Topology - Solutions Sections 51-54

7/21/2019 Topology - Solutions Sections 51-54

http://slidepdf.com/reader/full/topology-solutions-sections-51-54 1/29

§51. Homotopy of Paths

General note: every problem here uses Theorem 18.2, mostly part (c). You should probablybe on good terms with this Theorem.

1. Show that if h, h : X → Y are homotopic and k, k : Y → Z are homotopic, thenk ◦ h k ◦ h.

Since h h, we have a homotopy

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

and since k k , we have a homotopy

K : Y × I → Z with K (x, 0) = k(x), K (x, 1) = k (x).

Then define F : X × I → Z by

F (x, t) = k(H (x, 2t)), 0 ≤ t ≤ 12

K (h(x), 2t − 1), 12

≤ t ≤ 1.

Then check that F does what it oughta do, at different times t:1

t = 0 F (x, 0) = k(H (x, 0)) = k(h(x)) = k ◦ h(x)

t = 12

F

x, 12

= k(H (x, 1)) = k(h(x)) = K (h(x), 0)

t = 1 F (x, 1) = K (h(x), 1) = k (h(x)) = k ◦ h(x).

2. Given spaces X and Y , let [X, Y ] denote the set of homotopy classes of maps of X

into 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 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.

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

2.

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Topology (2nd ed.) — James R. Munkres

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 ≤ 12

γ (2t − 1), 12

≤ t ≤ 1.

Note that Φ does not depend on x once t ≥ 12

! 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 along

γ (I ) while 12

≤ t ≤ 1.Check that Φ does what it oughta do:

t = 0 Φ(x, 0) = ϕ(x)

t = 12

Φ

x, 12

= ϕ(0) = a = γ (0)

t = 1 Φ(x, 1) = γ (1) = p.

3. A space X is said to be contractible if the identity map iX : X → X is nulhomotopic.

(a) Show that I and R are contractible.

Define the constant map ϕ(x) = 0, for either space. Then we define the homo-topy H : X × I → X by

H (x, t) = (1 − t) · idX (x) = (1 − t)x.

This polynomial in x and t is clearly continuous, and

t = 0 H (x, 0) = idX (x)

t = 1 H (x, 1) = 0.

(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 a

path from z to p (recall that y, z are fixed ). Thus, define a path by

γ (t) =

H (y, 2t), 0 ≤ t ≤ 1

2

H (z, 2 − 2t), 12

≤ t ≤ 1.

And check it:

t = 0 γ (t) = H (y, 0) = idX (y) = y

t = 12

γ (t) = H (y, 1) = p = H (z, 1)

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Solutions by Erin P. J. Pearse

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 all

maps 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 needa 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.

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Now we can set up the requisite homotopy.

Φ(x, t) =

ϕ(H (x, 2t)), 0 ≤ t ≤ 1

2

γ (2t − 1), 12

≤ t ≤ 1.

And check it:t = 0 Φ(x, 0) = ϕ(idX (x)) = ϕ(x)

t = 12

Φ

x, 12

= ϕ( p) = γ (0)

t = 1 Φ(x, 1) = γ (1) = q = ϕ(x).

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§52. The Fundamental Group

1. A subset A of Rn is star convex iff for some point a0 ∈ A, all the line segments joininga0 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 I 2 = I × I ⊆ R2 and let X = {(x, y) ... y = 0} ⊆ R2. Then A = I 2 ∪ X isa 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 + ta

shows A is contractible (via the straight line homotopy). Thus, [X, A] consistsof single element, by §51, Exercise 3(c), for any space X . In particular, thisis true for [S 1, A]. Now let [S 1, A]a ⊆ [S 1, A] be those maps which send atleast one point of S 1 to a. Then [S 1, A]a also consists of a single element. But

[S

1

, 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 thatif γ = α ∗ β , then γ = β ◦ α.

We show γ = β ◦ α by showing that γ ([f ]) = (β ◦ α)([f ]), for every path f .

γ ([f ]) = [γ ] ∗ [f ] ∗ [γ ]

=

α ∗ β

∗ [f ] ∗ [α ∗ β ] def of γ

= β ∗ α ∗ [f ] ∗ [α ∗ β ] def of the reverse α

= β ∗ [α] ∗ [f ] ∗ [α] ∗ [β ] def of ∗, p.326

=

β

∗ α([f ]) ∗ [β ] def of α

= β (α([f ])) def of β

= (β ◦ α)([f ])

Note that the reverse of α ∗ β (α followed by β ) is the reverse of β followed by thereverse of α, in the third line above.

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3. Let x0 and x1 be points of the path-connected space X . Show that π1(X, x0) is

abelian iff for every pair α, β of paths from x0 to x1, we have α = β .

(⇒) Suppose π1(X, x0) is abelian, and let α, β be paths from x0 to x1. Since α andˆβ 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 α is an isomorphism, by Cor 52.2 (we are given that X is

path connected). Now we show α = β .

