^∆(eⁿ₊)⊕H_n^∆(eⁿ₋)→H_n^∆(Sⁿ)−→^∂n H_n^∆₋₁(Sⁿ⁻¹)→H_n^∆₋₁(eⁿ₊)⊕H_n^∆₋₁(eⁿ₋) implying that, for n ≥ 2, ∂n is an isomorphism (between infinite cyclic groups). Naturality of the Mayer-Vietoris sequence implies that for any map f :Sⁿ⁻¹→Sⁿ⁻¹the following diagram commutes:

H_n^∆(Sⁿ) −−−−→^(Σf)∗ H_n^∆(Sⁿ)

⏐⏐

^∂n ⏐⏐^∂n H_n^∆₋₁(Sⁿ⁻¹) −−−−→^f∗ H_n^∆₋₁(Sⁿ⁻¹)

This implies that deg(f) = deg(Σf). Sincefn,d=Σfn−1,dand deg(f1,d) =

d, this completes the argument.

We turn to Item 4 on our initial list.

Theorem 2.4.5. (Brouwer-Hopf Theorem) Two maps Sⁿ →Sⁿ are ho- motopic iff they have the same degree.

In view of the previous propositions the only part of this theorem not yet proved⁵is that forn≥1 every mapf :Sⁿ →Sⁿ of degreedis homotopic to fn,d. We begin with the casen= 1.

A map p:E →B is a covering projection if for everyb∈ B there is an open neighborhoodU of b such thatp⁻¹(U) can be written as the union of pairwise disjoint open subsets ofEeach mapped homeomorphically byponto U. Sets U ⊂B with this property are said to be evenly covered by p. The spaceB is thebase space andE is thecovering space. For example, the map exp :R→S¹,t→e^2πit, is a covering projection.

The following two theorems about covering spaces are fundamental. The proofs are elementary, and are to be found in numerous books on algebraic topology, for example, [74, Chap. 5].

5 A complete proof of the Brouwer-Hopf Theorem is included here because it is omitted from many books on algebraic topology. The reader may prefer to skip it and go to Theorem 2.4.19.

Theorem 2.4.6. (Homotopy Lifting Property) Let p:E →B be a cov- ering projection, let F :Y ×I→B andf :Y × {0} →E be maps such that p◦f =F |Y × {0}. Then there exists F¯ : Y ×I →E making the following diagram commute:

Y × {0} ^f //

E

p

Y ×I

F¯

;;w

ww ww ww ww

F //B

Theorem 2.4.7. (Unique Path Lifting) Let p : E → B be a covering projection. Ifωi:I→E are paths, for i= 1,2, such that p◦ω1=p◦ω2 and

ω1(0) =ω2(0), thenω1=ω2.

Consider a map f :S¹ →S¹, and the covering projection exp : R→S¹. Let t₀ ∈ R be such that f(1) = e^2πit0 (where 1 ∈ S¹ ⊂ C). By 2.4.6, the pathω:I→S¹, t→f(e^2πit), lifts throughpto a path ˜ω:I→Rsuch that

˜

ω(0) =t₀.

Proposition 2.4.8.The numberω(1)˜ −ω(0)˜ is an integer and is independent of the choice oft0∈exp⁻¹(f(1)).

Proof. Let ˜ω(1) =t₁. ω(0) =ω(1), so e^2πit0 =e^2πit1, hence t₀−t₁ ∈Z. For a∈R, letT_a :R→Rbe the translation homeomorphismx→x+a. If ˜ω is replaced by ¯ω where ¯ω(0) =t₀ and exp◦ ω¯ =ω, then T_(t

0−t₀)◦ω˜ = ¯ω, by 2.4.7. Hencet₁:= ¯ω(1) =t₁+t₀−t₀. Hencet₁−t₀=t₁−t₀.

The integer ˜ω(1)−ω(0) will be denoted by˜ δ(f).

Proposition 2.4.9.Two maps f, g:S¹→S¹ are homotopic iffδ(f) =δ(g).

