Topology - Solutions Sections 51-54

§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, h0 : X → Y are homotopic and k, k 0 : Y → Z are homotopic, then k ◦ h ' k 0 ◦ 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 ' k 0 , we have a homotopy K :Y ×I →Z with K(x, 0) = k(x), K(x, 1) = k 0 (x). Then define F : X × I → Z by ( k(H(x, 2t)), 0 ≤ t ≤ 21 F (x, t) = K(h0 (x), 2t − 1), 21 ≤ t ≤ 1. Then check that F does what it oughta do, at different times t:1 t=0 t= 1 2 t=1 F (x, 0) = k(H(x, 0)) = k(h(x)) = k ◦ h(x)
F x, 12 = k(H(x, 1)) = k(h0 (x)) = K(h0 (x), 0) F (x, 1) = K(h0 (x), 1) = k 0 (h0 (x)) = k 0 ◦ h0 (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 any continuous 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 equivalent under ', 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 constant map at p. Now let ϕ : I → Y be arbitrary; we will again show ϕ ' ϕ. ˜ Denote ϕ(0) = a and ϕ(1) = 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 = 12 . 1 Topology (2nd ed.) — James R. Munkres Since Y is path-connected, there is a path γ : I → Y with γ(0) = a = ϕ(0) and γ(1) = p. Now define a homotopy Φ : I × I → Y by (
ϕ (1 − 2t)x , 0 ≤ t ≤ 12 Φ(x, t) = 1 γ(2t − 1), ≤ t ≤ 1. 2 Note that Φ does not depend on x once t ≥ 12 ! This is because Φ is the constant map to the point γ(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 t= 1 2 t=1 Φ(x, 0) = ϕ(x)
Φ x, 12 = ϕ(0) = a = γ(0) 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 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) = 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 and some 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 a path 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 ( H(y, 2t), 0 ≤ t ≤ 21 γ(t) = H(z, 2 − 2t), 12 ≤ t ≤ 1. And check it: t=0 t= 1 2 γ(t) = H(y, 0) = idX (y) = y γ(t) = H(y, 1) = p = H(z, 1) 2 J. γ(1) = q.. 1)) = f (x) = p. with γ(0) = ϕ(p). So f (x) = p. call it f . call it f . if Y is contractible. Pick some other point q ∈ Y and define a constant map ϕ˜ : Y → Y by ϕ(y) ˜ = q. 3 . ˜ The plan is: use the contractibility of X to shrink it to a point (at p). ∀x ∈ X and H : idX ' f . then for any X. 1) = H(ϕ(x). ˜ so that all maps from X to Y are homotopic to ϕ. Y ] has a single element. Let X be a contractible space. Pearse t=1 γ(t) = H(z. So f (y) = p. i. ∀y ∈ Y . Take any arbitrary map ϕ : X → Y . t) = H(ϕ(x). 0)) = ϕ(x) t=1 Φ(x. So every map ϕ : X → Y is homotopic to the constant map at p. 1) = f (x) = p. (d) Show that if X is contractible and Y is path connected. t)).. Just as above. the set [X. then use the path-connectedness of Y to move ϕ(p) (which is in Y ) to q ∈ Y . We will show ϕ ' ϕ. 0) = H(ϕ(x). Then t=0 Φ(x. H :Y ×I →Y with H(y. ∀y ∈ Y and H : idY ' f . then we have a homotopy H between idY and some constant map.Solutions by Erin P.e. 0) = idY (y) and H(y.e. i. 0) = idX (x) and H(x. then [X. Then there is a homotopy H between idX and some constant map. By the path-connectedness of Y we have one: γ : I → Y. Take any arbitrary map ϕ : X → Y and define Φ(x. So we need a path γ from ϕ(p) to q. Y ] has a single element. 0) = idX (z) = z. 1) = f (y) = p. H :X ×I →X with H(x. (c) Show that if Y is contractible. 0) = ϕ(idX (x)) = ϕ(x)
Φ x. 1) = γ(1) = q = ϕ(x).) — James R. t) = 1 γ(2t − 1). 2 And check it: t=0 t= 1 2 t=1 Φ(x.Topology (2nd ed. 2t)). ( ϕ(H(x. ≤ t ≤ 1. 12 = ϕ(p) = γ(0) Φ(x. 0 ≤ t ≤ 12 Φ(x. ˜ 4 . Munkres Now we can set up the requisite homotopy. for any space X. Let α be a path in X from x0 to x1 . The convex hull of A (smallest convex set containing A. p. A subset A of Rn is star convex iff for some point a0 ∈ A.. Thus. A]a ⊆ [S 1 . Then [S 1 . i. A] be those maps which send at least one point of S 1 to a. y) . (Hence the name “star convex”.. A] consists of single element. Pearse §52. J. (a) Find a star convex set that is not convex. A six-pointed star like the Star of David. In particular. But [S 1 . The Fundamental Group 1. Or let I 2 = I × I ⊆ R2 and let X = {(x. by §51. (1 − λ)a + λa0 ∈ A. or intersection of all convex sets containing A) is conv(A) = {(x.e. A]. this is true for [S 1 .326 def of α ˆ def of βˆ Note that the reverse of α ∗ β (α followed by β) is the reverse of β followed by the reverse of α. y = 0} ⊆ R2 .. A]a = π1 (A. ˆ We show γˆ = βˆ ◦ α ˆ by showing that γˆ ([f ]) = (βˆ ◦ α)([f ˆ ]). Let a ∈ A be a point satisfying the definition of star convexity. γˆ([f ]) = [¯ γ ] ∗ [f ] ∗ [γ]   = α ∗ β ∗ [f ] ∗ [α ∗ β]   = β¯ ∗ α ¯ ∗ [f ] ∗ [α ∗ β]   = β¯ ∗ [α] ¯ ∗ [f ] ∗ [α] ∗ [β]   = β¯ ∗ α([f ˆ ]) ∗ [β] ˆ α([f = β( ˆ ])) = (βˆ ◦ α)([f ˆ ]) def of γ def of the reverse α ¯ def of ∗. x0 )! So π1 (A. Show that if γ = α ∗ β. in the third line above. all the line segments joining a0 to other points of A lie in A. let β be a path in X from x1 to x2 . ∀λ ∈ (0. then γˆ = βˆ ◦ α. y) . 