Sách giải bài tậpTopo Đại số 1

Given spaces X and Y, let X, Y denote the set of homotopy classes of maps of Xinto Y. Let I = 0, 1.(a) Show that for any X, the set X, I has a single element.Define ϕ˜ : X → I to be the zero map: ϕ˜(x) = 0, ∀x ∈ X. Let ϕ : X → I by anycontinuous map. We show ϕ ϕ˜: define Φ : X × I → I byΦ(x, t) = (1 − t)ϕ(x).This is evidently a homotopy ϕ ϕ˜.Since is an equivalence relation, this shows all maps ϕ : X → I are equivalentunder , i.e., there is only one equivalence class.

General note: every problem here uses Theorem 18.2, mostly part (c) You should probably

be on good terms with this Theorem

1 Show that if h, h0

: X → Y are homotopic and k, k0

: Y → Z are homotopic, then

k ◦ h ' k0

◦ h0

.Since h ' h0

, we have a homotopy

H : X × I → Y with H(x, 0) = h(x), H(x, 1) = h0

(x),and since k ' k0, we have a homotopy

(a) Show that for any X, the set [X, I] has a single element

Define ˜ϕ : X → I to be the zero map: ˜ϕ(x) = 0, ∀x ∈ X Let ϕ : X → I by anycontinuous map We show ϕ ' ˜ϕ: define Φ : X × I → I by

Φ(x, t) = (1 − t)ϕ(x)

This is evidently a homotopy ϕ ' ˜ϕ

Since ' is an equivalence relation, this shows all maps ϕ : X → I are equivalentunder ', i.e., there is only one equivalence class

(b) Show that if Y is path connected, the set [I, Y ] has a single element

Pick any point p ∈ Y and define ˜ϕ : I → Y by ˜ϕ(x) = p, so ˜ϕ is the constantmap at p Now let ϕ : I → Y be arbitrary; we will again show ϕ ' ˜ϕ Denoteϕ(0) = a and ϕ(1) = b

1 Here, and elsewhere, we are actually using the Pasting Lemma (Thm 18.3) to ensure this piecewise-defined function is actually continuous This is justified by the middle calculation, for t = 1

2

Since Y is path-connected, there is a path γ : I → Y with γ(0) = a = ϕ(0) andγ(1) = p Now define a homotopy Φ : I × I → Y by

Φ(x, t) =

(

ϕ (1 − 2t)x

, 0 ≤ t ≤ 12γ(2t − 1), 12 ≤ t ≤ 1

Note that Φ does not depend on x once t ≥ 1

2! This is because Φ is the constantmap to the point γ(2t − 1) from this point onwards Φ has the effect of shrinkingthe image of ϕ to a point while 0 ≤ t ≤ 1

2, then moving that point to p alongγ(I) while 12 ≤ t ≤ 1

Check that Φ does what it oughta do:

(b) Show that a contractible space is path connected

Let X be a contractible space Then there is a homotopy H between idX andsome constant map; call it f So f (x) = p, ∀x ∈ X and H : idX ' f , i.e.,

H : X × I → X with H(x, 0) = idX(x) and H(x, 1) = f (x) = p

To show X is path connected, we fix any two points y, z ∈ X and construct apath between them Note that H(y, t) is a path from y to p and H(z, t) is apath from z to p (recall that y, z are fixed ) Thus, define a path by

γ(t) =

(H(y, 2t), 0 ≤ t ≤ 12H(z, 2 − 2t), 12 ≤ t ≤ 1

And check it:

t = 0 γ(t) = H(y, 0) = idX(y) = y

t = 1

2 γ(t) = H(y, 1) = p = H(z, 1)

t = 1 γ(t) = H(z, 0) = idX(z) = z.

