§51. Homotopy
of Paths
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, h : X → Y Y are homotopic and k, k : Y → Z Z are homotopic, then k ◦ h k ◦ h . Since h h , we have a homotopy
× I → Y with H (x, 0) = h( H : X × h(x), H (x, 1) = h (x), and since k k , we have a homotopy K : Y × I → Z with K (x, 0) = k( k (x), K (x, 1) = k (x). Then define F : X × × I → Z by F ( F (x, t) =
k(H (x, 2t)), )), 0 ≤ t ≤ 21 K (h (x), 2t − 1), 1), 21 ≤ t ≤ 1. 1 .
Then check that F does F does what it oughta do, at different times t:1 t = 0 t =
1 2
t = 1
F ( F (x, 0) = k( k (H (x, 0)) = k = k((h(x)) = k ◦ h(x)
F x, 21 = k = k((H (x, 1)) = k = k((h (x)) = K K ((h(x), 0) F ( F (x, 1) = K K ((h (x), 1) = k (h (x)) = k ◦ h (x).
2. Given spaces X and Y , Y , let [X, [X, Y ] Y ] denote the set of homotopy classes of maps of X of X into Y into Y . . Let I = [0, [0, 1]. (a) Show that for any X , the set [X, [X, I ] has a single element. Define ϕ˜ : X : X → I to I to be the zero zero map: map: ϕ˜(x) = 0, ∀x ∈ X. X . Let ϕ : X : X → I by I by any continuous map. We show ϕ ϕ˜: define Φ : X : X × × I → I by Φ(x, Φ(x, t) = (1 − t)ϕ(x). This is evidently a homotopy ϕ ϕ˜. Since is an equivalence relation, this shows all maps ϕ maps ϕ : : X X → I are I are equivalent under , i.e., there is only one equivalence class. (b) Show that if Y if Y is is path connected, the set [I [I , Y ] Y ] has a single element. Pick any point p ∈ Y Y and defin definee ϕ˜ : I → Y by ϕ˜(x) = p, so so ϕ˜ is the constant map at p. Now let ϕ : I → Y Y be arbitrary; we will again show ϕ ϕ˜. Denot Denotee ϕ(0) = a = a and a nd ϕ ϕ(1) (1) = b = b.. 1Here, 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 = 21 .
1
Topology (2nd ed.) — James R. Munkres Since Y is Y is path-connected, there is a path γ : I → Y Y with γ (0) (0) = a = a = = ϕ ϕ(0) (0) and γ (1) (1) = p = p.. Now define a homotopy Φ : I : I × × I → Y by Φ(x, Φ(x, t) =
ϕ (1 − 2t)x , 0 ≤ t ≤ 21 1 ≤ t ≤ 1. γ (2t (2t − 1), 1), 1 . 2
Note that Φ does not depend on x on x once t once t ≥ 21 ! This is because Φ is the constant map to the point γ point γ (2t (2t − 1) from this point onwards. Φ has the effect of shrinking the image of ϕ to a point while 0 ≤ t ≤ 21 , then moving that point to p along γ (I ) while 21 ≤ t ≤ 1. Check that Φ does what it oughta do: t=0 1 2
t =
Φ(x, 0) = ϕ( ϕ (x)
Φ x, 21 = ϕ = ϕ(0) (0) = a = a = = γ γ (0) (0)
t=1
Φ(x, 1) = γ (1) (1) = p. = p.
3. A space X space X is is said to be contractible if the identity map i map i X : X → X is is nulhomotopic. (a) Show that I and R are contractible. Define Define the constant constant map ϕ˜(x) = 0, for either either space. space. Then Then we define define the homotopy 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) = id X (x)
t = 1
H (x, 1) = 0. 0.