Proof. Let F : S¹ ×I → S¹ be a homotopy. Define G : I×I → S¹ by G(t, s) = F(e^2πit, s). Let ω(t) = G(0, t) and let ˜ω : I → R be such that ω= exp◦ ω. By 2.4.6, there is a map ˜˜ G:I×I→Rsuch that ˜G0= ˜ω. Note that δ(Fs) = ˜Gs(1)−G˜s(0), a formula continuous in s and taking values in the discrete subspaceZ⊂R, hence constant. Thusδ(f) =δ(g).

Conversely, let δ(f) =δ(g) = d. Let ω, τ :I → S¹ be defined by ω(t) = f(e^2πit) and τ(t) = g(e^2πit). Let ˜ω and ˜τ be lifts of ω and τ such that d=

˜

ω(1)−ω(0) = ˜˜ τ(1)−τ(0). Let˜ β : I → R be a path from ˜ω(0) to ˜τ(0).

Then s → β(s) +d is a path from ˜ω(1) to ˜τ(1). By 1.3.17, there is a map H˜ : I×I → R such that ˜H(t,0) = ˜ω(t), ˜H(t,1) = ˜τ(t), ˜H(0, s) = β(s), and ˜H(1, s) =β(s) +d. Clearly, there is a function H making the following diagram commute:

I×I −−−−→^H˜ R

(exp|)×id

⏐⏐

⏐⏐^exp S¹×I −−−−→^H S¹

2.4 The degree of a map from a sphere to itself 47 By 1.3.11,H is continuous. One checks thatH0=f andH1=g.

Proposition 2.4.10.δ(f1,d) =d.

Proof. Let d ∈ Z ⊂ R. The map ˜ω : I → R, t → td makes the following diagram commute:

I ^ω˜ //

exp|

ω

B

BB BB BB

B R

exp

S¹ _f

1,d

//S¹

whereω is defined by the diagram. Thusδ(f1,d) = ˜ω(1)−ω(0) =˜ d.

This completes the proof of 2.4.5 when n= 1. Now we assumen≥2.

We need some preliminaries (2.4.11–2.4.18).

Let K and K be two CW complex structures on the same underlying spaceX. The complexK is asubdivision ofK if for each celle_β ofK there is a celldαofK such that^◦eβ⊂d^◦α.

Proposition 2.4.11.LetLbe a subcomplex ofK and letK be a subdivision ofK. Then there is a subcomplex L ofK which is a subdivision ofL.

Proof. LetL consist of all cellseofK such thate⊂L. We show thatL is a subcomplex and thatL is a subdivision ofL.

Let e be a cell of L. By 1.2.13 there are only finitely many cells of K, e1, . . . , ersuch that^•e∩e^◦i=∅. Sincee⊂L,^•e⊂L, so^◦ei∩L=∅for alli. Let di be a cell ofK such that e^◦i ⊂d^◦i. Then for 1≤i≤r, ^◦di∩L=∅, hencedi

is a cell of L, hencee_i ⊂L, hencee_i is a cell of L, hence ^•e⊂L. So L is a subcomplex ofL.

Given a cell e ofL, there is a celldof K such that ^◦e⊂^◦d. Sincee⊂L, d◦∩L=∅, sod⊂L. Thusdis a cell ofL. Finally, ifx∈L, there are cellse_x ofK anddxofLsuch thatx∈e^◦x⊂^◦dx. So^◦ex⊂L, so ex⊂L, so xlies in a

cell ofL. ThusL is a subdivision ofL.

LetI¹:= [−1,1] have the CW complex structure consisting of two vertices at −1 and +1, and one 1-cell. For k ≥ 1, the k^th barycentric subdivision of I¹ is the CW complex I_k¹ whose vertices are the points i/2^k−1, where

−2^k−1 ≤ i ≤ 2^k−1, and whose 1-cells are the closed intervals bounded by adjacent vertices. The k^th barycentric subdivision ofIⁿ⁺¹ is the (n+ 1)-fold product CW complexI_kⁿ⁺¹ :=I_k¹×. . .×I_k¹. This is obviously a subdivision ofIⁿ⁺¹. By 2.4.11, there is a subcomplexI^•n+1_k ofI_kⁿ⁺¹ subdividingI^•n+1. Proposition 2.4.12.For each open coverU ofIⁿ⁺¹there existsk0such that for everyk≥k0 every cell of I_kⁿ⁺¹ lies in some element ofU.