α([f ]) = [α] ∗ [f ] ∗ [α] def of α

= [α] ∗ [α] ∗ [f ] π1(X, x1) is abelian

= [ex1 ] ∗ [f ] Thm 51.2(3)

= [f ] ∗ [ex1 ] π1(X, x1) is abelian

= [f ] Thm 51.2(2)

Similarly, β ([f ]) = [f ]. So we have actually proven that α and β are the identity, i.e.,that if X is path connected, it must also be simply connected. <

Of course, this is all completely wrong! Why? It makes no sense: [α] is not anelement of π1(X, x1) because it isn’t even a loop! Therefore, the fact that π1(X, x0)is abelian doesn’t apply. The expression [α] ∗ [f ], which appears in line 2 isn’t evendefined, since α ends at x1, and f ends at x0.

Try again:

[f ] ∗ [α] = [f ] ∗ [α] ∗ [ex1] Thm 51.2(2)

= [f ] ∗ [α] ∗ β ∗ [β ] Thm 51.2(3)

= [f ] ∗ α ∗ β ∗ [β ] def of ∗, p.326

=

α ∗ β

∗ [f ] ∗ [β ] π1(X, x0) is abelian

= [α] ∗

β

∗ [f ] ∗ [β ] def of ∗, p.326

[α] ∗ [f ] ∗ [α] = [α] ∗ [α] ∗

β

∗ [f ] ∗ [β ] multiply both sides on the left

[α] ∗ [f ] ∗ [α] = [ex1] ∗

β

∗ [f ] ∗ [β ] Thm 51.2(3)

[α] ∗ [f ] ∗ [α] =

β

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

α([f ]) = β ([f ]) def of α, β

Whew! Note that the fourth equality above is justified because α ∗ β is a loop, sowe avoid the problem of the other (incorrect) solution above.

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(⇐) Now suppose that α = β 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 ]. Nowwe have the freedom to choose our α and β , so define

α = f, β = f ∗ g.

Then compute:

α([f ]) = f ([f ]) α = f

= [ f ] ∗ [f ] ∗ [f ] def of f

= [ex0 ] ∗ [f ] Thm 51.2(3), ex0 = ex1

= [f ] Thm 51.2(2)

and

β ([f ]) = f ∗ g([f ]) β = f ∗ g

= f ∗ g ∗ [f ] ∗ [f ∗ g] def of f

= [g] ∗

f

∗ [f ] ∗ [f ] ∗ [g] see Ex 2, at the end

= [g] ∗ [ex0 ] ∗ [f ] ∗ [g] Thm 51.2(3), ex0 = ex1

= [g] ∗ [f ] ∗ [g] Thm 51.2(2).

Since α = β , we know α([f ]) = β ([f ]), and so

[f ] = [g] ∗ [f ] ∗ [g]

[g] ∗ [f ] = [g] ∗ [g] ∗ [f ] ∗ [g]

[g] ∗ [f ] = [f ] ∗ [g],using Thm 51.2(2,3) to cancel the g’s. So π1(X, x0) is abelian.

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 atx0, 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.

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5. Let A ∈ Rn and 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 mapg : (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 ] = [ey0 ], ∀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 checkt = 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 toa 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 mapis independent of a base point, up to isomorphisms of the groups involved. More

precisely, 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

β ◦ (hx0)∗ = (hx1)∗ ◦ α,

so that the diagram commutes:

π1(X, x0) (hx0)∗−−−−−−−→ π1(Y, y0)

α

β π1(X, x1)

(hx1)∗−−−−−−−→ π1(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

β ◦ (hx0)∗([f ]) = h ◦ α ◦ (hx0)∗([f ]) def of β

= h ◦ α([h ◦ f ]) def of h∗

=

h ◦ α

∗ [h ◦ f ] ∗ [h ◦ α] def of β

=

h ◦ α

∗ [h ◦ f ] ∗ [h ◦ α] def of β.

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On the other hand, we have

(hx1)∗ ◦ α([f ]) = (hx1)∗ ([α] ∗ [f ] ∗ [α]) def of α

= (hx1)∗ ([α ∗ f ∗ α]) def of ∗

= [h ◦ (α ∗ f ∗ α)] def of h∗

= [(h ◦ α) ∗ (h ◦ f ) ∗ (h ◦ α))] k ◦ (f ∗ g) = (k ◦ f ) ∗ (k ◦ g)

= [h ◦ α] ∗ [h ◦ f ] ∗ [h ◦ α] def of ∗.

So we still need

h ◦ α

= [h ◦ α]. But this is true:

h ◦ α(t) = (h ◦ α)(1 − t) = h(α(1 − t)) = h(α(t)) = h ◦ α(t).

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§53. Covering Spaces

1. Let Y have the discrete topology. Show that if p : X × Y → X is projection on thefirst coordinate, then p is a covering map.

It is clear that p is continuous and surjective (if you have doubts, read pp. 107–110). Pick x ∈ X and let U be a neighbourhood of x. We will show that U isevenly covered by p; that is, that p−1(U ) can be written as a union of disjoint setsV α ⊆ X × Y such that for each α, p

α

: V α → 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

p−1(U ) =y∈Y

U × {y}.

