The Fundamental Group of the Circle

To show that Φ is injective, suppose Φ(m) = Φ(n), which means ωm ' ωn. Let f_t be a homotopy from ω_m = f₀ to ω_n = f₁. By (b) this homotopy lifts to a homotopy ef_t of paths starting at 0 . The uniqueness part of (a) implies that ef₀= fω_m and ef₁ = fω_n. Since ef_t is a homotopy of paths, the endpoint ef_t(1) is independent oft . For t= 0 this endpoint is m and for t = 1 it is n, so m = n.

It remains to prove (a) and (b). Both statements can be deduced from a more general assertion:

→

^S¹ and a map eF : Y×{0}

→

R lifting F|Y ×{0}, then there is a unique map eF : Y×I

→

R lifting F and restricting to the given eF on Y×{0}.

Statement (a) is the special case that Y is a point, and (b) is obtained by applying (c) with Y = I in the following way. The homotopy f_t in (b) gives a mapF : I×I

→

^S¹ ^by

setting F (s, t)= ft(s) as usual. A unique lift eF : I×{0}

→

R is obtained by an appli-cation of (a). Then (c) gives a unique lift eF : I×I

→

R. The restrictions eF|{0}×I and F|{1}×I are paths lifting the constant path at xe 0, hence they must also be constant by the uniqueness part of (a). So ef_t(s)= eF (s, t) is a homotopy of paths, and ef_t lifts f_t since p eF= F .

We shall prove (c) using just one special property of the projection p :R

→

^S¹^,

namely:

(∗)

There is an open cover {Uα} of S¹ such that for each α , p⁻¹(U_α) can be decomposed as a disjoint union of open sets each of which is mapped homeo-morphically onto U_α by p .

For example, we could take the cover {Uα} to consist of any two open arcs in S¹ whose union isS¹.

To prove (c) we will first construct a lift eF : N×I

→

R for N some neighborhood in Y of a given point y₀∈ Y . Since F is continuous, every point (y0, t)∈ Y ×I has a product neighborhood N_t×(a_t, b_t) such that F N_t×(a_t, b_t)

⊂ U_α for some α . By compactness of {y₀}×I , finitely many such products N_t×(a_t, b_t) cover {y₀}×I . This implies that we can choose a single neighborhood N of y₀ and a partition 0 = t0 < t₁ < ··· < tm = 1 of I so that for each i, F(N ×[ti, t_i+1]) is contained in some U_α, which we denote U_i. Assume inductively that eF has been constructed on N×[0, t_i] . We have F (N×[t_i, t_i+1])⊂ U_i, so by (∗) there is an open set eU_i ⊂ R projecting homeomorphically ontoU_i by p and containing the point eF (y₀, t_i) . After replacingN by a smaller neighborhood of y₀ we may assume that eF (N×{ti}) is con-tained in eU_i, namely, replaceN×{ti} by its intersection with ( eF ||N ×{ti})⁻¹( eU_i) . Now we can define eF on N×[ti, t_i+1] to be the composition of F with the homeomorphism p⁻¹:U_i

→

^U^eⁱ. After finitely many repetitions of this induction step we eventually get a lift eF : N×I

→

R for some neighborhood N of y0.

Next we show the uniqueness part of (c) in the special case thatY is a point. In this case we can omitY from the notation. So suppose eF and eF⁰are two lifts ofF : I

→

^S¹

such that eF (0)= eF⁰(0) . As before, choose a partition 0= t₀< t₁<··· < t_m = 1 of I so that for each i , F ([t_i, t_i+1]) is contained in some U_i. Assume inductively that Fe= eF⁰on[0, t_i] . Since [t_i, t_i+1] is connected, so is eF ([t_i, t_i+1]) , which must therefore lie in a single one of the disjoint open sets eU_i projecting homeomorphically to U_i as in(∗). By the same token, eF⁰([t_i, t_i+1]) lies in a single eU_i, in fact in the same one that contains eF ([t_i, t_i+1]) since eF⁰(t_i)= eF (t_i) . Because p is injective on eU_i andp eF = p eF⁰, it follows that eF = eF⁰on [t_i, t_i+1] , and the induction step is finished.