0 ≤ y < 1} ∪ X. A is simply connected.) The set {(x. Now let [S 1 . Exercise 3(c).. A]a also consists of a single element.Solutions by Erin P.. 2. t) = (1 − t)x + ta shows A is contractible (via the straight line homotopy).. x = 0 or y = 0} ⊆ R2 is star convex with respect to the origin. y) . 5 . [X. Then A = I 2 ∪ X is a star convex subset of R2 which is not convex. 1). Then H : A×I → A by H(x. x0 ) is trivial. for every path f . (b) Show that if A is star convex. or a pentacle will work if you let a0 be the center.. and f ends at x0 .2 (we are given that X is ˆ path connected). x0 ) to π1 (X. β([f ˆ and βˆ are the identity.. x0 ) is abelian.2(3) Thm 51.e. Try again: [f ] ∗ [α] = [f ] ∗ [α] ∗ [ex1 ]   = [f ] ∗ [α] ∗ β¯ ∗ [β]   = [f ] ∗ α ∗ β¯ ∗ [β]   = α ∗ β¯ ∗ [f ] ∗ [β]   = [α] ∗ β¯ ∗ [f ] ∗ [β]   [α] ¯ ∗ [f ] ∗ [α] = [α] ¯ ∗ [α] ∗ β¯ ∗ [f ] ∗ [β]   [α] ¯ ∗ [f ] ∗ [α] = [ex1 ] ∗ β¯ ∗ [f ] ∗ [β]   [α] ¯ ∗ [f ] ∗ [α] = β¯ ∗ [f ] ∗ [β] Thm 51. ˆ βˆ   Whew! Note that the fourth equality above is justified because α ∗ β¯ is a loop. β be paths from x0 to x1 .2(3) def of ∗. β of paths from x0 to x1 .2(2) ˆ ]) α ˆ ([f ]) = β([f def of α. The expression [α] ∗ [f ].2(2) and Thm 51. x1 ) must also be abelian. this is all completely wrong! Why? It makes no sense: [α] is not an element of π1 (X.) — James R. x1 ). Of course. Let x0 and x1 be points of the path-connected space X. since α ends at x1 . Now we show α ˆ = β. (⇒) Suppose π1 (X.2(2) Thm 51. i. and let α. x0 ) is abelian def of ∗. we have α ˆ = β. x0 ) is ˆ abelian iff for every pair α. α([f ˆ ]) = [α] ¯ ∗ [f ] ∗ [α] def of α ˆ = [α] ¯ ∗ [α] ∗ [f ] π1 (X. x1 ) is abelian = [f ] Thm 51. p. so we avoid the problem of the other (incorrect) solution above. it must also be simply connected. . Since α ˆ and ˆ β are both homomorphisms from π1 (X. x1 ) is abelian = [ex1 ] ∗ [f ] Thm 51. which appears in line 2 isn’t even defined. x0 ) to the same loop in π1 (X. x0 ) is abelian doesn’t apply. So we have actually proven that α Similarly.2(2) ¯ ]) = [f ]. < that if X is path connected.2(3) = [f ] ∗ [ex1 ] π1 (X. since α ˆ is an isomorphism. we need to prove that they both send a loop f ∈ π1 (X. the fact that π1 (X. x1 ). Note that π1 (X. 6 . Show that π1 (X. Munkres 3.Topology (2nd ed. p.326 multiply both sides on the left Thm 51. by Cor 52.326 π1 (X. x1 ) because it isn’t even a loop! Therefore. ex0 = ex1 = [f ] Thm 51. β = f ∗ g.2(3). x0 ) is abelian.e. Now we have the freedom to choose our α and β. such that r∗ ([g]) = [r ◦ g] = [f ]. based at x0 .2(2) and ˆ ]) = f[ β([f ∗ g([f ])   = f ∗ g ∗ [f ] ∗ [f ∗ g]   = [g] ∗ f ∗ [f ] ∗ [f ] ∗ [g] β =f ∗g def of fˆ see Ex 2. Let f : I → A be a loop in A. β of paths from x0 to x1 . 7 ∀t ∈ I. so define α = f. We must show π1 (X. We must find a loop g in X. a continuous map such that r(a) = a for each a ∈ A. show that r∗ : π1 (X. There is actually a much simpler way to do each direction of this proof. a0 ) → π1 (A. i. so pick f.. Let A ⊆ X. ˆ we know α ˆ ]). 4. since f (t) ∈ A and r is the identity on A. based at x0 . Then compute: α ˆ ([f ]) = fb([f ]) = [f¯] ∗ [f ] ∗ [f ] α=f def of fˆ = [ex0 ] ∗ [f ] Thm 51. we can let g = f . ˆ ([f ]) = β([f [f ] = [g] ∗ [f ] ∗ [g] [g] ∗ [f ] = [g] ∗ [g] ∗ [f ] ∗ [g] [g] ∗ [f ] = [f ] ∗ [g]. Can you find it? Hint: remember the old tricks of “adding 0” or “multiplying by 1” for the forward direction. x0 ) is abelian. So π1 (X. Then r ◦ f (t) = r(f (t)) = f (t). a0 ) is surjective. based at x0 . suppose r : X → A is a retraction.3) to cancel the g’s. at the end = [g] ∗ [ex0 ] ∗ [f ] ∗ [g] Thm 51. and so Since α ˆ = β. β for the backward direction. and choose a more clever α. Because f is also a loop in X. using Thm 51. ex0 = ex1 = [g] ∗ [f ] ∗ [g] Thm 51.2(2). x0 ) and show [f ] ∗ [g] = [g] ∗ [f ]. Pearse (⇐) Now suppose that α ˆ = βˆ for every pair α. . J.Solutions by Erin P. g ∈ π1 (X. If a0 ∈ A.2(2.2(3). ˜ Note: we don’t know that h((1 − t)f (x) + ta0 ) is defined. Show that βˆ ◦ (hx )∗ = (hx )∗ ◦ α. x0 ). Show that if X is path connected. Let α be a path in X from x0 to x1 . On the one hand. x0 ) −−−−− −−→ π1 (Y. Then define Φ : Rn × I → Y by Φ(x. x1 ) −−−−− −−→ π1 (Y. with h(x0 ) = y0 and h(x1 ) = y1 .Topology (2nd ed. y1 ) We need to see that these two operations do the same thing to any [f ] ∈ π1 (X. t) = g((1 − t)f (x) + ta0 ). let h : X → Y be continuous. y0 ) such that g(a) = a whenever a ∈ A. since (1 − t)f (x) + ta0 is just some point between f (x) ∈ A and a0 . 