(b) Show that a contractible space is path connected. Let X be X be a contrac contractib tible le space. Then Then there there is a homotop homotopy y H H between idX and some constant map; call it f . f . So f ( f (x) = p, ∀x ∈ X and H : id X f , f , i.e., H : X × f (x) = p. × I → X with H (x, 0) = id X (x) and H (x, 1) = f ( To show X is X is path connected, we fix any two points y, z ∈ X and X and construct a path betw between them. them. Note 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 ≤ 21 H (z, 2 − 2t), 21 ≤ t ≤ 1. 1 .
And check it: t = 0 t =
1 2
γ (t) = H ( H (y, 0) = id X (y ) = y γ (t) = H ( H (y, 1) = p = p = H H ((z, 1)
2
Solutions by Erin P. J. Pearse t = 1
γ (t) = H (z, 0) = id X (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 id Y and some constant map; call it f . So f (y) = p, ∀y ∈ Y and H : id Y f , i.e., H : Y × I → Y with H (y, 0) = id Y (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 single element. Let X be a contractible space. Then there is a homotopy H between idX and some constant map; call it f . So f (x) = p, ∀x ∈ X and H : id X f , i.e., H : X × I → X with H (x, 0) = id X (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 use the 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.
3
Topology (2nd ed.) — James R. Munkres Now we can set up the requisite homotopy. Φ(x, t) =
t=0
Φ(x, 0) = ϕ(idX (x)) = ϕ(x)
ϕ(H (x, 2t)), 0 ≤ t ≤ 21 1 ≤ t ≤ 1. γ (2t − 1), 2
And check it: t =
1 2
t=1
Φ x, 21 = ϕ( p) = γ (0) Φ(x, 1) = γ (1) = q = ϕ(x). ˜
4
Solutions by Erin P. J. Pearse
§52. The
Fundamental Group
1. A subset A of Rn is star convex iff for some point a 0 ∈ 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 a 0 be the center. (Hence the name “star convex”.) The set {(x, y) ... x = 0 or y = 0} ⊆ R 2 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 is a star convex subset of R2 which is not convex. The convex hull of A (smallest convex 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] consists of single element, by §51, Exercise 3(c), for any space X . In particular, this is true for [S 1 , A]. Now let [S 1 , A]a ⊆ [S 1 , A] be those maps which send at least one point of S 1 to a. Then [S 1 , A]a also consists of a single element. But [S 1 , A]a = π 1 (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 ˆ ◦ ˆ if γ = α ∗ β , then γˆ = β α. ˆ ◦ ˆ ˆ ◦ ˆ We show γˆ = β α by showing that γˆ ([f ]) = (β α)([f ]), for every path f . γˆ ([f ]) = [¯ γ ] ∗ [f ] ∗ [γ ]
= α ∗ β ∗ [f ] ∗ [α ∗ β ] ¯ ∗ ¯ = β α ∗ [f ] ∗ [α ∗ β ] ¯ ∗ [¯ = β α] ∗ [f ] ∗ [α] ∗ [β ]
def of γ
¯ ∗ α([f ]) = β ˆ ∗ [β ] ˆ(α([f ])) = β ˆ
def of αˆ ˆ def of β
def of the reverse α¯ def of ∗ , p.326
ˆ ◦ α)([f ]) = (β ˆ Note that the reverse of α ∗ β (α followed by β ) is the reverse of β followed by the reverse of α, in the third line above.