The last step in the proof of (c) is to observe that since the eF ’s constructed above on sets of the form N×I are unique when restricted to each segment {y}×I , they must agree whenever two such sets N×I overlap. So we obtain a well-defined lift eF on all of Y×I . This eF is continuous since it is continuous on each N×I , and it is

unique since it is unique on each segment {y}×I . tu

Now we turn to some applications of this theorem. Although algebraic topology is usually ‘algebra serving topology,’ the roles are reversed in the following proof of the Fundamental Theorem of Algebra.

T

heorem 1.8. Every nonconstant polynomial with coefficients in C has a root in C.

P

roof: We may assume the polynomial is of the form p(z)= zⁿ+ a1zⁿ⁻¹+ ··· + an. If p(z) has no roots in C, then for each real number r ≥ 0 the formula

f_r(s)= p(r e^{2π is})/p(r )

|p(r e^{2π is})/p(r )|

defines a loop in the unit circleS¹⊂ C based at 1. As r varies, fr is a homotopy of loops based at 1 . Since f₀ is the trivial loop, we deduce that the class [f_r]∈ π1(S¹) is zero for all r . Now fix a large value of r , bigger than|a₁| + ··· + |a_n| and bigger than 1 . Then for |z| = r we have

|zⁿ| = rⁿ= r · rⁿ⁻¹> (|a1| + ··· + |an|)|zⁿ⁻¹| ≥ |a1zⁿ⁻¹+ ··· + an|

From the inequality|zⁿ| > |a₁zⁿ⁻¹+ ··· + a_n| it follows that the polynomial p_t(z)= zⁿ+t(a1zⁿ⁻¹+···+an) has no roots on the circle|z| = r when 0 ≤ t ≤ 1. Replacing p by p_t in the formula forf_r above and letting t go from 1 to 0 , we obtain a homo-topy from the loop f_r to the loop ω_n(s)= e^{2π ins}. By Theorem 1.7, ω_n represents n times a generator of the infinite cyclic group π₁(S¹) . Since we have shown that [ω_n]= [f_r]= 0, we conclude that n = 0. Thus the only polynomials without roots

in C are constants. tu

Our next application is the Brouwer fixed point theorem in dimension 2 .

T

heorem 1.9. Every continuous map h : D²

→

^D² has a fixed point, that is, a point x with h(x)= x .

Here we are using the standard notation Dⁿ for the closed unit disk in Rⁿ, all vectorsx of length|x| ≤ 1. Thus the boundary of Dⁿ is the unit sphereSⁿ⁻¹.

P

roof: Suppose on the contrary that h(x)≠ x for all x ∈ D². Then we can define a map r : D²

→

^S¹ ^{by letting}r (x) be the point of S¹ where the ray inR² starting at h(x) and passing

throughx leaves D². Continuity ofr is clear since small per- x r(x)

h(x)

turbations of x produce small perturbations of h(x) , hence also small perturbations of the ray through these two points.

The crucial property of r , besides continuity, is that r (x)= x if x ∈ S¹. Thus r is a retraction ofD² ontoS¹. We will show that no such retraction can exist.

Let f₀ be any loop inS¹. InD² there is a homotopy of f₀ to a constant loop, for example the linear homotopy f_t(s)= (1 − t)f0(s)+ tx0 where x₀ is the basepoint of f₀. Since the retraction r is the identity on S¹, the composition r f_t is then a homotopy in S¹ from r f₀= f₀ to the constant loop at x₀. But this contradicts the

fact thatπ₁(S¹) is nonzero. tu

This theorem was first proved by Brouwer around 1910, one of the early triumphs of algebraic topology. Brouwer in fact proved the corresponding result for Dⁿ, and we shall obtain this generalization in Corollary 2.11 using homology groups in place of π₁. One could also use the higher homotopy group π_n. Brouwer’s original proof used neither homology nor homotopy groups, which had not been invented at the time. Instead it used the notion of degree for mapsSⁿ

→

^Sⁿ, which we shall define in

§2.2 using homology but which Brouwer defined directly in more geometric terms.

These proofs are all arguments by contradiction, and so they show just the exis-tence of fixed points without giving any clue as to how to find one in explicit cases.

Our proof of the Fundamental Theorem of Algebra was similar in this regard. There exist other proofs of the Brouwer fixed point theorem that are somewhat more con-structive, for example the elegant and quite elementary proof by Sperner in 1928, which is explained very nicely in [Aigner-Ziegler 1999].