0) = g(f (x)) = g ◦ f (x) = h ◦ f (x) t=1 Φ(x. then we would have h∗ ([f ]) = [h ◦ f ] = [ey0 ]. the homomorphism induced by a continuous map is independent of a base point. Define ϕ˜ : I → Y by ϕ(t) ˜ = y0 . y0 )    ˆ α ˆy yβ (hx )∗ 1 π1 (X. a0 ) so that f : I → A is a loop at a0 . The extendability of h says that we can find a continuous map g : (Rn . Strategy: if h ◦ f were nulhomotopic. we have βˆ ◦ (hx )∗ ([f ]) = h[ ◦ α ◦ (hx )∗ ([f ]) def of β 0 0 = h[ ◦ α([h ◦ f ])   = h ◦ α ∗ [h ◦ f ] ∗ [h ◦ α]   = h ◦ α ∗ [h ◦ f ] ∗ [h ◦ α] 8 def of h∗ def of βˆ ˆ def of β. we konw that g((1 − t)f (x) + ta0 ) is defined. Then (g ◦ f )(t) = g(f (t)) = h(f (t)) = (h ◦ f )(t). ∀t ∈ I. 6. So we show h ◦ f to be nulhomotopic. Let A ∈ Rn and let h : (A. so let f be a loop in X based at x0 . y0 ). Munkres 5. y0 ). Show that if h is extendable to a continuous map of Rn into Y . Let [f ] ∈ π1 (A. ∀f.) — James R. and consider h∗ ([f ]) = [h ◦ f ] ∈ π1 (Y. up to isomorphisms of the groups involved. More precisely. and check t=0 Φ(x. since f (t) ∈ A and g = h on A. then h∗ is the trivial homomorphism. ∀t. 1) = g(a0 ) = y0 = ϕ(x). ˆ 0 1 so that the diagram commutes: (hx )∗ 0 π1 (X. and let β = h ◦ α. a0 ) → (Y. so that h∗ is trivial. but since h is extendible to a continuous map on all of Rn . . a0 ) → (Y. Pearse On the other hand.   So we still need h ◦ α = [h ◦ α]. ¯ 9 . J. ¯ But this is true: h ◦ α(t) = (h ◦ α)(1 − t) = h(α(1 − t)) = h(α(t)) ¯ = h ◦ α(t).Solutions by Erin P. we have (hx1 )∗ ◦ α([f ˆ ]) = (hx1 )∗ ([¯ α] ∗ [f ] ∗ [α]) def of α ˆ = (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 ∗. We will show that U is evenly covered by p.Topology (2nd ed. that is. Covering Spaces 1. 107– 110).) — James R. Let Y have the discrete topology. then p is a covering map. It is clear that p is continuous and surjective (if you have doubts. read pp. Pick x ∈ X and let U be a neighbourhood of x. Munkres §53. that p−1 . Show that if p : X × Y → X is projection on the first coordinate. (U ) can be written as a union of disjoint sets Vα ⊆ X × Y such that for each α. p. Note that every Vα . For bβ ∈ Vβ ⊆ E.α : Vα → U is a homeomorphism. There will be such a Vβ0 . For injectivity. Show that if U is connected. Suppose we have two partitions of p−1 (U ) into slices: A = {Vα }α∈A and B = {Vβ }β∈B . Since Y is discrete. Fix b ∈ B. note that b β ∈ V α1 ∩ V α1 =⇒ by the disjointness of the partition.e. Then we show p−1 (U ) = G U × {y}. i. Suppose that U is an open set of B that is evenly covered by p. S Now to see that this is a bijection. by homeomorphism with U . Let x ∈ p−1 (U ) 2. ∃α such that bβ ∈ Vα because Vα contains E. We will show that there is a bijection between these partitions. Vβ is connected. Let p : E → B be continuous and surjective. Then for any α. y∈Y Note: this union is disjoint because (x. Define f : A → B as follows: find the unique β0 such that bα0 ∈ Vβ0 and define f (α0 ) = β0 . Fix α0 . and for any β. because E = ∪Vβ . and that Vβ0 will be unique by the disjointness of the Vβ . {y} is open in Y and so U × {y} is open in X × Y by the def of product topology. 10 α 1 = α2 . A is actually just a reindexing of B. we can find the unique point bβ ∈ Vβ such that p(bβ ) = b.. This shows surjectivity. y) ∈ U × {y1 } ∩ U × {y2 } =⇒ y = y 1 = y2 =⇒ U × {y1 } = U × {y2 }. then the partition of p−1 (U ) into slices is unique. we can find the unique point (by homeomorphism) bα ∈ Vα such that p(bα ) = b. then p−1 (b) has k elements for every b ∈ B. J. i. let B be connected. we can find U such that −1 p (U ) = k G Vi i=1 and . E is a k-fold covering of B. Since |p−1 (b0 )| = k. Pearse 3. Show that if p−1 (b0 ) has k elements for some b0 ∈ B.e.Solutions by Erin P. Let p : E → B be a covering map.. p. Vi = pi : Vi → U is a homeomorphism. so C 6= ∅. Also. D as a disconnection of B. and b1 ∈ D. |p−1 (b)| = k} and D = {b .. We have b0 ∈ C. Similarly. . For b ∈ C. Show that if r −1 (z) is finite for each z ∈ Z. Thus for x ∈ Ub . we can find Ub such that p−1 (Ub ) = ki=1 Vi . ∀i. D areFopen. We assume that ∃b1 such that |p−1 (b1 )| = j 6= k. D is < Hence no such b exists. so D 6= ∅.. This gives C. then p is a covering map. |p−1 (b)| 6= k}. 1 4. |p−1 (x)| = k. it is clear that C ∩ D = ∅ and C ∪ D = B. where the Vi are open. So it just remains to show C. pick z ∈ Z and find an open neighbourhood Z of z such . As r is a covering map. Hence b ∈ Ub ⊆ C shows C is open... Let q : X → Y and r : Y → Z be covering maps. let p = r ◦ q. Define C = {b . and we will contradict the fact that B is connected. open. F that p−1 (Z) = ni=1 Vi where Vi is open in Y and r . Define vi to be the single element of p−1 (z) ∩ Vi . As q is Fa covering map. for each i.Vi: Vi → Z is a homeomorphism. −1 we can find an open . neighbourhood Ui of vi such that q (Ui ) = α Aiα where Aiα is open in X and q . Ai: Aiα → Ui is a homeomorphism. which are also open. Now C is a neighbourhood of z ∈ Z for which G . i=1 Each Ui ∩ Vi is a neighbourhood of vi in Y which is evenly covered by q. We next define Diα = q −1 (Ui ∩ Vi ) ∩ Aiα . and has r(Ui ∩Vi ) ⊆ Z. α With so many sets. the fact that C is a finite intersection guarantees that C is open. we need to find some common ground. so define C= n \ r(Ui ∩ Vi ). Most importantly. So C is evenly covered by r (each slice is a Ui ∩Vi ). . p−1 (C) = Diα and p. Di = (r ◦ q). the Diα are disjoint because the Aiα are. α α i. 11 .