Proof. Letx∈U^(x)⊂Iⁿ⁺¹, whereU^(x)∈ U. There are intervalsV₁^(x), . . . , V_n+1^(x) which are open sets in I¹ such that x∈ V₁^(x)×. . .×V_n+1^(x) =:V^(x) ⊂U^(x). {Vx | x ∈ X} has a finite subcover V^(x1), . . . , V^(xr) since Iⁿ⁺¹ is compact.

Clearly, there is a subdivisionI_k¹

i ofI¹ every cell of which lies in someV_i^(xj). Letk0 = max{k1, . . . , kn+1}. Then for each k≥k0 every cell of I_kⁿ⁺¹ lies in

someV^(xj)⊂U^(xj).

Corollary 2.4.13.For each open cover U of I^•n+1 there exists k0 such that for everyk≥k0 each cell of I^•n+1_k lies in some element ofU.

Proof. For eachU ∈ U pickUan open subset ofIⁿ⁺¹ such thatU∩^•Iⁿ⁺¹= U. LetU consist of all the setsUand the setIⁿ⁺¹−I^•n+1. Apply 2.4.11 and

2.4.12 toU.

Proposition 2.4.14.Let X be an n-dimensional CW complex, let eⁿ be an n-cell ofX and letU ⊂e^◦n be a neighborhood ofz∈e^◦n. There is a homotopy H : X ×I → X such that H₀ = id, H_t = id on X −^◦eⁿ for all t, and H1(eⁿ−U)⊂^•eⁿ.

Proof. As in the proof of 1.4.2, pick a characteristic maph: (Bⁿ, Sⁿ⁻¹)→ (eⁿ,e^•n) such thath(0) =z.h⁻¹(U) is a neighborhood of 0 inBⁿ, so there is a >0 such thath(B)⊂U, where B={x∈Bⁿ | |x| ≤}. By 1.4.2, there is a homotopyH : (X−inth(B))×I →X such that H0= id, Ht= id on X−^◦eⁿ for allt, andH1(eⁿ−inth(Bⁿ))⊂^•eⁿ. ExtendH0to be the identity on all ofX× {0}. By 1.2.8, X is obtained fromX −inth(B) by attaching ann-cell. So, by 1.3.12,H further extends to mapX×ItoX as required.

Proposition 2.4.15.Let x1, . . . , xm, y1, . . . , ym be distinct points of Sⁿ, wheren≥2. There is a homeomorphism k:Sⁿ→Sⁿ, which is homotopic to id, such that k(xi) =yi for eachi.

Proof. For any z ∈ Sⁿ, Sⁿ − {z} is homeomorphic to Rⁿ. By choosing z different from each x_i and eachy_i, we may work inRⁿ. LetM >0 be such that every xi and every yi lies in (−M, M)ⁿ. We will show that there is a homeomorphism hof [−M, M]ⁿ which fixes every point of the frontier, and takesxi toyi for alli. Clearly this will be enough.

The proof requires two elementary lemmas whose proofs are left to the reader.

Lemma 2.4.16.Let N be a compact convex neighborhood of a point x∈Rⁿ and let y∈intN. There is a homotopyH :N×I →N such that H₀ = id, Ht= idon fr N for all t,Ht is a homeomorphism for all t, andH1(x) =y.

2.4 The degree of a map from a sphere to itself 49 Lemma 2.4.17.Ifn≥3, there exist compact convex pairwise disjoint neigh- borhoodsMi ofxi and Ni of yi, and pointsx_i∈intMi andy_i ∈intNi such that no four points in{x₁, . . . , x_m, y₁, . . . , y_m } are coplanar.

We now complete the proof of 2.4.15 for n≥3. By 2.4.16 and 2.4.17, we may assume that the line segmentsL_i := [x_i, y_i] inRⁿ are pairwise disjoint.

Since eachLi is compact and convex, Ci :={x∈Rⁿ | |x−u| ≤ for some u∈Li}is a compact convex neighborhood of every point ofLi, and ifis small enough, the setsCiare pairwise disjoint and lie in (−M, M)ⁿ. Applying 2.4.16 separately to eachCi to movexitoyi, and extending the resulting homotopy to be the identity onRⁿ−

m i=1

Ci, we obtain the desired homeomorphismk.