The techniques used to calculateπ₁(S¹) can be applied to prove the Borsuk–Ulam theorem in dimension two:

T

heorem 1.10. For every continuous map f : S²

→

^R²there exists a pair of antipodal points x and−x in S² with f (x)= f (−x).

It may be that there is only one such pair of antipodal points x ,−x , for example iff is simply orthogonal projection of the standard sphere S²⊂ R³ onto a plane.

The Borsuk–Ulam theorem holds also for maps Sⁿ

→

^Rⁿ, as we show in Proposi-tion 2B.6. The proof forn= 1 is easy since the difference f (x)−f (−x) changes sign as x goes halfway around the circle, hence this difference must be zero for some x . For n ≥ 2 the theorem is certainly less obvious. Is it apparent, for example, that at every instant there must be a pair of antipodal points on the surface of the earth having the same temperature and the same barometric pressure?

The theorem says in particular that there is no one-to-one continuous map from S² toR², soS² is not homeomorphic to a subspace of R², an intuitively obvious fact that is not easy to prove directly.

P

roof: If the conclusion is false for f : S²

→

^R², we can define a map g : S²

→

^S¹ ^by

g(x) = f (x) − f (−x)

/|f (x) − f (−x)|. Define a loop η circling the equator of S²⊂ R³ by η(s)= (cos 2πs, sin 2πs, 0), and let h : I

→

^S¹ be the composed loop gη . Sinceg(−x) = −g(x), we have the relation h(s +¹/₂)= −h(s) for all s in the interval [0,¹/₂]. As we showed in the calculation of π₁(S¹) , the loop h can be lifted to a path eh : I

→

R. The equation h(s +¹/₂) = −h(s) implies that eh(s +¹/₂) = eh(s) +^q/₂ for some odd integer q that might conceivably depend on s ∈ [0,¹/₂]. But in fact q is independent ofs since by solving the equation eh(s+¹/₂)= eh(s)+^q/₂forq we see that q depends continuously on s∈ [0,¹/₂], so q must be a constant since it is constrained to integer values. In particular, we have eh(1)= eh(¹/₂)+^q/₂ = eh(0) + q. This means thath represents q times a generator of π₁(S¹) . Since q is odd, we conclude that h is not nullhomotopic. Buth was the composition gη : I

→

^S²

→

^S¹^{, and}η is obviously nullhomotopic in S², sogη is nullhomotopic in S¹ by composing a nullhomotopy of η with g . Thus we have arrived at a contradiction. tu

C

orollary 1.11. Whenever S² is expressed as the union of three closed sets A₁, A₂, and A₃, then at least one of these sets must contain a pair of antipodal points{x, −x}.

P

roof: Let d_i:S²

→

R measure distance to A_i, that is, d_i(x)= inf_y∈A_i|x − y|. This is a continuous function, so we may apply the Borsuk–Ulam theorem to the map S²

→

^R²^, ^x

,

^d¹^{(x), d}²^(x), obtaining a pair of antipodal points x and −x with d₁(x)= d1(−x) and d2(x)= d2(−x). If either of these two distances is zero, then x and−x both lie in the same set A₁ or A₂ since these are closed sets. On the other hand, if the distances from x and −x to A₁ and A₂ are both strictly positive, then x and −x lie in neither A1 nor A₂ so they must lie in A₃. tu

To see that the number ‘three’ in this result is best possible, consider a sphere inscribed in a tetrahedron. Projecting the four faces of the tetrahedron radially onto the sphere, we obtain a cover ofS² by four closed sets, none of which contains a pair of antipodal points.

Assuming the higher-dimensional version of the Borsuk–Ulam theorem, the same arguments show that Sⁿ cannot be covered by n+ 1 closed sets without antipodal pairs of points, though it can be covered byn+2 such sets, as the higher-dimensional analog of a tetrahedron shows. Even the case n= 1 is somewhat interesting: If the circle is covered by two closed sets, one of them must contain a pair of antipodal points. This is of course false for nonclosed sets since the circle is the union of two disjoint half-open semicircles.

The relation between the fundamental group of a product space and the funda-mental groups of its factors is as simple as one could wish:

P

roposition 1.12. π₁(X×Y ) is isomorphic to π₁(X)×π₁(Y ) if X and Y are path-connected.