5
Topology (2nd ed.) — James R. Munkres 3. Let x0 and x1 be points of the path-connected space X . Show that π1 (X, x0) is ˆ. abelian iff for every pair α, β of paths from x0 to x 1 , we have α ˆ = β (⇒) Suppose π 1 (X, x0 ) is abelian, and let α, β be paths from x 0 to x 1 . Since α and ˆ ˆ are both homomorphisms from π 1 (X, x0 ) to π1 (X, x1 ), we need to prove that they β both 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 ˆ an isomorphism, by Cor 52.2 (we are given that X is ˆ. path connected). Now we show α = ˆ β α ˆ ([f ]) = [¯ α] ∗ [f ] ∗ [α]
def of αˆ
= [¯ α] ∗ [α] ∗ [f ]
π1 (X, x1 ) is abelian
= [ex ] ∗ [f ]
Thm 51.2(3)
= [f ] ∗ [ex ]
π1 (X, x1 ) is abelian
= [f ]
Thm 51.2(2)
1
1
¯([f ]) = [f ]. So we have actually proven that α and ˆ are the identity, i.e., Similarly, β ˆ β < 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 an element 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 even defined, since α ends at x1 , and f ends at x0 . Try again: [f ] ∗ [α] = [f ] ∗ [α] ∗ [ex ] ¯ ∗ [β ] = [f ] ∗ [α] ∗ β ¯ ∗ [β ] = [f ] ∗ α ∗ β
Thm 51.2(2)
1
Thm 51.2(3) def of ∗ , p.326
¯ ∗ [f ] ∗ [β ] = α ∗ β ¯ ∗ [f ] ∗ [β ] = [α] ∗ β
π1 (X, x0 ) is abelian def of ∗ , p.326
¯ ∗ [f ] ∗ [β ] [¯ α] ∗ [f ] ∗ [α] = [¯ α] ∗ [α] ∗ β ¯ ∗ [f ] ∗ [β ] [¯ α] ∗ [f ] ∗ [α] = [ex ] ∗ β ¯ ∗ [f ] ∗ [β ] [¯ α] ∗ [f ] ∗ [α] = β 1
ˆ([f ]) α ˆ ([f ]) = β
multiply both sides on the left Thm 51.2(3) Thm 51.2(2) and Thm 51.2(2) ˆ def of α, ˆ β
¯ is a loop, so Whew! Note that the fourth equality above is justified because α ∗ β we avoid the problem of the other (incorrect) solution above.
6
Solutions by Erin P. J. Pearse ˆ for every pair α, β of paths from x 0 to x1 . We must (⇐) Now suppose that α = ˆ β show π1 (X, x0) is abelian, so pick f, g ∈ π 1 (X, x0 ) and show [f ] ∗ [g] = [g] ∗ [f ]. Now we have the freedom to choose our α and β , so define α = f, β = f ∗ g. Then compute:
α ˆ ([f ]) = f ([f ]) ¯ ∗ [f ] ∗ [f ] = [f ]
α = f def of f ˆ
= [ex ] ∗ [f ]
Thm 51.2(3), ex = e x
= [f ]
Thm 51.2(2)
0
0
1
and ˆ([f ]) = f β ∗ g([f ]) = f ∗ g ∗ [f ] ∗ [f ∗ g]
β = f ∗ g def of f ˆ
= [g] ∗ f ∗ [f ] ∗ [f ] ∗ [g]
see Ex 2, at the end
= [g] ∗ [ex ] ∗ [f ] ∗ [g]
Thm 51.2(3), e x = e x
= [g] ∗ [f ] ∗ [g]
Thm 51.2(2).
0
0
1
ˆ, we know α ˆ([f ]), and so Since α = ˆ β ˆ ([f ]) = β [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 you find it? Hint: remember the old tricks of “adding 0” or “multiplying by 1” for the forward 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 that r(a) = a for each a ∈ A. If a0 ∈ A, show that r∗ : π 1 (X, a0 ) → π 1 (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 , we can let g = f . Then r ◦ f (t) = r(f (t)) = f (t), since f (t) ∈ A and r is the identity on A.