Note: this union is disjoint because

(x, y) ∈ U × {y1} ∩ U × {y2} =⇒ y = y1 = y2

=⇒ U × {y1} = U × {y2}.

Let x ∈ p−1(U )

2. Let p : E → B be continuous and surjective. Suppose that U is an open set of Bthat is evenly covered by p. Show that if U is connected, then the partition of p−1(U )into slices is unique.

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

A = {V α}α∈A and B = {V β }β ∈B.

Fix b ∈ B. Then for any α, we can find the unique point (by homeomorphism)bα ∈ V α such that p(bα) = b, and for any β , we can find the unique point bβ ∈ V β such that p(bβ ) = b. Note that every V α, V β is connected, by homeomorphism withU . We will show that there is a bijection between these partitions; i.e., A is actually just a reindexing of B .

Fix α0. Define f : A → B as follows: find the unique β 0 such that bα0 ∈ V β 0 anddefine f (α0) = β 0. There will be such a V β 0, because E = ∪V β , and that V β 0 will beunique by the disjointness of the V β .

Now to see that this is a bijection. For bβ ∈ V β ⊆ E , ∃α such that bβ ∈ V α becauseV α contains E . This shows surjectivity. For injectivity, note that

bβ ∈ V α1 ∩ V α1 =⇒ α1 = α2

by the disjointness of the partition.

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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

p−1(U ) =ki=1

V i and pV i

= pi : V i → U is a homeomorphism, ∀i.

We assume that ∃b1 such that | p−1(b1)| = j = k, and we will contradict the factthat B is connected. Define

C = {b ... | p−1(b)| = k} and D = {b ... | p−1(b)| = k}.

We have b0 ∈ C , so C = ∅, and b1 ∈ D, so D = ∅. Also, it is clear that C ∩ D = ∅

and C ∪ D = B. So it just remains to show C, D are open.For b ∈ C , we can find U b such that p−1(U b) = k

i=1 V i, where the V i are open.Thus for x ∈ U b, | p−1(x)| = k. Hence b ∈ U b ⊆ C shows C is open. Similarly, D is

open. 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 ) =

n

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

V i: V i → Z is a homeomorphism.

Define vi to be the single element of p−1

(z ) ∩ V i, for each i. As q is a covering map,we can find an open neighbourhood U i of vi such that q −1(U i) = α Aiα where Aiα

is open in X and q Aiα

: Aiα → U i is a homeomorphism.

With so many sets, we need to find some common ground, so define

C =ni=1

r(U i ∩ V i).

Each U i ∩ V i is a neighbourhood of vi in Y which is evenly covered by q , and hasr(U i∩V i) ⊆ Z . So C is evenly covered by r (each slice is a U i∩V i). Most importantly,the fact that C is a finite intersection guarantees that C is open. We next define

Diα = q −1(U i ∩ V i) ∩ Aiα,

which are also open. Now C is a neighbourhood of z ∈ Z for which

p−1(C ) =i,α

Diα and pDiα

= (r ◦ q )Diα

is a homeomorphism.

Also, the Diα are disjoint because the Aiα are.

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5. Show that the map p : S 1 → S 1 given by p(z ) = z 2 is a covering map. Generalize tothe map p(z ) = z n.

Here we consider S 1 ⊆ C, so that for z ∈ S 1 we may write

z = eiθ

and p(z ) = z 2

= e2iθ

.For z ∈ S 1, 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 → eiθ so that U is the open semicircle centered at z . Then p−1(U )consists of the “quarter circle” centered at z

V 1 = exp

θ − π4

, θ + π4

and the “quarter circle” centered at −z

V 2 = exp

−θ + π4

, −θ − π4

.

Clearly, U = V 1 V 2 and U is homeomorphic to each of V 1, V 2 by p(z ) = z 2. Since z has the requisite neighbourhood, we are done.

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

i=1 V i, where each V i goes12n

of the way around the circle S 1.

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

p−1(U ) =α∈A

V α, where the V α are open.

Since the union of the V α is disjoint,

V α ∩ p−1(b) = {xα},

where xα is the unique point in V α such that p(xα) = b. We have just represented {xα} asan intersection of an open set V α with the subspace p−1(b) ⊆ E , which shows that {xα} isopen in the subspace topology of p−1(b). Since

p

−1

(b) = α∈A{xα},

this shows p−1(b) is discrete.

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§54. The Fundamental Group of the Circle

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

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

α and β be liftings of them such that α(1) = β (0). Show that α ∗ β is a lifting of α ∗ β .

We need to show p ◦ (α ∗ β ) = α ∗ β . By definition, p ◦ (α ∗ β ) : I → E by

p ◦ (α ∗ β )(t) =

p ◦ α(2t), 0 ≤ t ≤ 1

2,

p ◦ β (2t − 1), 12

≤ t ≤ 1, def of ∗

= α(2t), 0 ≤ t ≤ 1

2,

β (2t − 1), 1

2 ≤ t ≤ 1,

def of α, β

= α ∗ β (t) def of ∗.