P

roof: A basic property of the product topology is that a map f : Z

→

^X^{×Y is}

con-tinuous iff the mapsg : Z

→

X and h : Z

→

Y defined by f (z)= (g(z), h(z)) are both continuous. Hence a loopf in X×Y based at (x0, y₀) is equivalent to a pair of loops g in X and h in Y based at x₀andy₀respectively. Similarly, a homotopyf_t of a loop in X×Y is equivalent to a pair of homotopies gt and h_t of the corresponding loops in X and Y . Thus we obtain a bijection π₁ X×Y , (x0, y₀)

≈ π1(X, x₀)×π1(Y , y₀) , [f ]

,

([g], [h]) . This is obviously a group homomorphism, and hence an

isomor-phism. tu

E

xample 1.13: The Torus. By the proposition we have an isomorphism π₁(S¹×S¹)≈ Z×Z. Under this isomorphism a pair (p, q) ∈ Z×Z corresponds to a loop that winds p times around one S¹ factor of the torus andq times around the

other S¹ factor, for example the loop ω_pq(s)= (ωp(s), ω_q(s)) . Interestingly, this loop can be knotted, as the figure shows for the case p= 3, q = 2. The knots that arise in this fashion, the so-calledtorus knots, are studied in Example 1.24.

More generally, the n dimensional torus, which is the product of n circles, has fundamental group isomorphic to the product of n copies of Z. This follows by induction onn .

Induced Homomorphisms

Supposeϕ : X

→

Y is a map taking the basepoint x₀∈ X to the basepoint y0∈ Y . For brevity we write ϕ : (X, x₀)

→

^{(Y , y}⁰) in this situation. Then ϕ induces a homo-morphismϕ_∗:π₁(X, x₀)

→

^π¹^{(Y , y}⁰) , defined by composing loops f : I

→

^{X based at}

x₀ with ϕ , that is, ϕ_∗[f ] = [ϕf ]. This induced map ϕ_∗ is well-defined since a homotopyf_t of loops based at x₀ yields a composed homotopy ϕf_t of loops based aty₀, so ϕ_∗[f₀]= [ϕf0]= [ϕf1]= ϕ_∗[f₁] . Furthermore, ϕ_∗ is a homomorphism since ϕ(f g)= (ϕf ) (ϕg), both functions having the value ϕf (2s) for 0 ≤ s ≤¹/₂ and the value ϕg(2s− 1) for¹/₂≤ s ≤ 1.

Two basic properties of induced homomorphisms are:

(ϕψ)_∗= ϕ_∗ψ_∗ for a composition (X, x₀)

---→

^ψ ^{(Y , y}⁰⁾

---→

^ϕ ^{(Z, z}⁰^{) .}

11_∗=¹1 , which is a concise way of saying that the identity map11 :X

→

^{X induces}

the identity map 11 :π₁(X, x₀)

→

^π¹^{(X, x}⁰^{) .}

The first of these follows from the fact that composition of maps is associative, so (ϕψ)f = ϕ(ψf ), and the second is obvious. These two properties of induced homo-morphisms are what makes the fundamental group a functor. The formal definition

of a functor requires the introduction of certain other preliminary concepts, however, so we postpone this until it is needed in§2.3.

Ifϕ is a homeomorphism with inverse ψ then ϕ_∗is an isomorphism with inverse ψ_∗ sinceϕ_∗ψ_∗= (ϕψ)_∗=¹1_∗=¹1 and similarly ψ_∗ϕ_∗=¹1 . We will use this fact in the following calculation of the fundamental groups of higher-dimensional spheres:

P

roposition 1.14. π₁(Sⁿ)= 0 if n ≥ 2.

P

roof: Let f be a loop in Sⁿ at a chosen basepoint x₀. If the image off is disjoint from some other point x ∈ Sⁿ then f is nullhomotopic since Sⁿ− {x} is homeo-morphic to Rⁿ, which is simply-connected. So it will suffice to homotope f to be nonsurjective. To do this we will look at a small open ball B in Sⁿ about any point x ≠ x₀ and see that the number of times that f enters B , passes

throughx , and leaves B is finite, and each of these portions of f can be pushed offx without changing the rest of f . At first glance this might appear to be a difficult task to achieve since the parts off in B could be quite complicated geometrically, for example space-filling curves. But in fact it turns out to be rather easy.