α Also.Di is a homeomorphism. Topology (2nd ed. where each Vi goes 1 of the way around the circle S 1 . V2 by p(z) = z 2 . You will get p−1 (U ) = ni=1 Vi .. Since p is a covering map. so that for z ∈ S 1 we may write z = eiθ p(z) = z 2 = e2iθ . α∈A Since the union of the Vα is disjoint. Here we consider S 1 ⊆ C. Let U be the image of θ − π2 . We have just represented {xα } as an intersection of an open set Vα with the subspace p−1 (b) ⊆ E. we can find a neighbourhood U of b which is evenly covered by p.e. where the Vα are open. F For p(z) = z n . Since G p−1 (b) = {xα }. Consider the topology that p−1 (b) inherits from E. U = V1 t V2 and U is homeomorphic to each of V1 . show that p−1 (b) ⊆ E has the discrete topology. Show that the map p : S 1 → S 1 given by p(z) = z 2 is a covering map. Munkres 5. α∈A this shows p (b) is discrete. we are done. −θ − π 4
.) — James R. Clearly. let z = eiθ so θ = arg(z) ∈ [0. Generalize to the map p(z) = z n . which shows that {xα } is open in the subspace topology of p−1 (b). 2π). θ + π4 and the “quarter circle” centered at −z V2 = exp −θ + π4 . for any b ∈ B. i. where xα is the unique point in Vα such that p(xα ) = b. θ + π2 under the map θ 7→ eiθ so that U is the open semicircle centered at z. Since z has the requisite neighbourhood. 2n Bonus Problem: If p : E → B is a covering map. Vα ∩ p−1 (b) = {xα }. and
For z ∈ S 1 . G p−1 (U ) = Vα . use the same U . Then p−1 (U ) consists of the “quarter circle” centered at z
V1 = exp θ − π4 . −1 12 . p ◦ β(2t 2 ( α(2t). 1 ≤ t ≤ 1.1 for the local homeomorphism of Example 2 of §53? The very first step: you cannot find an open set U which contains b0 = (1.2. ˜ = α ∗ β. 2. 21 ≤ t ≤ 1. let ˜ α ˜ and β˜ be liftings of them such that α(1) ˜ = β(0). In defining the map F˜ in the proof of Lemma 54. p ◦ (α ˜ : I → E by We need to show p ◦ (α ˜ ∗ β) ˜ ∗ β) ( p ◦ α(2t). ˜ p ◦ (α ˜ ∗ β)(t) = def of ∗ ˜ − 1). J.Solutions by Erin P. 3. ˜ 0 ≤ t ≤ 12 . Show that α ˜ ∗ β˜ is a lifting of α ∗ β. Pearse §54. If you performed the lifting for the squares in random order. What goes wrong with the path-lifting lemma 54. 0) and is evenly covered by p. = α ∗ β(t) def of ∗. there is no guarantee that the lifts would “connect” or “match up” along the boundaries of the little squares. By definition. p is not a covering map. 13 . β˜ β(2t − 1). The Fundamental Group of the Circle 1. 0 ≤ t ≤ 21 . Let α and β be paths in B with α(1) = β(0). = def of α ˜ . why were we so careful about the order in which we considered the small rectangles? To maintain the continuity of the lift. Let p : E → B be a covering map. g(t) = ((1 + t) cos 2πt. h(t) = f ∗ g.) — James R. 0)}. Find liftings of the paths f (t) = (2 − t. t 2 ~ f f 1 -1 p 0 1 0 1 2 x Figure 1. sin 2πx). 14 2 . t) 7→ ((cos 2πx. 0). Three liftings of f . Munkres 4. 2I use the shorthand R20 = R2 \{(0. t g 2 ~ f -1 f ~ g 1 0 p 1 -2 -1 0 1 x Figure 3. Consider the covering map p : R × R+ → R20 of Example 6 of §53:2 (x. f˜ ∗ g˜ is a lifting of f ∗ g. f˜ is the one in the centre. g˜ is the one in the centre.Topology (2nd ed. (1 + t) sin 2πt). The point here is that f ∗ g is a closed loop. Three liftings of g. t g 2 1 ~ g -1 0 p 1 -2 -1 0 1 2 x Figure 2. Sketch the paths and their liftings. t) 7→ t(cos 2πx. sin 2πx). but no lifting of it is. Consider the maps g. t ∈ I under g. γ(t) = e . Since

n g ◦ γ(t) = g e2πit = e2πit = e2πint 2πit is a loop which goes n times around the circle. The path (loop) f begins at the white dot and traverses T2 . we have that the loop

−n h ◦ γ(t) = h e2πit = e2πit = e−2πint goes n times around the circle. J. Sketch what f looks like when S 1 × S 1 is identified with the doughnut surface D. in the direction opposite to γ. sin 2πt) × (cos 4πt. 6. Find a lifting f˜ of f to R × R.Solutions by Erin P. Pearse 5. the second S 1 axis is sketched in front. T2 = S 1 × S 1 is represented two different ways on the right. This gives h∗ ([γ]) = [h ◦ γ] = [γ] ∗ · · · ∗ [γ] = [γ]∗n = [γ]∗(−n) | {z } n times 15 or h : γ(t) 7→ γ(−nt). 2 2π ~ f p 1 = S1 f f 0 1 2 0 S1 2π S1 × S1 Figure 4. For visualization. e2it ). Consider the covering map p × p : R × R → S 1 × S 1 of Example 4 of §53. sin 4πt) = (eit . in the direction of γ. we have g∗ ([γ]) = [g ◦ γ] = [γ] ∗ · · · ∗ [γ] = [γ]∗n | {z } or g : γ(t) 7→ γ(nt). The group is cyclic. Consider the path in S 1 × S 1 given by f (t) = (cos 2πt. it traverses the vertical S 1 twice. b0 ) into itself. so it suffices to determine the image of its generator. 