7
∀t ∈ I ,
Topology (2nd ed.) — James R. Munkres 5. Let A ∈ Rn and let h : (A, a0 ) → (Y, y0). Show that if h is extendable to a continuous map 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 0, ∀t. Then define Φ : Rn × I → Y by → Y by ϕ(t) 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) = y 0 = 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) + ta 0 ) 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. More precisely, let h : X → Y be continuous, with h(x0 ) = y 0 and h(x1 ) = y 1 . Let α be a path in X from x0 to x 1 , and let β = h ◦ α. Show that ˆ ◦ (hx )∗ = (hx )∗ ◦ α, β ˆ 0
1
so that the diagram commutes: (hx0 )
∗
π1 (X, x0 ) −−−−−−−→ π1 (Y, y0)
α ˆ
(hx1 )
∗
ˆ β
π1 (X, x1 ) −−−−−−−→ π1 (Y, y1) We need to see that these two operations do the same thing to any [f ] ∈ π 1 (X, x0 ), so let f be a loop in X based at x 0 . On the one hand, we have ˆ ◦ (hx )∗([f ]) = h β def of β ◦ α ◦ (hx )∗ ([f ]) 0
0
= h ◦ α([h ◦ f ]) = h ◦ α ∗ [h ◦ f ] ∗ [h ◦ α]
def of h∗ ˆ def of β
= h ◦ α ∗ [h ◦ f ] ∗ [h ◦ α]
ˆ def of β.
8
Solutions by Erin P. J. Pearse On the other hand, we have (hx )∗ ◦ α([f ]) ˆ = (hx )∗ ([¯ α] ∗ [f ] ∗ [α]) 1
def of α ˆ
1
= (hx )∗ ([¯ 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 ∗ .
1
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).
9
Topology (2nd ed.) — James R. Munkres
§53.
Covering Spaces
1. Let Y have the discrete topology. Show that if p : X × Y → X is projection on the first 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 is evenly covered by p; that is, that p−1 (U ) can be written as a union of disjoint sets V 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
−1
p (U ) =
U × {y }.
y ∈Y
Note: this union is disjoint because (x, y) ∈ U × {y1 } ∩ U × {y2}
=⇒
y = y 1 = y 2
=⇒
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 B that 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 with U . 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 bα ∈ V β and define f (α0 ) = β 0 . There will be such a V β , because E = ∪ V β , and that V β will be unique by the disjointness of the V β . Now to see that this is a bijection. For b β ∈ V β ⊆ E , ∃ α such that b β ∈ V α because V α contains E . This shows surjectivity. For injectivity, note that 0
0
bβ ∈ V α ∩ V α 1
=⇒
1
by the disjointness of the partition.
10
0
0
α1 = α 2
Solutions by Erin P. J. Pearse 3. Let p : E → B be a covering map; let B be connected. Show that if p−1(b0 ) has k elements for some b0 ∈ B, then p−1(b) has k elements for every b ∈ B, i.e., E is a k-fold covering of B. Since | p−1 (b0 )| = k, we can find U such that k
−1
p (U ) =
V i
p
and
i=1
Vi
= p i : V i → U is a homeomorphism, ∀i.
k, and we will contradict the fact We assume that ∃b1 such that | p−1(b1 )| = j = that B is connected. Define C = { b ... | p−1(b)| = k }
D = { b ... | p−1(b)| = k }.
and
We have b0 ∈ C , so C = = ∅. Also, it is clear that C ∩ D = ∅ ∅, and b1 ∈ D, so 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 ) = ki=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 < Hence no such b exists. open. This gives C, D as a disconnection of B. 1
4. Let q : X → Y and r : Y → Z be covering maps; let p = r ◦ q . Show that if r −1(z ) is finite 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 such that p−1(Z ) = ni=1 V i where V i is open in Y and r V : 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 Ai where Ai is open in X and q A : A i → U i is a homeomorphism. With so many sets, we need to find some common ground, so define
iα
α
i
α
α
n
C =
r(U i ∩ V i ).
i=1
Each U i ∩ V i is a neighbourhood of vi in Y which is evenly covered by q , and has r(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 Di = q −1(U i ∩ V i ) ∩ Ai , α
α
which are also open. Now C is a neighbourhood of z ∈ Z for which −1
p (C ) =
i,α
Di
α
and
p
Diα
= (r ◦ q )
Also, the D i are disjoint because the Ai are. α
α
11
Diα
is a homeomorphism.