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4. Consider the covering map p : R× R+ → R20 of Example 6 of §53:2

(x, t) → ((cos 2πx, sin2πx), t) → t(cos 2πx, sin2πx).

Find liftings of the paths

f (t) = (2 − t, 0),g(t) = ((1 + t)cos2πt, (1 + t)sin2πt),

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.

p

f f

x

t

1

0-1 1

0 1 2

2

~

Figure 1. Three liftings of f . f is the one in the centre.

pg

g

x

t

1

0-1 1

0 1-1 2

2

-2

~

Figure 2. Three liftings of g. g is the one in the centre.

pg

g

x

t

1

0-1 1

0 1-1 2

2

-2~

f ~

f

Figure 3. f ∗ g is a lifting of f ∗ g.

2I use the shorthand R20 = R

2\{(0, 0)}.

14




5. Consider the covering map p × p : R× R → S 1 × S 1 of Example 4 of §53. Considerthe path in S 1 × S 1 given by

f (t) = (cos 2πt, sin2πt) × (cos4πt, sin4πt) = (eit, e2it),

if we consider T2

= S 1

× S 1

as a subspace of C× C. Sketch what f looks like whenS 1 × S 1 is identified with the doughnut surface D. Find a lifting f of f to R × R,and sketch it.

2π

2π

0S 1

S

1

×

S

1

S 11

1 2

2

0

f

p

= f

f ~

Figure 4. T2 = S 1 × S 1 is represented two different ways on the right. Thepath (loop) f begins at the white dot and traverses T2. For visualization, thefirst S 1 axis is sketched on top of the doughnut, the second S 1 axis is sketchedin front, with an extra copy in back to indicate where f goes “through thetop” of the square representation of T2. Note that as the path traverses thehorizontal S 1 once, it traverses the vertical S 1 twice.

6. Consider the maps g, h : S 1 → S 1 given g(z ) = z n and h(z ) = z −n. Compute theinduced homomorphisms g∗, h∗ of the infinite cyclic group π1(S 1, 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

g ◦ γ (t) = g

e2πit

=

e2πitn

= e2πint

is a loop which goes n times around the circle, in the direction of γ , we have

g∗([γ ]) = [g ◦ γ ] = [γ ] ∗ · · · ∗ [γ ]

n times

= [γ ]∗n or g : γ (t) → γ (nt).

Similarly for h, we have that the loop

h ◦ γ (t) = h

e2πit

=

e2πit−n

= e−2πint

goes n times around the circle, in the direction opposite to γ . This gives

h∗([γ ]) = [h ◦ γ ] = [γ ] ∗ · · · ∗ [γ ] n times

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

15




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 → S 1 by the covering map given by p(x) = (cos 2πx, sin2πx). 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) andb0 = 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. Thenφ([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

P ◦ ˜g(t) = P ((m, n) + g(t))

= ( p(m + g1(t)), p(n + g2(t)))

= ( p(g1(t)), p(g2(t)))

= P ◦ g(t)

= g(t),

so ˜g is a lifting of g . The third equality here follows because p(m + x) = p(x) by theformula for p, and the last line follows because g is a lifting of g. Then the productf ∗ ˜g is a well-defined path, and it is the lifting of f ∗ g that begins at (0, 0). It isclear that it begins at (0, 0), since f does. We check that it is a lifting:

P ◦

f ∗ ˜g

(t) =

P ( f (2t)), 0 ≤ t ≤ 1

2,

P (˜g(2t − 1)), 12

≤ t ≤ 1, def of ∗

=

f (2t), 0 ≤ t ≤ 1

2,

g(2t − 1)), 12

≤ t ≤ 1, by above

= f ∗ g(t).

The end point of ˜g is

˜g(1) = (m, n) + g(1) = (m + j, n + k).

Since [f ] ∗ [g] begins at b0, φ([f ] ∗ [g]) is well-defined, and we have

φ([f ] ∗ [g]) = f ∗ ˜g(1) = (m + j, n + k).

Thus φ([f ] ∗ [g]) = φ([f ]) + φ([g]), and we are done.

16




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., E and B are homeomorphic.

17




§55. Retractions and Fixed Points

Note: I use the following notation for the image of a map f : X → Y .

Im(f ) = {y ∈ Y ...

y = f (x) for some x ∈ X }

1. Show that if A is a retract of B2, then every continuous map f : A → A has a fixedpoint.

If A is a retract, then A ⊆ B2 and there is a retraction i.e., a continuous mapr : B2 → A with r(a) = a, ∀a ∈ A. Let j : A → B2 by inclusion. Then j ◦ f ◦ r :B2 → B2 is a continuous map, so it has a fixed point by Brouwer’s Theorem (55.6).Denote it by c. Since Im(f ◦ r) ⊆ A, it must be that c ∈ A. Then

c = j ◦ f ◦ r(c) = j ◦ f (c) rA is the identity= f (c) j is inclusion.