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. and sketch it. Compute the induced homomorphisms g∗ . if we consider T2 = S 1 × S 1 as a subspace of C × C. . n times Similarly for h. h∗ of the infinite cyclic group π1 (S 1 . the first S 1 axis is sketched on top of the doughnut. h : S 1 → S 1 given g(z) = z n and h(z) = z −n . and it is the lifting of f ∗ g that begins at (0. ( f (2t). 0 ≤ t ≤ 12 . x. since f does.) — James R. It is clear that it begins at (0. Given [f ]. Since R2 is simply connected. n) ∈ R2 . Now take g˜˜ = (m. y) ∈ R2 . we know m. Then P −1 (b0 ) = {(x. Thus φ([f ] ∗ [g]) = φ([f ]) + φ([g]). 0). Since these points are in the preimage of b0 . Then φ([f ]) + φ([g]) = (m. We check that it is a lifting: (   P (f˜(2t)). [g] ∈ π1 (T2 . k) = g˜(1). sin 2πx). as in Example 4 of §53. and we have φ([f ] ∗ [g]) = f˜ ∗ g˜˜(1) = (m + j. Take e0 = (0. Munkres 7. and we are done. def of ∗ P ◦ f˜ ∗ g˜˜ (t) = P (g˜˜(2t − 1)). and the last line follows because g˜ is a lifting of g. The third equality here follows because p(m + x) = p(x) by the formula for p. k ∈ Z. Let (m. n + k) ∈ Z2 . so we just need to show φ is a homomorphism. 16 . n) + g˜(t)) = (p(m + g˜1 (t)). k) = (m + j. φ([f ] ∗ [g]) is well-defined. 0). Generalize the proof of Theorem 54. so g˜˜ is a lifting of g. = by above g(2t − 1)). n) + (j. Since [f ] ∗ [g] begins at b0 . the lifting correspondence φ gives a bijection (by Thm 54.5 to show that the fundamental group of the torus is isomorphic to the group Z2 = Z × Z. Then by Thm 53. we may define P = p × p : R 2 → T2 . 0) and b0 = P (e0 ). b0 ).. Then the product f˜ ∗ g˜˜ is a well-defined path. = f ∗ g(t).. n) = f˜(1) and (j. the standard cover of the torus.3. n + k). 0 ≤ t ≤ 21 . 0). j. n. y ∈ Z} = Z2 . 21 ≤ t ≤ 1. p(˜ g2 (t))) = P ◦ g˜(t) = g(t).4). let f˜ and g˜ be their respective liftings to paths in R2 beginning at e0 = (0.Topology (2nd ed. n) + g˜(1) = (m + j. 12 ≤ t ≤ 1. n) + g˜ to be the translate of g˜ which begins at (m. The end point of g˜˜ is g˜˜(1) = (m. n + k). Let p : R → S 1 by the covering map given by p(x) = (cos 2πx. p(n + g˜2 (t))) = (p(˜ g1 (t)). Then P ◦ g˜˜(t) = P ((m. Show that if B is simply connected. and p∗ is actually an isomorphism. e0 ) → π1 (B. p∗ : π1 (E. which shows p−1 (b0 ) = e0 by the lifting correspondence — if you have a surjection (by Thm 54. E is a 1-fold covering of B.Solutions by Erin P. Pearse 8. Since π1 (B. Then every loop f at b0 is in H. then the range must also consists of only one element. it must also be the case that π1 (E. and hence lifts to a loop f˜ at e0 .3. with E path connected. b0 ) is 1-1. Let p : E → B be a covering map.6(a). Thus by Ex 53.e. By Thm 54. then p is a homeomorphism.. i. b0 ) is trivial by hypothesis. e0 ) is also trivial. J. E and B are homeomorphic.4) where the domain has only one element. 17 . then A ⊆ B 2 and there is a retraction i. Let j : A → B 2 by inclusion. Im(f ) = {y ∈ Y . a continuous map r : B 2 → A with r(a) = a. ∀a ∈ A.Topology (2nd ed.. Then .6). then every continuous map f : A → A has a fixed point.) — James R. so it has a fixed point by Brouwer’s Theorem (55.. it must be that c ∈ A.. Show that if A is a retract of B 2 . Munkres §55. Since Im(f ◦ r) ⊆ A. Retractions and Fixed Points Note: I use the following notation for the image of a map f : X → Y . Denote it by c. Then j ◦ f ◦ r : B 2 → B 2 is a continuous map.e. y = f (x) for some x ∈ X} 1. If A is a retract. c = j ◦ f ◦ r(c) = j ◦ f (c) r . and x 6= 0.7.6). as an extension.A is the identity = f (c) j is inclusion. call it a. then A has a positive real eigenvalue. h extends to k : B 2 → S 1 by Lemma 55. Suppose h : S 1 → S 1 is nulhomotopic. Since A is nonsingular. k(b) = b implies that b ∈ S 1 . let B = S 2 ∩ O1 ⊆ R3 . But S 1 ⊆ B 2 .3 with X = S 1 . In fact. All the components of x ∈ B are nonnegative. call it b. Since Im(k) ⊆ S 1 . 55. Then a = f ◦ h(a) = f (h(a)) = −h(a) =⇒ h(a) = −a. Thus h(b) is defined. Then f ◦ h : S 1 → S 1 is nulhomotopic. 3. so we can actually consider k as a mapping k : B 2 → B 2 . (a) Show that h has a fixed point. Show that if A is a nonsingular 3 × 3 matrix having nonnegative entries. 2. Then k must have a fixed point by Brouwer’s Theorem (55. and the map S:B→B by S(x) = Ax/kAxk 18 . so it must have a fixed point by (a). Define f : S 1 → S 1 by f (x) = −x. k = h on S 1 b is a fixed point of k. Ax 6= 0. Following Cor. h(b) = k(b) =b (b) Show that h maps some point x to its antipode −x. Assume: for each n. 4. Pearse is well-defined. 