Topology (2nd ed.) — James R. Munkres 5. Show that the map p : S 1 → S 1 given by p(z ) = z 2 is a covering map. Generalize to the map p(z ) = z n . Here we consider S 1 ⊆ C, so that for z ∈ S 1 we may write z = eiθ
p(z ) = z 2 = e2iθ .
and
For z ∈ S 1, let z = e iθ so θ = arg(z ) ∈ [0, 2π). Let U be the image of θ − π2 , θ + π2 e iθ so that U is the open semicircle centered at z . Then p−1 (U ) under the map θ → consists of the “quarter circle” centered at z
V 1 = exp θ − π4 , θ +
π
4
and the “quarter circle” centered at −z
V 2 = exp −θ + π4 , −θ −
π
4
.
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 ) = 1 of the way around the circle S 1 . 2n
n i=1
V i , where each V i goes
If p : E → B is a covering map, show that p −1(b) ⊆ E has the discrete topology, for any b ∈ B. Bonus Problem:
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., −1
p (U ) =
V α, where the V α are open.
α∈A
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 } as an intersection of an open set V α with the subspace p−1 (b) ⊆ E , which shows that {xα } is open in the subspace topology of p−1 (b). Since −1
p (b) =
α∈A
this shows p−1 (b) is discrete.
12
{xα },
Solutions by Erin P. J. Pearse
§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) and is evenly covered by p; p is not a covering map. ˜ the proof of Lemma 54.2, why were we so careful about the 2. In defining the map F in order 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 ˜ be liftings of them such that α ˜( 0). Show that α˜ ∗ β is ˜ a lifting of α ˜ and β ˜ (1) = β α ∗ β . ˜) = α ∗ β . By definition, p ◦ (α ˜) : I → E by We need to show p ◦ (α ˜ ∗ β ˜ ∗ β
p ◦ ˜ α(2t), 0 ≤ t ≤ 21 , ˜ p ◦ (˜ α ∗ β )(t) = ˜(2t − 1), 1 ≤ t ≤ 1, p ◦ β 2 =
α(2t), 0 ≤ t ≤ 21 , β (2t − 1), 21 ≤ t ≤ 1,
= α ∗ β (t)
def of ∗ ˜ def of α, ˜ β def of ∗.
13
Topology (2nd ed.) — James R. Munkres 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. t
2 ~
f
f
1
-1
p
0
1
Figure 1. Three
0
1
2
x
˜ the one in the centre. liftings of f . f is
t g
2
~
1
-1
g
0
p
1
Figure 2. Three
-2
-1
0
1
2
x
liftings of g. g˜ is the one in the centre.
t g
2 ~
f
-1
f
~
1
g
0
p
1
use the shorthand
2
2
R0 = R
-1
0
1
x
Figure 3.
2I
-2
f ˜ ∗ g˜ is a lifting of f ∗ g.
\{(0, 0)}.
14
2
Solutions by Erin P. J. Pearse 5. Consider the covering map p × p : R × R → S 1 × S 1 of Example 4 of §53. Consider the 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 when S 1 × S 1 is identified with the doughnut surface D. Find a lifting f ˜ of f to R × R, and sketch it.
2
2π
~
p
f 1
=
S 1
f
f 0
1
2
0
S 1
2π
S 1 × S 1
T2 = S 1 × S 1 is represented two different ways on the right. The path (loop) f begins at the white dot and traverses T2 . For visualization, the Figure 4.
first S 1 axis is sketched on top of the doughnut, the second S 1 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 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 the induced homomorphisms g∗ , h∗ of the infinite cyclic group π 1 (S 1 , b0 ) into itself. The group is cyclic, so it suffices to determine the image of its generator, γ (t) = e , t ∈ I under g. Since 2πit
g ◦ γ (t) = g e2πit = e2πit
n
= e 2πint
is a loop which goes n times around the circle, in the direction of γ , we have g∗([γ ]) = [g ◦ γ ] = [γ ] ∗ · · · ∗ [γ ] = [γ ]∗n
g : γ (t) → γ (nt).
or
n times
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 ] = [γ ]∗n = [γ ]∗(−n) n times
15
or
h : γ (t) → γ (−nt).