The setf⁻¹(B) is open in (0, 1) , hence is the union of a possibly infinite collection of disjoint open intervals(a_i, b_i) . The compact set f⁻¹(x) is contained in the union of these intervals, so it must be contained in the union of finitely many of them. Consider one of the intervals(a_i, b_i) meeting f⁻¹(x) . The path f_i obtained by restrictingf to the closed interval[a_i, b_i] lies in the closure of B , and its endpoints f (a_i) and f (b_i) lie in the boundary of B . If n≥ 2, we can choose a path g_i from f (a_i) to f (b_i) in the closure of B but disjoint from x . For example, we could choose g_i to lie in the boundary ofB , which is a sphere of dimension n− 1, hence path-connected if n ≥ 2.

Since the closure of B is homeomorphic to a convex set in Rⁿ and hence simply-connected, the pathf_i is homotopic to g_i by Proposition 1.6, so we may homotopef by deformingf_i to g_i. After repeating this process for each of the intervals (a_i, b_i) that meet f⁻¹(x) , we obtain a loop g homotopic to the original f and with g(I)

disjoint fromx . tu

E

xample 1.15. For a point x in Rⁿ, the complementRⁿ− {x} is homeomorphic to Sⁿ⁻¹×R, so by Proposition 1.12 π1(Rⁿ− {x}) is isomorphic to π1(Sⁿ⁻¹)×π1(R).

Hence π₁(Rⁿ− {x}) is Z for n = 2 and trivial for n > 2.

Here is an application of this calculation:

C

orollary 1.16. R² is not homeomorphic to Rⁿ for n≠ 2.

P

roof: Suppose f :R²

→

^Rⁿ is a homeomorphism. The casen= 1 is easily disposed of since R²− {0} is path-connected but the homeomorphic space Rⁿ− {f (0)} is not path-connected when n = 1. When n > 2 we cannot distinguish R²− {0} from

Rⁿ− {f (0)} by the number of path-components, but by the preceding calculation of π₁(Rⁿ− {x}) we can distinguish them by their fundamental groups. tu

The more general statement that R^m is not homeomorphic to Rⁿ if m≠ n can be proved in the same way using either the higher homotopy groups or homology groups. In fact, nonempty open sets in R^m and Rⁿ can be homeomorphic only if m= n, as we will show in Theorem 2.19 using homology.

Induced homomorphisms allow relations between spaces to be transformed into relations between their fundamental groups. Here is an illustration of this principle:

P

roposition 1.17. If a space X retracts onto a subspace A , then the homomorphism i_∗:π₁(A, x₀)

→

^π¹^{(X, x}⁰) induced by the inclusion i : A

>

X is injective. If A is a deformation retract of X , then i_∗ is an isomorphism.

P

roof: If r : X

→

A is a retraction, then r i=¹1 , hencer_∗i_∗=¹1 , which implies thati_∗ is injective. Ifr_t:X

→

X is a deformation retraction of X onto A , so r₀=¹1 ,r_t|A =¹1 , andr₁(X)⊂ A, then for any loop f : I

→

X based at x₀∈ A the composition rtf gives a homotopy of f to a loop in A , so i_∗ is also surjective. tu

This gives another way of seeing thatS¹ is not a retract ofD², a fact we showed earlier in the proof of the Brouwer fixed point theorem, since the inclusion-induced map π₁(S¹)

→

^π¹^(D²) is a homomorphismZ

→

0 that cannot be injective.

The exact group-theoretic analog of a retraction is a homomorphismρ of a group G onto a subgroup H such that ρ restricts to the identity on H . In the notation above, if we identifyπ₁(A) with its image under i_∗, thenr_∗is such a homomorphism fromπ₁(X) onto the subgroup π₁(A) . The existence of a retracting homomorphism ρ : G

→

H is quite a strong condition on H . If H is a normal subgroup, it implies that G is the direct product of H and the kernel of ρ . If H is not normal, then G is what is called in group theory the semi-direct product of H and the kernel of ρ .