The casen= 2 is left as an exercise.

We remark that 2.4.15 and 2.4.17 are false whenn= 1. The fundamental geometric Proposition 2.4.15 will be used in the proof of Theorem 2.4.5. It is also the core of the proof that higher homotopy groups are abelian (compare 4.4.5).

For n≥2, giveSⁿ the CW structure consisting of one 0-cell e⁰, lying in Sⁿ⁻¹⊂Sⁿ, one (n−1)-celleⁿ⁻¹ =Sⁿ⁻¹, and twon-cells eⁿ_± =Sⁿ∩Rⁿ⁺¹_± . A mapf :Sⁿ →Sⁿ preserves hemispheres iff is cellular with respect to the given CW structure,f(eⁿ₊)⊂eⁿ₊, and f(eⁿ₋)⊂eⁿ₋.

Proposition 2.4.18.Forn≥2, every mapSⁿ→Sⁿ is homotopic to a map which preserves hemispheres.

Proof. Recall that we may identify I^•n+1 with Sⁿ via the homeomorphism an+1 |:x→x/|x|. By the upper [resp.lower] open hemisphere we mean the set of points whose last coordinate is positive [resp. negative]; their closures are closed hemispheres. We are given a mapf1:I^•n+1→I^•n+1. Using the Cellular Approximation Theorem with respect to various CW complex structures, we will obtain a map homotopic tof1 which preserves hemispheres.

Define

U+:={(x1, . . . , xn+1)∈I^•n+1|xn+1>−1 2} U₋:={(x₁, . . . , x_n+1)∈I^•n+1|x_n+1<1

2}.

By 2.4.13 there existsk such that every cell of the subdivision I^•n+1_k lies in f₁⁻¹(U₊) orf₁⁻¹(U₋).

Letφ:I¹→I¹be the piecewise linear map which fixes 0, 1 and−1, which takes±2¹ to 0, and which is affine on [−1,−¹2], [−¹2,¹₂], [¹₂,1]. The homotopy G:I^•n+1_k ×I→I^•n+1_k which takes (x₁, . . . , x_n+1, t) to (x₁, . . . , x_n,(1−t)x_n+1+ tφ(x_n+1)) has the property that G₀ = id, G₁(U₊) lies in the upper closed hemisphere, andG1(U₋) lies in the lower closed hemisphere. Letf2=G1◦f1.

Then f2 takes eachn-cell of I^•n+1_k into either the upper or the lower closed hemisphere.

We now regard f2 as a map ^•Iⁿ⁺¹_k → Sⁿ. Here, Sⁿ has one vertex, one (n−1)-cell (which is Sⁿ⁻¹), and the two closed hemispheres as n-cells. By 1.4.3 and 1.4.4,f2 is homotopic to a cellular mapf3 which maps the (n−1)- skeleton of I^•n+1_k into Sⁿ⁻¹ and which maps eachn-cell of ^•Iⁿ⁺¹_k into one of the closed hemispheres.

Let eⁿ₁, . . . , eⁿ_m be the n-cells of I^•n+1_k which are not mapped into Sⁿ⁻¹ by f3. Pick xi ∈ ^◦eⁿ_i such that f3(xi) ∈/ Sⁿ⁻¹. Pick yi in the upper [resp.

lower] open hemisphere of I^•n+1 iff f3(xi) is in the upper [resp. lower] open hemisphere ofSⁿ, and such thaty1, . . . , ymare distinct. By 2.4.15, there is a homeomorphismk:^•Iⁿ⁺¹ →I^•n+1, homotopic to id, takingxi toyi for alli.

LetVi ⊂^◦eⁿ_i be an open neighborhood ofxi such thatk(Vi) is entirely in the open hemisphere of^•Iⁿ⁺¹containingyi. See Fig. 2.2. By 2.4.14,f3is homotopic to a mapf4:I^•n+1→Sⁿ such that, for 1≤i≤m,f4(eⁿ_i −Vi) =f4(^•eⁿ_i)⊂ Sⁿ⁻¹, f4(eⁿ_i) =f3(eⁿ_i), while f4=f3 outside

m i=1

eⁿ_i. Sincek is homotopic to id, so is k⁻¹, hencef4 is homotopic to f5 :=f4◦k⁻¹. If yi is in the upper [resp. lower] open hemisphere, thenf5maps the neighborhoodk(Vi) ofyi into the upper [resp. lower] open hemisphere ofSⁿ.f5mapsI^•n+1−

m i=1

k(Vi) into

Sⁿ⁻¹. Thusf₅preserves hemispheres.