Topology (2nd ed.) — James R. Munkres 7. Generalize the proof of Theorem 54.5 to show that the fundamental group of the torus 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 by Thm 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 Thm 54.4), so we just need to show φ is a homomorphism. ˜ and g˜ be their respective liftings to paths in R2 Given [f ], [g] ∈ π1 (T2 , b0 ), let f ˜ and ( j, k) = g˜(1). Since these points are beginning at e0 = (0, 0). Let (m, n) = f (1) 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 P ◦ g˜˜(t) = P ((m, n) + g˜(t)) = ( p(m + g˜1 (t)), p(n + g˜2(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 the formula for p, and the last line follows because g˜ is a lifting of g. Then the product f ˜ ∗ ˜g˜ is a well-defined path, and it is the lifting of f ∗ g that begins at (0, 0). It is clear 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 ≤ 21 , P (g˜˜(2t − 1)), 21 ≤ t ≤ 1,
def of ∗
f (2t), 0 ≤ t ≤ 21 , g(2t − 1)), 21 ≤ 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 b 0 , 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.
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Solutions by Erin P. J. Pearse 8. Let p : E → B be a covering map, with E path connected. Show that if B is simply connected, 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 by hypothesis, 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 also consists of only one element. Thus by Ex 53.3, E is a 1-fold covering of B, i.e., E and B are homeomorphic.
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Topology (2nd ed.) — James R. Munkres
§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 B 2 , then every continuous map f : A → A has a fixed point. If A is a retract, then A ⊆ B 2 and there is a retraction i.e., a continuous map r : B 2 → A with r(a) = a, ∀a ∈ A. Let j : A → B 2 by inclusion. Then j ◦ f ◦ r : B 2 → B 2 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)
r A is the identity j is inclusion.
= f (c) 2. Suppose h : S 1 → S 1 is nulhomotopic. (a) Show that h has a fixed point.
h extends to k : B 2 → S 1 by Lemma 55.3 with X = S 1 . But S 1 ⊆ B 2, so we can actually consider k as a mapping k : B 2 → B 2. Then k must have a fixed point by Brouwer’s Theorem (55.6), call it b. Since Im(k) ⊆ S 1 , k(b) = b implies that b ∈ S 1 . Thus h(b) is defined. In fact, as an extension, k = h on S 1 b is a fixed point of k.
h(b) = k(b) = b
(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 it must 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 has a positive real eigenvalue. Following Cor. 55.7, let B = S 2 ∩ O1 ⊆ R3 . All the components of x ∈ B are nonnegative, and x = 0. Since A is nonsingular, Ax = 0, and the map S : B → B
by S (x) = Ax/Ax
18
Solutions by Erin P. J. Pearse 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 = Ax0 x0 ,
i.e., A has an eigenvector x0 with corresponding eigenvalue Ax0 > 0. 4. Assume: for each n, there is no retraction r : B n+1 → S n . (a) The identity map i : S n → S n is not nulhomotopic. Suppose i = idS 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 a constant map. Define π : S n × I → B n+1 by n
π(x, t) = (1 − t)x. Then π is a quotient map and H induces a continuous map k : B n+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 B n+1 points directly outward at some point, and directly inward at some point. (d) Every continuous map f : B n+1 → B n+1 has a fixed point. (e) Every (n + 1) × (n + 1) matrix with positive real entries has a positive real eigenvalue. (f) If h : S n → S n is nulhomotopic, then h has a fixed point, and h maps some point x to its antipode −x.