(c) Show that if Y is contractible, then for any X, the set [X, Y ] has a single element.Just as above, if Y is contractible, then we have a homotopy H between idY andsome constant map; call it f So f (y) = p, ∀y ∈ Y and H : idY ' f , i.e.,

H : Y × I → Y with H(y, 0) = idY(y) and H(y, 1) = f (y) = p

Take any arbitrary map ϕ : X → Y and define

Φ(x, t) = H(ϕ(x), t))

Then

t = 0 Φ(x, 0) = H(ϕ(x), 0)) = ϕ(x)

t = 1 Φ(x, 1) = H(ϕ(x), 1)) = f (x) = p

So every map ϕ : X → Y is homotopic to the constant map at p

(d) Show that if X is contractible and Y is path connected, then [X, Y ] has a singleelement

Let X be a contractible space Then there is a homotopy H between idX andsome constant map; call it f So f (x) = p, ∀x ∈ X and H : idX ' f , i.e.,

H : X × I → X with H(x, 0) = idX(x) and H(x, 1) = f (x) = p

Pick some other point q ∈ Y and define a constant map ˜ϕ : Y → Y by ˜ϕ(y) =

q, ∀y ∈ Y Take any arbitrary map ϕ : X → Y We will show ϕ ' ˜ϕ, so that allmaps from X to Y are homotopic to ˜ϕ

The plan is: use the contractibility of X to shrink it to a point (at p), then usethe path-connectedness of Y to move ϕ(p) (which is in Y ) to q ∈ Y So we need

a path γ from ϕ(p) to q By the path-connectedness of Y we have one:

γ : I → Y, with γ(0) = ϕ(p), γ(1) = q

Now we can set up the requisite homotopy.

Φ(x, t) =

(ϕ(H(x, 2t)), 0 ≤ t ≤ 1

2

γ(2t − 1), 1

2 ≤ t ≤ 1.And check it:

§52 The Fundamental Group

1 A subset A of Rnis star convex iff for some point a0 ∈ A, all the line segments joining

a0 to other points of A lie in A, i.e., (1 − λ)a + λa0 ∈ A, ∀λ ∈ (0, 1)

(a) Find a star convex set that is not convex

A six-pointed star like the Star of David, or a pentacle will work if you let a0 bethe center (Hence the name “star convex”.)

The set {(x, y) .x = 0 or y = 0} ⊆ R2 is star convex with respect to the origin

Or let I2 = I × I ⊆ R2 and let X = {(x, y) y = 0} ⊆ R2 Then A = I2∪ X is

a star convex subset of R2 which is not convex The convex hull of A (smallestconvex set containing A, or intersection of all convex sets containing A) is

conv(A) = {(x, y) 0 ≤ y < 1} ∪ X

(b) Show that if A is star convex, A is simply connected

Let a ∈ A be a point satisfying the definition of star convexity Then H : A×I →

A by

H(x, t) = (1 − t)x + tashows A is contractible (via the straight line homotopy) Thus, [X, A] consists

of single element, by §51, Exercise 3(c), for any space X In particular, this

is true for [S1, A] Now let [S1, A]a ⊆ [S1, A] be those maps which send atleast one point of S1 to a Then [S1, A]a also consists of a single element But[S1, A]a = π1(A, x0)! So π1(A, x0) is trivial

2 Let α be a path in X from x0 to x1; let β be a path in X from x1 to x2 Show that

3 Let x0 and x1 be points of the path-connected space X Show that π1(X, x0) isabelian iff for every pair α, β of paths from x0 to x1, we have ˆα = ˆβ.

(⇒) Suppose π1(X, x0) is abelian, and let α, β be paths from x0 to x1 Since ˆα andˆ

β are both homomorphisms from π1(X, x0) to π1(X, x1), we need to prove that theyboth send a loop f ∈ π1(X, x0) to the same loop in π1(X, x1) Note that π1(X, x1)must also be abelian, since ˆα is an isomorphism, by Cor 52.2 (we are given that X ispath connected) Now we show ˆα = ˆβ