19 . Then x0 = Ax0 /kAx0 k =⇒ Ax0 = kAx0 kx0 . t) = (1 − t)x. S must have a fixed point x0 by Brouwer’s Thm (55. and directly inward at some point. Suppose i = idS n were nulhomotopic.e. (d) Every continuous map f : B n+1 → B n+1 has a fixed point. then h has a fixed point.6). (f) If h : S n → S n is nulhomotopic. (a) The identity map i : S n → S n is not nulhomotopic. i. there is no retraction r : B n+1 → S n . A has an eigenvector x0 with corresponding eigenvalue kAx0 k > 0. and h maps some point x to its antipode −x. (c) Every nonvanishing vector field on B n+1 points directly outward at some point. 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.3. J.. We follow the proof of Lemma 55. Define π : S n × I → B n+1 by π(x. Let H : S n × I → S n be a homotopy between i and a constant map. Since B is homeomorphic to the ball.Solutions by Erin P. (1) =⇒ (2) =⇒ (3). (e) Every (n + 1) × (n + 1) matrix with positive real entries has a positive real eigenvalue. as in §51 Exercise 1. 1) = r(x). 1) = s(x). ∂ I 1 \{( 12 . 2. 0 ≤ t ≤ 21 . Also. I 1 \{( 12 . we have s 'K idA by some homotopy K which keeps every point of B fixed. For each of the following spaces. Show that if A is a deformation retract of X and B is a deformation retract of A. t) = ((1 − t)x. Munkres §56. t) = K(r(x). (d) The infinite cylinder IC = S 1 × R. y) shows that S 1 is a deformation retract of the solid torus. by the straight-line homotopy (from
( 21 . This is a standard abuse of language. 1. 2t). so π1 (B 2 × S 1 ) = π1 (S 1 ) = Z. Under the quotient map which makes I 2 into the torus. 21 ≤ t ≤ 1. H(x. 0) = idX . Then s ◦ r : X → B and for b ∈ B. 0) = idA . 12 )). because the cylinder has S 1 as a deformation retract by the homotopy H((x. K(x. (1 − t)y). t) = (x. (c) The cylinder C = S 1 × I. 20 . Deformation Retracts and Homotopy Type Note: I am occasionally sloppy and saying “x0 ” when I mean “the constant map at x0 ”. π1 (IC) = Z by the same reason (and same homotopy) as (c). the homotopy H((x. π1 (C) = Z. K(x. because b ∈ B ⊆ A ⊆ X. Then s ◦ r 'F idX . F (x. (b) The torus with a point removed. H(x.Topology (2nd ed. s ◦ r(b) = s(r(b)) = s(b) = b. 12 )} becomes a figure-eight. So we have some r : X → A which is the identity on A. H keeps every point of B ⊆ A fixed. 2t − 1). For (x. or (3) isomorphic to the fundamental group of the figure eight. then B is a deformation retract of X. y). y). (a) The solid torus B 2 × S 1 . y) ∈ B 2 × S 1 . the fundamental group is either (1) trivial.) — James R. We also have r 'H idX by some homotopy H which keeps every point of A fixed. (2) infinite cyclic. Define F : X × I → X by ( H(x. In particular. 12 )} has its boundary as a deformation retract. and s : A → B which is the identity on B. . t) = (1 − t)x + tx/kxk. Consider that any loop in this space is homotopic to some combination of α..Solutions by Erin P. Note that α ∗ β = γ. (h) {x .. S1 ∪ (R+×0) S1 ∪ (R+×R) S1 ∪ (R×0) Figure 6. so π1 ((f)) = Z. kxk > 1}.. 2S 1 is obviously homeomorphic to S 1 by x 7→ x/2. This space has 2S 1 = {x ∈ R2 . so fundamental group is not abelian. π1 ((i)) is trivial by H(x. as depicted in Figure 5. so π1 ((g)) = Z. J. t) = (1 − t)x + 2tx/kxk. (i) S 1 ∪ (R+ × 0). t) = (1 − t)x + tx/kxk.. The spaces in exercises (i)-(l).. fundamental group can be generated without γ. β. kxk = 2} as a deformation retract by the straightline homotopy H(x. kxk ≥ 1}. (f) {x . This space has S 1 as a deformation retract by the straight-line homotopy H(x. By H(x. Also. kxk < 1}. so that the β γ α x0 Figure 5. 21 R2 − (R+×0) . note that β ∗ α ' γ 6= γ. Loops around the axes in R3 . γ (and their inverses). t) = (1 − t)x. Pearse (e) R3 with the nonnegative axes deleted. π1 ((h)) is trivial... (g) {x . 58. H(x. This has the fundamental group of the figure eight. So X u X.3 to Example 2). π1 ((j)) is trivial by H(x. Show that given a collection C of spaces. But this is just the same as Y u X. Describe maps f : X → Y and g : Y → X that are homotopy inverse to each other. f maps all the points inside the dotted line to the vertical bar of the theta. (k) S 1 ∪ (R × 0). π1 ((l)) is trivial. 0). 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. We need to show that h ◦ f : X → Z and g ◦ k : Z → X are homotopy inverses of each other. We show the three properties. g : Y → X such that g ◦ f ' idX and f ◦ g ' idY . (l) R2 \(R+ × 0). Define the maps as indicated in Figure 7. Then h. as both are retracts of the doubly punctured plane (apply Thm. 3. (ii) Let X u Y . (iii) Let X u Y and Y u Z. t) = . By the homotopy ( x kxk ≤ 1. t) = (1 − t)x + tx/kxk. the relation of homotopy equivalence is an equivalence relation on C. we obtain the theta space of Example 3. g ◦ f are both homotopic to the identity on X (since they are the identity on X). (i) Let f = g = idX : X → X. Munkres (j) S 1 ∪ (R+ × R). Let X be the figure eight and let Y be the theta space.1 (transitivity). so it cannot have an inverse. By H(x. g 22 . Note: it is not 1-1. 