Recall from Chapter 0 the general definition of a homotopy as a familyϕ_t:X

→

^{Y ,}

t∈ I , such that the associated map Φ : X×I

→

^{Y ,}Φ(x, t) = ϕ_t(x) , is continuous. If ϕ_t takes a subspaceA⊂ X to a subspace B ⊂ Y for all t , then we speak of a homotopy of maps of pairs, ϕ_t:(X, A)

→

(Y , B) . In particular, a basepoint-preserving homotopy ϕ_t:(X, x₀)

→

^{(Y , y}⁰) is the case that ϕ_t(x₀)= y0 for all t . Another basic property of induced homomorphisms is their invariance under such homotopies:

If ϕ_t:(X, x₀)

→

^{(Y , y}⁰) is a basepoint-preserving homotopy, then ϕ_0∗= ϕ_1∗. This holds since ϕ_0∗[f ] = [ϕ₀f ]= [ϕ₁f ]= ϕ_1∗[f ] , the middle equality coming from the homotopy ϕ_tf .

There is a notion of homotopy equivalence for spaces with basepoints. One says (X, x₀) ' (Y , y0) if there are maps ϕ : (X, x₀)

→

^{(Y , y}⁰) and ψ : (Y , y₀)

→

^{(X, x}⁰⁾

with homotopies ϕψ ' ¹1 and ψϕ ' ¹1 through maps fixing the basepoints. In this case the induced maps on π₁ satisfy ϕ_∗ψ_∗ = (ϕψ)_∗ =¹1_∗ =¹1 and likewise ψ_∗ϕ_∗ =¹1 , so ϕ_∗ and ψ_∗ are inverse isomorphisms π₁(X, x₀) ≈ π₁(Y , y₀) . This somewhat formal argument gives another proof that a deformation retraction induces an isomorphism on fundamental groups, since ifX deformation retracts onto A then (X, x₀)' (A, x0) for any choice of basepoint x₀∈ A.

Having to pay so much attention to basepoints when dealing with the fundamental group is something of a nuisance. For homotopy equivalences one does not have to be quite so careful, as the conditions on basepoints can actually be dropped:

P

roposition 1.18. If ϕ : X

→

Y is a homotopy equivalence, then the induced homo-morphism ϕ_∗:π₁(X, x₀)

→

^π¹ ^{Y , ϕ(x}⁰⁾is an isomorphism for all x₀∈ X .

The proof will use a simple fact about homotopies that do not fix the basepoint:

L

emma 1.19. If ϕ_t:X

→

Y is a homotopy and h is the path ϕ_t(x₀) formed by the images of a basepoint x₀∈ X , then the three maps in the diagram at the right satisfy ϕ_0∗= β_hϕ_1∗.

π1(X,x0)

π1(Y,ϕ1(x0))

ϕ1

π1(Y,ϕ0(x0))

ϕ0

β_h

∗

−−−−−→

−−−−−−−−→ −−−−−−−−→

P

roof: Let h_t be the restriction ofh to the interval [0, t] , with a reparametrization so that the domain ofh_t is still [0, 1] . Explicitly, we can take h_t(s)= h(ts). Then if f is a loop inX at the basepoint x₀, the producth_t (ϕ_tf ) h_t

gives a homotopy of loops at ϕ₀(x₀) . Restricting this _x

ϕ0

0f ϕ

tf ϕ

1f ϕ

( ) x₀

h_t

ϕ ( )

x₀ ϕ ( )

homotopy to t = 0 and t = 1, we see that ϕ_0∗([f ])= β_h ϕ_1∗([f ])

. tu

P

roof of 1.18: Let ψ : Y

→

X be a homotopy-inverse for ϕ , so that ϕψ ' ¹1 and ψϕ'¹1 . Consider the maps

π₁(X, x₀)

---→

^ϕ^∗ ^π¹ ^{Y , ϕ(x}⁰⁾

---→

^ψ^∗ ^π¹ ^{X, ψϕ(x}⁰⁾

---→

^ϕ^∗ ^π¹ ^{Y , ϕψϕ(x}⁰⁾

The composition of the first two maps is an isomorphism sinceψϕ'¹1 implies that ψ_∗ϕ_∗= βh for someh , by the lemma. In particular, since ψ_∗ϕ_∗ is an isomorphism, ϕ_∗ is injective. The same reasoning with the second and third maps shows that ψ_∗ is injective. Thus the first two of the three maps are injections and their composition is an isomorphism, so the first mapϕ_∗ must be surjective as well as injective. tu

Exercises

1. Show that composition of paths satisfies the following cancellation property: If f₀ g₀' f1 g₁ and g₀' g1 then f₀' f1.

2. Show that the change-of-basepoint homomorphism β_h depends only on the homo-topy class of h .

3. For a path-connected space X , show that π₁(X) is abelian iff all basepoint-change homomorphismsβ_h depend only on the endpoints of the path h .