∗ [f ] ∗ [β] multiply both sides on the left[ ¯α] ∗ [f ] ∗ [α] = [ex 1] ∗ ¯β

∗ [f ] ∗ [β] Thm 51.2(3)[ ¯α] ∗ [f ] ∗ [α] = ¯β

∗ [f ] ∗ [β] Thm 51.2(2) and Thm 51.2(2)ˆ

(⇐) Now suppose that ˆα = ˆβ for every pair α, β of paths from x0 to x1 We mustshow π1(X, x0) is abelian, so pick f, g ∈ π1(X, x0) and show [f ] ∗ [g] = [g] ∗ [f ] Now

we have the freedom to choose our α and β, so define

There is actually a much simpler way to do each direction of this proof Can youfind it? Hint: remember the old tricks of “adding 0” or “multiplying by 1” for theforward direction, and choose a more clever α, β for the backward direction

4 Let A ⊆ X; suppose r : X → A is a retraction, i.e., a continuous map such thatr(a) = a for each a ∈ A If a0 ∈ A, show that

r∗ : π1(X, a0) → π1(A, a0)

is surjective

Let f : I → A be a loop in A, based at x0 We must find a loop g in X, based at

x0, such that r∗([g]) = [r ◦ g] = [f ] Because f is also a loop in X, based at x0, wecan let g = f Then

r ◦ f (t) = r(f (t)) = f (t), ∀t ∈ I,since f (t) ∈ A and r is the identity on A

5 Let A ∈ Rnand let h : (A, a0) → (Y, y0) Show that if h is extendable to a continuousmap of Rn into Y , then h∗ is the trivial homomorphism.

Let [f ] ∈ π1(A, a0) so that f : I → A is a loop at a0, and consider h∗([f ]) =[h ◦ f ] ∈ π1(Y, y0) The extendability of h says that we can find a continuous map

g : (Rn, a0) → (Y, y0) such that g(a) = a whenever a ∈ A Then

(g ◦ f )(t) = g(f (t)) = h(f (t)) = (h ◦ f )(t), ∀t ∈ I,since f (t) ∈ A and g = h on A

Strategy: if h ◦ f were nulhomotopic, then we would have

h∗([f ]) = [h ◦ f ] = [ey 0], ∀f,

so that h∗ is trivial So we show h ◦ f to be nulhomotopic

Define ˜ϕ : I → Y by ˜ϕ(t) = y0, ∀t Then define Φ : Rn× I → Y by

Φ(x, t) = g((1 − t)f (x) + ta0),and check

t = 0 Φ(x, 0) = g(f (x)) = g ◦ f (x) = h ◦ f (x)

t = 1 Φ(x, 1) = g(a0) = y0 = ˜ϕ(x)

Note: we don’t know that h((1 − t)f (x) + ta0) is defined, but since h is extendible to

a continuous map on all of Rn, we konw that g((1 − t)f (x) + ta0) is defined, since(1 − t)f (x) + ta0 is just some point between f (x) ∈ A and a0

6 Show that if X is path connected, the homomorphism induced by a continuous map

is independent of a base point, up to isomorphisms of the groups involved Moreprecisely, let h : X → Y be continuous, with h(x0) = y0 and h(x1) = y1 Let α be apath in X from x0 to x1, and let β = h ◦ α Show that



y





yβˆ

π1(X, x1) −−−−−−−→ π(hx1)∗ 1(Y, y1)

We need to see that these two operations do the same thing to any [f ] ∈ π1(X, x0),

so let f be a loop in X based at x0 On the one hand, we have

On the other hand, we have

Vα ⊆ X × Y such that for each α, p

α : Vα → U is a homeomorphism

Since Y is discrete, {y} is open in Y and so U × {y} is open in X × Y by the def

of product topology Then we show

Suppose we have two partitions of p− 1(U ) into slices:

A = {Vα}α∈A and B = {Vβ}β∈B.Fix b ∈ B Then for any α, we can find the unique point (by homeomorphism)

bα ∈ Vα such that p(bα) = b, and for any β, we can find the unique point bβ ∈ Vβ

such that p(bβ) = b Note that every Vα, Vβ is connected, by homeomorphism with

U We will show that there is a bijection between these partitions; i.e., A is actuallyjust a reindexing of B

Fix α0 Define f : A → B as follows: find the unique β0 such that bα 0 ∈ Vβ 0 anddefine f (α0) = β0 There will be such a Vβ 0, because E = ∪Vβ, and that Vβ 0 will beunique by the disjointness of the Vβ

Now to see that this is a bijection For bβ ∈ Vβ ⊆ E, ∃α such that bβ ∈ Vα becauseS

Vα contains E This shows surjectivity For injectivity, note that

bβ ∈ Vα 1 ∩ Vα 1 =⇒ α1 = α2

by the disjointness of the partition

3 Let p : E → B be a covering map; let B be connected Show that if p−1(b0) has kelements for some b0 ∈ B, then p− 1(b) has k elements for every b ∈ B, i.e., E is ak-fold covering of B.