4. Then f ◦ g. k are homotopy inverses. (1 − t)x + tx/kxk kxk > 1.Topology (2nd ed.) — James R. By associativity. (g ◦ k) ◦ (h ◦ f ) = g ◦ (k ◦ h) ◦ f. t) = (1 − t)x + t(−1. The argument is identical for (h ◦ f ) ◦ (g ◦ k) = h ◦ (f ◦ g) ◦ k. with corresponding inverses g : Y → X and k : Z → Y . Then ∃f : X → Y. Then we have homotopy equivalences f : X → Y and h : Y → Z. 3(c) to [X. There is a homotopy between idX and g ◦ f which maps the figure eight to itself. Similarly. H(x. Recall that a space X is contractible iff the identity map idX is nulhomotopic. . Show that X is contractible iff X has the homotopy type of a one-point space. X]. 0) = x. but they are homotopy inverses. 5. maps are not inverses. but continuously scrunches all the points inside the dotted line down into the centre point. The spaces in exercises (i)-(l). J. ∀x ∈ X. Show that a retract of a contractible space is contractible. there is a homotopy from idY to f ◦ g which acts by collapsing the vertical bar to the centre point of the vertical bar. This is just an application of Exercise 51. 1) = x0 .Solutions by Erin P. 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. If X is contractible. These f g Figure 7. 6. as t goes from 0 to 1. and continuously slurping in part of the boundary of the circle to replace the bar. then we have H : X × I → X with H(x. Suppose r : X → A is a retraction. so that r . A = idA . Since it doesn’t matter what point . let x0 ∈ A. Then H .3(b)). x0 we chose above (X is path connected by Exercise 51. The integer d = deg h is the degree of h. and define γ(x0 ) = α ˆ (γ) so 1 that γ(x0 ) generates π1 (S . Define the degree of a continuous map h : S 1 → S 1 as follows: Let b0 = (1.A×I defines a homotopy from idA to x0 . 9. x0 ). 0) ∈ S 1 . Consider the homomorphism h∗ : π1 (S 1 . x1 ). Given h : S 1 → S 1 . choose a generator γ for the infinite cyclic group π1 (S 1 . b0 ). For x0 ∈ S 1 . let α : b0 /o /o /o / x0 be a path from b0 to x0 . Then h∗ (γ(x0 )) = d · γ(x1 ) for some d ∈ Z. 23 . choose x0 ∈ S 1 and let h(x0 ) = x1 . x0 ) → π1 (S 1 . Munkres (a) Show that d is independent of the choice of x0 . where z is a complex number. −x2 ). shoe the identity map is homotopic to the antipodal map.) — James R. y0 ∈ S 1 with h(x0 ) = x1 and h(y0 ) = y1 . (b) If h : S n → S n has degree different from (−1)n+1 . . and the reflection map ρ(x1 . and the map h(z) = z n . (d) If S n has nonvanishing tangent vector field v. and take the map g : S 1 → S 1 to be rotation by −θ. . then h maps some point x to its antipode −x. (iii) The identity map has degree 1. k : S 1 → S 1 have the same degree. the reflection map ρ(x1 .Topology (2nd ed. . xn . (d) Compute the degrees of the constant map. (c) Show that deg(h ◦ k) = (deg h) · (deg k). (ii) deg(h ◦ k) = (deg h) · (deg k). . Consider x0 = θ ∈ S 1 as being given by an angle θ ∈ [0. and constant map has degree 0. (c) If h : S n → S n has degree different from 1. 10. then h has a fixed point. xn+1 ) = (x1 . such that (i) Homotopic maps have the same degree. k : S 1 → S 1 are homotopic. then n is odd. Suppose that to every map h : S n → S n we have defined deg h ∈ Z. and that α : b0 /o /o /o / x0 and β : b0 /o /o /o / y0 (b) Show that if h. xn . 2π). −xn+1 ) has degree −1. . they are homotopic. the identity map. (Hint: If v exists. . Suppose x0 .) 24 . (e) Show that if h. they have the same degree. . Then prove the following: (a) There is no retraction r : B n+1 → S n . . x2 ) = (x1 . so Cor. The projection of X to the x-axis would then be [−2. Choose a point p ∈ S 2 not lying in the image of f . J. 1) × R2 . You may not be able to pick a point p in the image of f . 272). U ∩ V is clearly nonempty and contractible. 44. use the straight line path xy. It is possible to have a continuous surjection from I to S 2 (of course. Rn0 has S n−1 as a deformation retract by H(x. Let X be the union of two copies of S 2 having a single point in common. 2. ∞) is not connected.Solutions by Erin P. For every n. Pearse §57. Criticize the following “proof” that S 2 is simply connected: Let f be a loop in S 2 based at x0 . In fact. In this case. y and take the straight line path xz followed by the straight line path zy. 1]. For x. Since S 2 \p is homeomorphic with R2 and R2 is simply connected. 