4. A subspace X⊂ Rⁿ is said to bestar-shaped if there is a point x₀∈ X such that, for each x ∈ X , the line segment from x₀ to x lies in X . Show that if a subspace X ⊂ Rⁿ is locally star-shaped, in the sense that every point of X has a star-shaped neighborhood in X , then every path in X is homotopic in X to a piecewise linear path, that is, a path consisting of a finite number of straight line segments traversed at constant speed. Show this applies in particular when X is open or when X is a union of finitely many closed convex sets.

5. Show that for a space X , the following three conditions are equivalent:

(a) Every map S¹

→

X is homotopic to a constant map, with image a point.

(b) Every map S¹

→

X extends to a map D²

→

^{X .}

Deduce that a space X is simply-connected iff all maps S¹

→

X are homotopic. [In this problem, ‘homotopic’ means ‘homotopic without regard to basepoints.’]

6. We can regard π₁(X, x₀) as the set of basepoint-preserving homotopy classes of maps (S¹, s₀)

→

^{(X, x}⁰^{) . Let [S}¹, X] be the set of homotopy classes of maps S¹

→

^{X ,}

with no conditions on basepoints. Thus there is a natural mapΦ : π1(X, x₀)

→

^[S¹^{, X]}

obtained by ignoring basepoints. Show thatΦ is onto if X is path-connected, and that Φ([f ]) = Φ([g]) iff [f ] and [g] are conjugate in π1(X, x₀) . Hence Φ induces a one-to-one correspondence between [S¹, X] and the set of conjugacy classes in π₁(X) , whenX is path-connected.

7. Define f : S¹×I

→

^S¹×I by f (θ, s) = (θ + 2πs, s), so f restricts to the identity on the two boundary circles of S¹×I . Show that f is homotopic to the identity by a homotopy f_t that is stationary on one of the boundary circles, but not by any ho-motopyf_t that is stationary on both boundary circles. [Consider what f does to the path s

,

^(θ⁰, s) for fixed θ₀∈ S¹.]

8. Does the Borsuk–Ulam theorem hold for the torus? In other words, for every map f : S¹×S¹

→

^R² must there exist(x, y)∈ S¹×S¹ such thatf (x, y)= f (−x, −y)?

9. Let A₁, A₂, A₃ be compact sets in R³. Use the Borsuk–Ulam theorem to show that there is one planeP ⊂ R³ that simultaneously divides eachA_i into two pieces of equal measure.

10. From the isomorphism π₁ X×Y , (x0, y₀)

≈ π1(X, x₀)×π1(Y , y₀) it follows that loops inX×{y₀} and {x₀}×Y represent commuting elements of π₁ X×Y , (x₀, y₀)

. Construct an explicit homotopy demonstrating this.

11. If X₀ is the path-component of a spaceX containing the basepoint x₀, show that the inclusion X₀

>

X induces an isomorphism π₁(X₀, x₀)

→

^π¹^{(X, x}⁰^{) .}

12. Show that every homomorphism π₁(S¹)

→

^π¹^(S¹) can be realized as the induced homomorphismϕ_∗ of a map ϕ : S¹

→

^S¹^.

13. Given a space X and a path-connected subspace A containing the basepoint x₀, show that the mapπ₁(A, x₀)

→

^π¹^{(X, x}⁰) induced by the inclusion A

>

X is surjective iff every path in X with endpoints in A is homotopic to a path in A .

14. Show that the isomorphism π₁(X×Y ) ≈ π₁(X)×π₁(Y ) in Proposition 1.12 is given by [f ]

,

^(p^1∗^{([f ]), p}^2∗([f ])) where p₁ and p₂ are the projections of X×Y onto its two factors.

15. Given a map f : X

→

Y and a path h : I

→

from x₀ to x₁, show that f_∗β_h = βf hf_∗ in the diagram at the right.