Since |p− 1(b0)| = k, we can find U such that

For b ∈ C, we can find Ub such that p−1(Ub) = Fk

i=1Vi, where the Vi are open.Thus for x ∈ Ub, |p−1(x)| = k Hence b ∈ Ub ⊆ C shows C is open Similarly, D isopen This gives C, D as a disconnection of B < Hence no such b1 exists

4 Let q : X → Y and r : Y → Z be covering maps; let p = r ◦ q Show that if r− 1(z) isfinite for each z ∈ Z, then p is a covering map

As r is a covering map, pick z ∈ Z and find an open neighbourhood Z of z suchthat p− 1(Z) =Fn

i=1Vi where Vi is open in Y and r

V i: Vi → Z is a homeomorphism.Define vi to be the single element of p− 1(z) ∩ Vi, for each i As q is a covering map,

we can find an open neighbourhood Ui of vi such that q− 1(Ui) =F

Di α = q− 1(Ui∩ Vi) ∩ Ai α,which are also open Now C is a neighbourhood of z ∈ Z for which

5 Show that the map p : S1 → S1 given by p(z) = z2 is a covering map Generalize tothe map p(z) = zn.

Here we consider S1 ⊆ C, so that for z ∈ S1 we may write

z = eiθ and p(z) = z2 = e2iθ.For z ∈ S1, let z = eiθ so θ = arg(z) ∈ [0, 2π) Let U be the image of θ − π

2, θ + π 2

under the map θ 7→ eiθ so that U is the open semicircle centered at z Then p− 1(U )consists of the “quarter circle” centered at z

V1 = exp θ − π

4, θ + π 4

and the “quarter circle” centered at −z

V2 = exp −θ + π

4, −θ − π

4

.Clearly, U = V1t V2 and U is homeomorphic to each of V1, V2 by p(z) = z2 Since zhas the requisite neighbourhood, we are done

For p(z) = zn, use the same U You will get p− 1(U ) =Fn

i=1Vi, where each Vi goes

1

2n of the way around the circle S1

Bonus Problem: If p : E → B is a covering map, show that p− 1(b) ⊆ E has the discretetopology, for any b ∈ B

Consider the topology that p− 1(b) inherits from E Since p is a covering map, we can find

a neighbourhood U of b which is evenly covered by p, i.e.,

p− 1(U ) = G

α∈A

Vα, where the Vα are open

Since the union of the Vα is disjoint,

Vα∩ p− 1(b) = {xα},where xα is the unique point in Vα such that p(xα) = b We have just represented {xα} as

an intersection of an open set Vα with the subspace p− 1(b) ⊆ E, which shows that {xα} isopen in the subspace topology of p− 1(b) Since

p− 1(b) = G

α∈A

{xα},this shows p−1(b) is discrete

§54 The Fundamental Group of the Circle

1 What goes wrong with the path-lifting lemma 54.1 for the local homeomorphism ofExample 2 of §53?

The very first step: you cannot find an open set U which contains b0 = (1, 0) and

is evenly covered by p; p is not a covering map

2 In defining the map ˜F in the proof of Lemma 54.2, why were we so careful about theorder in which we considered the small rectangles?