2. Suppose h : S 1 → S 1 is nulhomotopic.

(a) Show that h has a fixed point.

h extends to k : B2 → S 1 by Lemma 55.3 with X = S 1. But S 1 ⊆ B2, so we canactually consider k as a mapping k : B2 → B2. Then k must have a fixed pointby Brouwer’s Theorem (55.6), call it b. Since Im(k) ⊆ S 1, k(b) = b implies thatb ∈ S 1. Thus h(b) is defined. In fact,

h(b) = k(b) as an extension, k = h on S 1

= b b is a fixed point of k.

(b) Show that h maps some point x to its antipode −x.

Define f : S 1 → S 1 by f (x) = −x. Then f ◦ h : S 1 → S 1 is nulhomotopic, so itmust have a fixed point by (a), call it a. Then

a = f ◦ h(a) = f (h(a)) = −h(a)

=⇒ h(a) = −a.

3. Show that if A is a nonsingular 3 × 3 matrix having nonnegative entries, then A hasa positive real eigenvalue.

Following Cor. 55.7, let B = S 2 ∩ O1 ⊆ R3. All the components of x ∈ B arenonnegative, and x = 0. Since A is nonsingular, Ax = 0, and the map

S : B → B by S (x) = Ax/Ax

18




is well-defined. Since B is homeomorphic to the ball, S must have a fixed point x0

by Brouwer’s Thm (55.6). Then

x0 = Ax0/Ax0 =⇒ Ax0 = Ax0x0,

i.e., A has an eigenvector x0 with corresponding eigenvalue Ax0 > 0.

4. Assume: for each n, there is no retraction r : Bn+1 → S n.

(a) The identity map i : S n → S n is not nulhomotopic.

Suppose i = idS n were nulhomotopic. We follow the proof of Lemma 55.3,(1) =⇒ (2) =⇒ (3). Let H : S n × I → S n be a homotopy between i and aconstant map. Define π : S n × I → Bn+1 by

π(x, t) = (1 − t)x.

Then π is a quotient map and H induces a continuous map k : Bn+1 → S n (via

π) that is an extension

(b) The inclusion map j : S n → Rn+1 is not nulhomotopic.

(c) Every nonvanishing vector field on Bn+1 points directly outward at some point,and directly inward at some point.

(d) Every continuous map f : Bn+1 → Bn+1 has a fixed point.

(e) Every (n + 1) × (n + 1) matrix with positive real entries has a positive realeigenvalue.

(f) If h : S n → S n is nulhomotopic, then h has a fixed point, and h maps some pointx to its antipode −x.

19




§56. Deformation Retracts and Homotopy Type

Note: I am occasionally sloppy and saying “x0” when I mean “the constant map at x0”.This is a standard abuse of language.

1. Show that if A is a deformation retract of X and B is a deformation retract of A,then B is a deformation retract of X .

So we have some r : X → A which is the identity on A, and s : A → B which isthe identity on B. Then s ◦ r : X → B and for b ∈ B,

s ◦ r(b) = s(r(b)) = s(b) = b,

because b ∈ B ⊆ A ⊆ X . We also have r H idX by some homotopy H which keepsevery point of A fixed. In particular, H keeps every point of B ⊆ A fixed. Also,

we have s K idA by some homotopy K which keeps every point of B fixed. DefineF : X × I → X by

F (x, t) =

H (x, 2t), 0 ≤ t ≤ 1

2, H (x, 0) = idX , H (x, 1) = r(x),

K (r(x), 2t − 1), 12

≤ t ≤ 1, K (x, 0) = idA, K (x, 1) = s(x),

as in §51 Exercise 1. Then s ◦ r F idX .

2. For each of the following spaces, the fundamental group is either (1) trivial, (2)infinite cyclic, or (3) isomorphic to the fundamental group of the figure eight.

(a) The solid torus B 2 × S 1.For (x, y) ∈ B2 × S 1, the homotopy

H ((x, y), t) = ((1 − t)x, y)

shows that S 1 is a deformation retract of the solid torus, so π1(B2 × S 1) =π1(S 1) = Z.

(b) The torus with a point removed.I 1\{(1

2, 12

)} has its boundary as a deformation retract, by the straight-line ho-motopy (from (1

2, 12

)). Under the quotient map which makes I 2 into the torus,

∂ I 1

\{(12 ,

12)} becomes a figure-eight.

(c) The cylinder C = S 1 × I .π1(C ) = Z, because the cylinder has S 1 as a deformation retract by the homotopy

H ((x, y), t) = (x, (1 − t)y).

(d) The infinite cylinder IC = S 1 × R.π1(IC ) = Z by the same reason (and same homotopy) as (c).

20




(e) R3 with the nonnegative axes deleted.Consider that any loop in this space is homotopic to some combination of α, β, γ (and their inverses), as depicted in Figure 5. Note that α ∗ β = γ , so that the

x 0

β

α

γ

Figure 5. Loops around the axes in R3.

fundamental group can be generated without γ . Also, note that β ∗ α γ = γ ,so fundamental group is not abelian.