π₁( x1) π1(X,x0)

Y, π₁( f(x0)) Y,

π₁( f(x1))

β_h

β_fh

∗

−−−−−→ −−−−−−−−−−−→ −−−−−→

f_∗

− − − − − − − − − − − →

16. Show that there are no retractions r : X

→

A in the following cases:

(a) X = R³ with A any subspace homeomorphic to S¹. (b) X = S¹×D² with A its boundary torus S¹×S¹. (c) X = S¹×D² and A the circle shown in the figure.

(d) X = D²∨ D² with A its boundary S¹∨ S¹.

(e) X a disk with two points on its boundary identified and A its boundary S¹∨ S¹. (f) X the M¨obius band andA its boundary circle.

17. Construct infinitely many nonhomotopic retractions S¹∨ S¹

→

^S¹^.

18. Using the technique in the proof of Proposition 1.14, show that if a space X is obtained from a path-connected subspace A by attaching a cell eⁿ with n≥ 2, then the inclusion A

>

X induces a surjection on π₁. Apply this to show:

(a) The wedge sumS¹∨ S² has fundamental group Z.

(b) For a path-connected CW complexX the inclusion map X¹

>

X of its 1 skeleton induces a surjection π₁(X¹)

→

^π¹(X) . [For the case that X has infinitely many cells, see Proposition A.1 in the Appendix.]

19. Modify the proof of Proposition 1.14 to show that if X is a path-connected 1 dimensional CW complex with basepoint x₀ a 0 cell, then every loop in X is ho-motopic to a loop consisting of a finite sequence of edges traversed monotonically.

[This gives an elementary proof that π₁(S¹) is cyclic, generated by the standard loop winding once around the circle. The more difficult part of the calculation of π₁(S¹) is therefore the fact that no iterate of this loop is nullhomotopic.]

20. Suppose f_t:X

→

X is a homotopy such that f₀ andf₁ are each the identity map.

Use Lemma 1.19 to show that for anyx₀∈ X , the loop f_t(x₀) represents an element of the center ofπ₁(X, x₀) . [One can interpret the result as saying that a loop represents an element of the center ofπ₁(X) if it extends to a loop of maps X

→

^{X .]}

The van Kampen theorem gives a method for computing the fundamental groups of spaces that can be decomposed into simpler spaces whose fundamental groups are already known. By systematic use of this theorem one can compute the fundamental groups of a very large number of spaces. We shall see for example that for every group G there is a space X_G whose fundamental group is isomorphic toG .

To give some idea of how one might hope to compute fundamental groups by decomposing spaces into simpler pieces, let us look at an example. Consider the space X formed by two circles A and B intersecting in a single point, which we choose as the basepoint x₀. By our preceding calculations we know that π₁(A) is infinite cyclic, generated by a loopa that goes once around A .

Similarly, π₁(B) is a copy of Z generated by a loop b going b a once aroundB . Each product of powers of a and b then gives

an element of π₁(X) . For example, the product a⁵b²a⁻³ba² is the loop that goes five times aroundA , then twice around B , then three times around A in the opposite direction, then once around B , then twice around A . The set of all words like this consisting of powers ofa alternating with powers of b forms a group usually denoted Z ∗ Z. Multiplication in this group is defined just as one would expect, for example (b⁴a⁵b²a⁻³)(a⁴b⁻¹ab³)= b⁴a⁵b²ab⁻¹ab³. The identity element is the empty word, and inverses are what they have to be, for example (ab²a⁻³b⁻⁴)⁻¹ = b⁴a³b⁻²a⁻¹. It would be very nice if such words in a and b corresponded exactly to elements of π₁(X) , so that π₁(X) was isomorphic to the group Z ∗ Z. The van Kampen theorem will imply that this is indeed the case.

Similarly, if X is the union of three circles touching at a single point, the van Kampen theorem will imply that π₁(X) is Z ∗ Z ∗ Z, the group consisting of words in powers of three letters a , b , c . The generalization to a union of any number of circles touching at one point will also follow.

The group Z ∗ Z is an example of a general construction called the free product of groups. The statement of van Kampen’s theorem will be in terms of free products, so before stating the theorem we will make an algebraic digression to describe the construction of free products in some detail.

在文檔中 Allen Hatcher (頁 38-50)