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Topology (2nd ed.) — James R. Munkres
§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 is the 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 keeps every 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. Define F : X × I → X by F (x, t) =
H (x, 2t), 0 ≤ t ≤ 21 , H (x, 0) = id X , H (x, 1) = r(x), K (r(x), 2t − 1), 21 ≤ t ≤ 1, K (x, 0) = id A , 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) ∈ B 2 × 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 (B 2 × S 1 ) = π1 (S 1 ) = Z. (b) The torus with a point removed. I 1 \{( 12 , 21 )} has its boundary as a deformation retract, by the straight-line homotopy (from ( 12 , 21 )). Under the quotient map which makes I 2 into the torus, ∂ I 1 \{( 12 , 21 )} 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
Solutions by Erin P. J. Pearse (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
β γ α
x 0
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 straightline homotopy H (x, t) = (1 − t)x + 2tx/x. 2S 1 is obviously homeomorphic to S 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 π 1 ((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)
Figure 6. The
S 1 ∪ (R×0)
spaces in exercises (i)-(l).
(i) S 1 ∪ (R+ × 0). π1 ((i)) is trivial by H (x, t) = (1 − t)x + tx/x.
21
R2 − (R+×0)
Topology (2nd ed.) — James R. Munkres (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 the figure 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 an equivalence relation on C . We show the three properties. (i) Let f = g = id X : X → X . Then f ◦ g, g ◦ f are both homotopic to the identity on 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 id X and f ◦ g id Y . 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 need to show that h ◦ f : X → Z and g ◦ k : Z → X are homotopy inverses of each other. By associativity, (g ◦ k) ◦ (h ◦ f ) = g ◦ (k ◦ h) ◦ f. Then h, k are homotopy inverses, so k ◦ h id Y and g ◦ (k ◦ h) g ◦ idY
§51 Exercise 1 §51 Exercise 1 again by trivial homotopy Lemma 51.1 (transitivity).
g ◦ (k ◦ h) ◦ f g ◦ idY ◦ f (x)
g ◦ f (x) id X (x)
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 dotted line to the vertical bar of the theta. Note: it is not 1-1, so it cannot have an inverse. g
22
Solutions by Erin P. J. Pearse maps all the points of the vertical bar to the “centre point” of the figure-eight where the 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 between idX and g ◦ f which maps the figure eight to itself, but continuously scrunches all the points inside the dotted line down into the centre point, as t goes from 0 to 1. Similarly, there is a homotopy from id Y to f ◦ g which acts by collapsing the vertical bar to the centre point of the vertical bar, and continuously slurping in part of the boundary of the circle to replace the bar. 5. Recall that a space X is contractible iff the identity map id X is nulhomotopic. Show that 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) = x 0,
∀x ∈ X.
Suppose r : X → A is a retraction, so that r A = id A . Since it doesn’t matter what point x 0 we chose above (X is path connected by Exercise 51.3(b)), let x 0 ∈ A. Then H A×I defines a homotopy from idA to x 0 .
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 ˆ ) so that γ (x0 ) generates π1 (S 1, x0). Given h : S 1 → S 1 , choose x0 ∈ S 1 and let h(x0 ) = x1 . Consider the homomorphism h∗ : π 1(S 1 , x0 ) → π 1 (S 1, x1 ). Then h ∗ (γ (x0 )) = d · γ (x1 ) for some d ∈ Z. The integer d = deg h is the degree of h.