(f) {x ... x > 1}.This space has 2S 1 = {x ∈ R2 ... x = 2} as a deformation retract by the straight-line homotopy H (x, t) = (1 − t)x + 2tx/x. 2S 1 is obviously homeomorphic toS 1 by x → x/2, so π1((f)) = Z.

(g) {x... x ≥ 1}.This space has S 1 as a deformation retract by the straight-line homotopy H (x, t) =

(1 − t)x + tx/x, so π1((g)) = Z.

(h) {x ... x < 1}.By H (x, t) = (1 − t)x, π1((h)) is trivial.

S 1∪ (R

+×0) S

1∪ (R

+×R) S

1∪ (R×0) R2

− (R+×0)

Figure 6. The spaces in exercises (i)-(l).

(i) S 1 ∪ (R+ × 0).π1((i)) is trivial by H (x, t) = (1 − t)x + tx/x.

21




(j) S 1 ∪ (R+ ×R).π1((j)) is trivial by H (x, t) = (1 − t)x + tx/x.

(k) S 1 ∪ (R× 0).By the homotopy

H (x, t) =

x x ≤ 1,

(1 − t)x + tx/x x > 1. ,

we obtain the theta space of Example 3. This has the fundamental group of thefigure eight, as both are retracts of the doubly punctured plane (apply Thm.58.3 to Example 2).

(l) R2\(R+ × 0).By H (x, t) = (1 − t)x + t(−1, 0), π1((l)) is trivial.

3. Show that given a collection C of spaces, the relation of homotopy equivalence is anequivalence relation on C .

We show the three properties.

(i) Let f = g = idX : X → X . Then f ◦ g, g ◦ f are both homotopic to the identityon X (since they are the identity on X ). So X X .

(ii) Let X Y . Then ∃f : X → Y, g : Y → X such that g ◦f idX and f ◦ g idY .But this is just the same as Y X .

(iii) Let X Y and Y Z . Then we have homotopy equivalences f : X → Y and

h : Y → Z , with corresponding inverses g : Y → X and k : Z → Y . We needto show that h ◦ f : X → Z and g ◦ k : Z → X are homotopy inverses of eachother. By associativity,

(g ◦ k) ◦ (h ◦ f ) = g ◦ (k ◦ h) ◦ f.

Then h, k are homotopy inverses, so k ◦ h idY and

g ◦ (k ◦ h) g ◦ idY §51 Exercise 1

g ◦ (k ◦ h) ◦ f g ◦ idY ◦ f (x) §51 Exercise 1 again

g ◦ f (x) by trivial homotopy

idX (x) Lemma 51.1 (transitivity).The argument is identical for (h ◦ f ) ◦ (g ◦ k) = h ◦ (f ◦ g) ◦ k.

4. Let X be the figure eight and let Y be the theta space. Describe maps f : X → Y and g : Y → X that are homotopy inverse to each other.

Define the maps as indicated in Figure 7. f maps all the points inside the dottedline to the vertical bar of the theta. Note: it is not 1-1, so it cannot have an inverse. g

22




maps all the points of the vertical bar to the “centre point” of the figure-eight wherethe two loops connect. It is similarly noninjective and hence noninvertible. These

f

g

Figure 7. The spaces in exercises (i)-(l).

maps are not inverses, but they are homotopy inverses. There is a homotopy betweenidX and g ◦ f which maps the figure eight to itself, but continuously scrunches allthe points inside the dotted line down into the centre point, as t goes from 0 to 1.Similarly, there is a homotopy from idY to f ◦ g which acts by collapsing the verticalbar to the centre point of the vertical bar, and continuously slurping in part of theboundary of the circle to replace the bar.

5. Recall that a space X is contractible iff the identity map idX is nulhomotopic. Showthat X is contractible iff X has the homotopy type of a one-point space.

This is just an application of Exercise 51.3(c) to [X, X ].

6. Show that a retract of a contractible space is contractible.

If X is contractible, then we have H : X × I → X with

H (x, 0) = x, H (x, 1) = x0, ∀x ∈ X.

Suppose r : X → A is a retraction, so that rA

= idA. Since it doesn’t matter whatpoint x0 we chose above (X is path connected by Exercise 51.3(b)), let x0 ∈ A. ThenH A×I

defines a homotopy from idA to x0.

9. Define the degree of a continuous map h : S 1 → S 1 as follows:Let b0 = (1, 0) ∈ S 1; choose a generator γ for the infinite cyclic group π1(S 1, b0).

For x0 ∈ S 1, let α : b0 x0 be a path from b0 to x0, and define γ (x0) = α(γ ) sothat γ (x0) generates π1(S 1, x0).

Given h : S 1 → S 1, choose x0 ∈ S 1 and let h(x0) = x1. Consider the homomor-phism

h∗ : π1(S 1, x0) → π1(S 1, x1).

Then h∗(γ (x0)) = d · γ (x1) for some d ∈ Z. The integer d = deg h is the degree of h.

23




(a) Show that d is independent of the choice of x0.