Vi

k (V )_i yi

x

.

i

3

.

.

^{f (x )i}

Fig. 2.2.

Proof (of Theorem 2.4.5 (concluded)).Letf :Sⁿ→Sⁿ be a map. By 2.4.18, it may be assumed thatf preserves hemispheres. Letg:Sⁿ⁻¹→Sⁿ⁻¹be the restriction off. Then for eachx∈Sⁿ, (Σg)(x) andf(x) are not diametrically opposite points ofSⁿ. Thus the line segment inRⁿ⁺¹,{ωt(x) := (1−t)Σg(x)+

2.4 The degree of a map from a sphere to itself 51 tf(x) | 0 ≤ t ≤ 1}, does not contain 0. The map (x, t) → ωt(x)/|ωt(x)| is a homotopy from Σg to f. By induction, g is homotopic to Σⁿ⁻²h where h:S¹→S¹is a map. By 2.4.9 and 2.4.10,his homotopic tof1,dfor somed.

Thusf Σⁿ⁻¹hΣⁿ⁻¹f1,d=:fn,d.

From the definition of degree one deduces:

Theorem 2.4.19. (Product Theorem for Degree)Let f, g:Sⁿ→Sⁿ be maps,n≥0. Then deg(g◦f) = deg(g)·deg(f).

Corollary 2.4.20.A homotopy equivalence f :Sⁿ→Sⁿ has degree±1.

Together with 2.4.4 and 2.4.5, this corollary implies that a homeomorphism h: Sⁿ → Sⁿ is homotopic either to id_Sn =f_n,1 or to the homeomorphism f_n,−1. If the former,hisorientation preserving; if the latter his orientation reversing. (For this to make sense when n = 0, define f_0,−1 to be the map which interchanges the two points ofS⁰.)

There is another important theorem concerning degree, the Sum Theorem.

For this we need the notion of the “wedge” of a family of spaces. If X is a space andx∈X, we write (X, x) for the pair (X,{x}), we call (X, x) apointed space⁶, and we callxthebase point. Thewedge

α

Xα of a family{(Xα, xα)} of pointed spaces is the quotient space

α∈A

Xα/∼where ∼ identifies all the base points xα. The wedge is also called the one-point union. The point of

α

Xα corresponding to the equivalence class of {xα} is the wedge point. If each Xαis a CW complex and if for allαthe pointxαis a vertex ofXα, then

α∈A

Xα acquires a quotient CW complex structure as in 1.2.22. The wedge point is a vertex. When eachXαis ann-sphere,

α

Xαis also called abouquet ofn-spheres.

Let the base pointv ofSⁿ be (1,0, . . . ,0). Thenv∈S⁰⊂S¹⊂. . ..

Consider

α∈A

S_αⁿ where (S_αⁿ, v_α) is a copy of (Sⁿ, v) for eachα. There are canonical maps S_βⁿ

←−qβ

−→i_β

α∈A

S_αⁿ for each β ∈ A; iβ is the inclusion of S_βⁿ as the β-summand of the wedge; qβ is the identity of S_βⁿ and maps all other summands tovβ ∈S_βⁿ.

Exercises

1. Fill in the details in the proof of 2.4.2.

6 By a pointed CW complex we mean a pointed space (X, v) where X is a CW complex andvis a vertex.

2. In the definition of “suspension” prove that Σg is continuous when gis; and thatΣg₁ andΣg₂ are homotopic wheng₁ andg₂ are.

3. Prove that homeomorphic CW complexes have the same dimension.

4. Describe a CW complex structure forRⁿ.

5. (for those who know category theory) Show that “wedge” is the category theoret- ical coproduct in the category Pointed Spaces. What is the category theoretical coproduct in the category Spaces?