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Topology (2nd ed.) — James R. Munkres (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 map g : S 1 → S 1 to be rotation by −θ. Suppose x0 , y0 ∈ S 1 with h(x0 ) = x1 and h(y0) = y 1, and that α : b0 x0 and β : 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 map ρ(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 reflection map ρ(x1 , . . . , xn , xn+1 ) = (x1 , . . . , xn , −xn+1 ) has degree −1. Then prove the following: (a) There is no retraction r : B n+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 its antipode −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
Solutions by Erin P. J. Pearse
§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 is the fundamental group of X ? Consider X as sitting in R3 along the x-axis, for purposes of description. The projection of X to the x-axis would then be [−2, 2], with the single point in common projecting 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 by collapsing 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, hence path-connected, so Cor. 59.2 applies and π 1 (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 is homeomorphic with R2 and R2 is simply connected, the loop f is path homotopic to the constant loop. You may not be able to pick a point p in the image of f . It is possible to have a 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 R n0 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 connected instead of connected. For every n, Rn0 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
Topology (2nd ed.) — James R. Munkres For n ≥ 3, this gives π1 (Rn0 ) = π 1 (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 homomorphism? If both i∗ and j∗ are trivial? If both are trivial, then the image of each is just the identity element. By Theorem 59.1, π 1 (X ) will be the group generated by the identity; but that’s just the 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, each of which has positive x-coordinate. (b) Give an example where i∗ and j∗ are trivial, but neither U nor V have trivial fundamental 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 group Z. However, X = U ∪ V is S 2 , which has trivial fundamental group. Thus, each of i∗, j∗ must send every loop class to the identity class (since there’s nothing else in π1 (S 2 ) to send it to) and is hence a trivial homomorphism.
26
Solutions by Erin P. J. Pearse
§58. The
Fundamental Group of Some Surfaces
1. Compute the fundamental group of the solid torus S 1 × B 2 and the product space S 1 × S 2. Applying Thm. 60.1, π1 (S 1 × B 2) ∼ = π 1 (S 1) × π1 (B 2 ) ∼ = Z × {e} ∼ = Z, and π1 (S 1 × S 2) ∼ = π 1 (S 1) × π1 (S 2) ∼ = Z × {e} ∼ = Z. 2. Let X be the quotient space obtained from B 2 by identifying each point x of S 1 with its 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 which identifies antipodal points of B 2 . Define a map f : P 2 → B 2 / ∼ 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 third coordinate 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 the equivalence 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,
g ◦ f ({x, −x}) =
g( p(x)) z (x) ≥ 0 = g( p(−x)) z (x) ≤ 0
27
and
{x, −x} z (x) ≥ 0 = {−x, x}. {−x, x} z (x) ≤ 0
Topology (2nd ed.) — James R. Munkres Note that B 2 / ∼ 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 (and hence also f ) is a homeomorphism. If f : X → Y is a continuous bijection with X compact and Y Hausdorff, then f is a homeomorphism. Lemma.
Proof. We need to show f is open. Let U ⊆ X be open, so that U C is closed. f is a
closed 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 the subspace of E that is the union of the x-axis and the y-axis. Show that p E is not a covering 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, as well as a small open horizontal interval around every point (n, 0) and a small vertical interval around every point (0, n), where n ∈ Z. p E 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 any map 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.
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Solutions by Erin P. J. Pearse 5. Consider the covering map indicated in Figure 8. Here, p wraps A1 around A twice and wraps B1 around B twice; p maps A0 homeomorphically onto A and B, respectively. Use this covering space to show that the fundamental group of the figure eight is not abelian. g ~
B
0
A
1
e 1
e 0
1
e 2
B0
~
f
p g B
Figure 8. An
f x 0
A
alternative cover for the figure-eight.
Let f loop once around A counterclockwise and let g loop once around B counterclockwise. By Thm. 54.3,
⇐⇒ (f ∗ g)∼ , (g ∗ f )∼ have the same endpoint. ˜ The lifting of f is f running counterclockwise from e0 to e 2 , so the lifting of f ∗ g ˜ looks 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 e 1. Since these liftings have different endpoints, Thm. 54.3 indicates that f ∗ g and g ∗ f are not path homotopic. Hence f ∗ g p g ∗ f
[f ] ∗ [g] = [f ∗ g] = [g ∗ f ] = [g] ∗ [f ], and the fundamental group is not abelian.
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