Consider x0 = θ ∈ S 1 as being given by an angle θ ∈ [0, 2π), and take the mapg : S 1 → S 1 to be rotation by −θ.Suppose x0, y0 ∈ S 1 with h(x0) = x1 and h(y0) = y1, and that α : b0 x0 andβ : b0 y0

(b) Show that if h, k : S 1 → S 1 are homotopic, they have the same degree.

(c) Show that deg(h ◦ k) = (deg h) · (deg k).

(d) Compute the degrees of the constant map, the identity map, the reflection mapρ(x1, x2) = (x1, −x2), and the map h(z ) = z n, where z is a complex number.

(e) Show that if h, k : S 1 → S 1 have the same degree, they are homotopic.

10. Suppose that to every map h : S n → S n we have defined deg h ∈ Z, such that

(i) Homotopic maps have the same degree.

(ii) deg(h ◦ k) = (deg h) · (deg k).

(iii) The identity map has degree 1, and constant map has degree 0, and the reflectionmap ρ(x1, . . . , xn, xn+1) = (x1, . . . , xn, −xn+1) has degree −1.

Then prove the following:

(a) There is no retraction r : Bn+1 → S n.

(b) If h : S n → S n has degree different from (−1)n+1, then h has a fixed point.

(c) If h : S n → S n has degree different from 1, then h maps some point x to itsantipode −x.

(d) If S n has nonvanishing tangent vector field v, then n is odd. (Hint: If v exists,shoe the identity map is homotopic to the antipodal map.)

24




§57. The Fundamental Group of S n

1. Let X be the union of two copies of S 2 having a single point in common. What isthe fundamental group of X ?

Consider X as sitting in R3 along the x-axis, for purposes of description. Theprojection of X to the x-axis would then be [−2, 2], with the single point in commonprojecting to 0. Let

U = X ∩ (−1, ∞) × R2 and V = X ∩ (−∞, 1) × R2,

so that U and V are clearly open. U has a copy of S 2 as a deformation retract bycollapsing the left hemisphere along the meridians leading to the origin, so π1(U ) =π1(S n) is trivial. Similarly for V . U ∩ V is clearly nonempty and contractible, hencepath-connected, so Cor. 59.2 applies and π1(X ) is trivial.

2. Criticize the following “proof” that S 2 is simply connected: Let f be a loop in S 2

based at x0. Choose a point p ∈ S 2 not lying in the image of f . Since S 2\ p ishomeomorphic with R2 and R2 is simply connected, the loop f is path homotopic tothe constant loop.

You may not be able to pick a point p in the image of f . It is possible to havea continuous surjection from I to S 2 (of course, the inverse will not be continuous).Consider a composition with the the Peano map (see Thm. 44.1 on p. 272).

3. (I use the shorthand Rn0 = Rn\{0}.)

(a) Show that R and Rn are not homeomorphic if n > 1.

R0 = (−∞, 0) ∪ (0, ∞) is not connected, but Rn0 is. In fact, it is path connected.

For x, y ∈ Rn0 , use the straight line path xy, unless (1 − t)x + ty = 0 for some

t ∈ [0, 1]. In this case, pick any z ∈ Rn0 that does not lie on the line through x, y

and take the straight line path xz followed by the straight line path zy .Now suppose we had a homeomorphism f : R → Rn. Then the restriction of f to R0 would give a homeomorphism between a connected and a disconnected

set. <

(b) Show that R2 and Rn are not homeomorphic if n > 2.

The solution is similar to the previous problem, but we use simply connectedinstead of connected. For every n, Rn

0 has S n−1 as a deformation retract by

H (x, t) = (1 − t)x + tx/x.

For n = 2 this gives

π1(R20) = π1(S 1) = Z.

25




For n ≥ 3, this gives

π1(Rn0 ) = π1(S n−1) = 0.

Since they have different fundamental groups, they cannot be homeomorphic.This is the essential point of this course, as presented formally (and more

strongly) in Thm. 58.7.

4. Assume the hypotheses of Theorem 59.1.

(a) What can you say about the fundamental group of X if j∗ is the trivial homo-morphism? If both i∗ and j∗ are trivial?

If both are trivial, then the image of each is just the identity element. ByTheorem 59.1, π1(X ) will be the group generated by the identity; but that’s justthe trivial group.If just j∗ is trivial, there’s not much you can say. It is not hard to come up with

cases where U does not have a trivial fundamental group, and neither does X .For example, let

U = {(x , y, z ) ∈ S 2 ... − 1 < x < 0},

so that U is a punctured hemisphere, and let V be S 2 with two punctures, eachof which has positive x-coordinate.

(b) Give an example where i∗ and j∗ are trivial, but neither U nor V have trivialfundamental groups.

Let U be S 2 with two punctures, each of which has negative x-coordinate. Let V

be S 2 with two punctures, each of which has positive x-coordinate. Then each of U , V has a copy of S 1 as a deformation retract, and thus has fundamental groupZ. However, X = U ∪ V is S 2, which has trivial fundamental group. Thus, eachof i∗, j∗ must send every loop class to the identity class (since there’s nothingelse in π1(S 2) to send it to) and is hence a trivial homomorphism.