To maintain the continuity of the lift If you performed the lifting for the squares

in random order, there is no guarantee that the lifts would “connect” or “match up”along the boundaries of the little squares

3 Let p : E → B be a covering map Let α and β be paths in B with α(1) = β(0); let

4 Consider the covering map p : R × R+→ R2

0 of Example 6 of §53:2

(x, t) 7→ ((cos 2πx, sin 2πx), t) 7→ t(cos 2πx, sin 2πx).Find liftings of the paths

f (t) = (2 − t, 0),g(t) = ((1 + t) cos 2πt, (1 + t) sin 2πt),h(t) = f ∗ g

Sketch the paths and their liftings

The point here is that f ∗ g is a closed loop, but no lifting of it is

5 Consider the covering map p × p : R × R → S1× S1 of Example 4 of §53 Considerthe path in S1× S1 given by

f (t) = (cos 2πt, sin 2πt) × (cos 4πt, sin 4πt) = (eit, e2it),

if we consider T2 = S1× S1 as a subspace of C × C Sketch what f looks like when

S1× S1 is identified with the doughnut surface D Find a lifting ˜f of f to R × R,and sketch it

2π 2π

Figure 4 T2 = S1× S1 is represented two different ways on the right The

path (loop) f begins at the white dot and traverses T2 For visualization, the

first S1 axis is sketched on top of the doughnut, the second S1 axis is sketched

in front, with an extra copy in back to indicate where f goes “through the

top” of the square representation of T2 Note that as the path traverses the

horizontal S1 once, it traverses the vertical S1 twice

6 Consider the maps g, h : S1 → S1 given g(z) = zn and h(z) = z−n Compute theinduced homomorphisms g∗, h∗ of the infinite cyclic group π1(S1, b0) into itself.The group is cyclic, so it suffices to determine the image of its generator, γ(t) =

e2πit, t ∈ I under g Since

h∗([γ]) = [h ◦ γ] = [γ] ∗ · · · ∗ [γ]

n times

= [γ]∗n = [γ]∗(−n) or h : γ(t) 7→ γ(−nt)

7 Generalize the proof of Theorem 54.5 to show that the fundamental group of thetorus is isomorphic to the group Z2 = Z × Z.

Let p : R → S1 by the covering map given by p(x) = (cos 2πx, sin 2πx) Then byThm 53.3, we may define

P = p × p : R2 → T2,the standard cover of the torus, as in Example 4 of §53 Take e0 = (0, 0) and

b0 = P (e0) Then

P− 1(b0) = {(x, y) ∈ R2

.x, y ∈ Z} = Z2.Since R2 is simply connected, the lifting correspondence φ gives a bijection (by Thm54.4), so we just need to show φ is a homomorphism

Given [f ], [g] ∈ π1(T2, b0), let ˜f and ˜g be their respective liftings to paths in R2

beginning at e0 = (0, 0) Let (m, n) = ˜f (1) and (j, k) = ˜g(1) Since these points are

in the preimage of b0, we know m, n, j, k ∈ Z Then

φ([f ]) + φ([g]) = (m, n) + (j, k) = (m + j, n + k) ∈ Z2.Now take ˜˜g = (m, n) + ˜g to be the translate of ˜g which begins at (m, n) ∈ R2 Then

8 Let p : E → B be a covering map, with E path connected Show that if B is simplyconnected, then p is a homeomorphism.

By Thm 54.6(a), p∗ : π1(E, e0) → π1(B, b0) is 1-1 Since π1(B, b0) is trivial byhypothesis, it must also be the case that π1(E, e0) is also trivial, and p∗ is actually

an isomorphism Then every loop f at b0 is in H, and hence lifts to a loop ˜f at e0,which shows p− 1(b0) = e0 by the lifting correspondence — if you have a surjection(by Thm 54.4) where the domain has only one element, then the range must alsoconsists of only one element Thus by Ex 53.3, E is a 1-fold covering of B, i.e., Eand B are homeomorphic

... p∗ : π1< /sub>(E, e0) → π1< /sub>(B, b0) is 1- 1 Since π1< /sub>(B, b0) is trivial byhypothesis, it must also be the case that π1< /sub>(E, e0)... class="text_page_counter">Trang 15

5 Consider the covering map p × p : R × R S1< /small>ì S1< /small> of Example of Đ53 Considerthe path in S1< /small>×... ∈ Vα 1< /small> ∩ Vα 1< /small> =⇒ α1< /small> = α2

by the disjointness of the partition