26




§58. The Fundamental Group of Some Surfaces

1. Compute the fundamental group of the solid torus S 1 × B2 and the product spaceS 1 × S 2.

Applying Thm. 60.1,

π1(S 1 × B2) ∼= π1(S 1) × π1(B2) ∼= Z× {e} ∼= Z, and

π1(S 1 × S 2) ∼= π1(S 1) × π1(S 2) ∼= Z× {e} ∼= Z.

2. Let X be the quotient space obtained from B2 by identifying each point x of S 1 withits antipode −x. Show that X is homeomorphic to the projective plane P 2.

An element of P 2 looks like {x, −x} where x ∈ S 2. Consider the copy of B 2 which

is the projection of S 2

into the xy-plane, and let ∼ be the equivalence relation whichidentifies antipodal points of B2. Define a map f : P 2 → B2/ ∼ by

f ({x, −x}) =

p(x) z (x) ≥ 0

p(−x) z (x) ≤ 0. ,

where p(x) is the orthogonal projection of x onto the xy-plane, and z (x) is the thirdcoordinate of x. In other words, if x ∈ S 2 lies above the xy-plane, then map {x, −x}onto the projection of x, and if x ∈ S 2 lies below the xy-plane, then map {x, −x}onto the projection of −x.

Projection is continuous, so we can use the Pasting Lemma if we verify that f is continuous for x such that z (x) = 0. But for such an x, p(x) = p(−x) by theequivalence relation, since p(x) and p(−x) are antipodal points of S 2.

Now note that f is bijective, by constructing the inverse. For b ∈ B 2/ ∼, let

x = S 2+ ∩ {b + (0, 0, t) ... t ∈ R}

be the point in the upper hemisphere of S 2 which lies in the vertical line through b,and then define

g(b) = {x, −x}.

Now by construction,

f ◦ g(b) = f ({x, −x}) = p(x) = b, and

g ◦ f ({x, −x}) =

g( p(x)) z (x) ≥ 0

g( p(−x)) z (x) ≤ 0 =

{x, −x} z (x) ≥ 0

{−x, x} z (x) ≤ 0 = {−x, x}.

27




Note that B2/ ∼ is compact, by Compactness Mantra (1), and that P 2 is Hausdorff by Thm. 60.3. Now apply the following lemma to f −1, and conclude that f −1 (andhence also f ) is a homeomorphism.

Lemma. If f : X → Y is a continuous bijection with X compact and Y Hausdorff,then f is a homeomorphism.

Proof. We need to show f is open. Let U ⊆ X be open, so that U C is closed. f is aclosed map, by Exercise 2 from the Fundamental Mantras of Compactness, so f (U C )is closed. Now using the Basic Survival Tools #3(d),

f (U C ) = f (X \U ) def complement

= f (X )\f (U ) BST 3(d)

= Y \f (U ) f is onto

= f (U )C ,

so f (U ) is open.

3. Let p : E → X be the map constructed in the proof of Lemma 60.5. Let E be thesubspace of E that is the union of the x-axis and the y-axis. Show that p

E

is not acovering map.

Consider a tiny open disc centered at x0. Its intersection with X is a small open

“X” shape. Its preimage looks like a similar open “X” centered at the origin, aswell as a small open horizontal interval around every point (n, 0) and a small verticalinterval around every point (0, n), where n ∈ Z.

pE

is not a covering map because while these open intervals are open and disjoint,they are not homeomorphic to the “X” in the base space. To see this, note that anymap from an “X” onto an interval is necessarily not injective.

4. The space P 1 and the covering map p : S 1 → P 1 are familiar ones. What are they?

For z ∈ S 1, define

p(z ) = z 2.

Check that this maps antipodal points to the same point:

p(−z ) = (−z )2 = z 2 = p(z ).

We already know this is a covering map, by previous problems (and homework).So P 1 = S 1.

28




5. Consider the covering map indicated in Figure 8. Here, p wraps A1 around A twiceand wraps B1 around B twice; p maps A0 homeomorphically onto A and B, respec-tively. Use this covering space to show that the fundamental group of the figure eightis not abelian.

e0

e2

e1

x 0

B

0

B1 A

1

B0

A

p

f

f

g

g~

~

Figure 8. An alternative cover for the figure-eight.

Let f loop once around A counterclockwise and let g loop once around B counter-clockwise. By Thm. 54.3,

f ∗ g p g ∗ f ⇐⇒ (f ∗ g)∼, (g ∗ f )∼ have the same endpoint.

The lifting of f is f running counterclockwise from e0 to e2, so the lifting of f ∗ glooks like f followed by a counterclockwise loop around B0 ending at e2.

The lifting of g is g running counterclockwise from e0 to e1, so the lifting of g ∗ f looks like g followed by a counterclockwise loop around A0 ending at e1.

Since these liftings have different endpoints, Thm. 54.3 indicates that f ∗ g andg ∗ f are not path homotopic. Hence

[f ] ∗ [g] = [f ∗ g] = [g ∗ f ] = [g] ∗ [f ],

and the fundamental group is not abelian.

Topology - Solutions Sections 51-54

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