2].1 on p. pick any z ∈ Rn0 that does not lie on the line through x. For n = 2 this gives π1 (R20 ) = π1 (S 1 ) = Z. What is the fundamental group of X? Consider X as sitting in R3 along the x-axis. The solution is similar to the previous problem. U has a copy of S 2 as a deformation retract by collapsing the left hemisphere along the meridians leading to the origin. but we use simply connected instead of connected. The Fundamental Group of S n 1. but Rn0 is. 25 . Consider a composition with the the Peano map (see Thm. so that U and V are clearly open. Now suppose we had a homeomorphism f : R → Rn . (b) Show that R2 and Rn are not homeomorphic if n > 2. for purposes of description. 3.) (a) Show that R and Rn are not homeomorphic if n > 1. ∞) × R2 and V = X ∩ (−∞. t) = (1 − t)x + tx/kxk. Then the restriction of f to R0 would give a homeomorphism between a connected and a disconnected < set. hence path-connected. the loop f is path homotopic to the constant loop. the inverse will not be continuous). with the single point in common projecting to 0. R0 = (−∞.2 applies and π1 (X) is trivial. so π1 (U ) = π1 (S n ) is trivial. 0) ∪ (0. unless (1 − t)x + ty = 0 for some t ∈ [0. it is path connected. . Let U = X ∩ (−1. y ∈ Rn0 . Similarly for V . (I use the shorthand Rn0 = Rn \{0}. 59. ) — James R. each of which has positive x-coordinate. However. each of which has negative x-coordinate. X = U ∪ V is S 2 . but that’s just the trivial group. By Theorem 59. Then each of U . − 1 < x < 0}. which has trivial fundamental group. so that U is a punctured hemisphere. and thus has fundamental group Z. Since they have different fundamental groups.1. they cannot be homeomorphic. π1 (X) will be the group generated by the identity. V has a copy of S 1 as a deformation retract. (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. there’s not much you can say.Topology (2nd ed. y. 4. let U = {(x. Munkres For n ≥ 3.7. z) ∈ S 2 . Thus. This is the essential point of this course. each of which has positive x-coordinate. Assume the hypotheses of Theorem 59. and neither does X. 26 . but neither U nor V have trivial fundamental groups. For example. this gives π1 (Rn0 ) = π1 (S n−1 ) = 0. It is not hard to come up with cases where U does not have a trivial fundamental group. each of i∗ . as presented formally (and more strongly) in Thm. If just j∗ is trivial. then the image of each is just the identity element. Let U be S 2 with two punctures. 58..1. 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. (b) Give an example where i∗ and j∗ are trivial. Let V be S 2 with two punctures. and let V be S 2 with two punctures.. For b ∈ B 2 / ∼. Now by construction. −x} onto the projection of −x. let x = S+2 ∩ {b + (0. and if x ∈ S 2 lies below the xy-plane. But for such an x. −x}) = p(x) = b. −x} onto the projection of x. if x ∈ S 2 lies above the xy-plane. Define a map f : P 2 → B 2 / ∼ by ( p(x) z(x) ≥ 0 f ({x. J. Pearse §58. then map {x. p(x) = p(−x) by the equivalence relation. and 1 2 ∼ 1 2 ∼ π1 (S × S ) = π1 (S ) × π1 (S ) = Z × {e} ∼ = Z. where p(x) is the orthogonal projection of x onto the xy-plane. Consider the copy of B 2 which is the projection of S 2 into the xy-plane. f ◦ g(b) = f ({x. 2. −x} where x ∈ S 2 . π1 (S 1 × B 2 ) ∼ = π1 (S 1 ) × π1 (B 2 ) ∼ = Z × {e} ∼ = Z. Let X be the quotient space obtained from B 2 by identifying each point x of S 1 with its antipode −x. p(−x) z(x) ≤ 0. −x}. t) . −x} z(x) ≥ 0 g ◦ f ({x. x} z(x) ≤ 0 27 . In other words. 0. Show that X is homeomorphic to the projective plane P 2 . An element of P 2 looks like {x. since p(x) and p(−x) are antipodal points of S 2 . The Fundamental Group of Some Surfaces 1. 60. −x}) = . and ( ( g(p(x)) z(x) ≥ 0 {x. Now note that f is bijective. and z(x) is the third coordinate of x.Solutions by Erin P. so we can use the Pasting Lemma if we verify that f is continuous for x such that z(x) = 0. 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.1. Applying Thm.. Projection is continuous. by constructing the inverse. x}. Compute the fundamental group of the solid torus S 1 × B 2 and the product space S 1 × S 2. g(p(−x)) z(x) ≤ 0 {−x. −x}) = = = {−x. then map {x. and let ∼ be the equivalence relation which identifies antipodal points of B 2 . ) — James R. If f : X → Y is a continuous bijection with X compact and Y Hausdorff. by Compactness Mantra (1). Let p : E → X be the map constructed in the proof of Lemma 60.Topology (2nd ed. Proof. so f (U C ) is closed. 60. We need to show f is open. Now using the Basic Survival Tools #3(d).5. so f (U ) is open.  3. Let U ⊆ X be open. then f is a homeomorphism. Let. and that P 2 is Hausdorff by Thm. so that U C is closed. f is a closed map.3. Lemma. by Exercise 2 from the Fundamental Mantras of Compactness. Munkres Note that B 2 / ∼ is compact. and conclude that f −1 (and hence also f ) is a homeomorphism. f (U C ) = f (X\U ) def complement = f (X)\f (U ) BST 3(d) = Y \f (U ) f is onto C = f (U ) . Now apply the following lemma to f −1 . E 0 be the subspace of E that is the union of the x-axis and the y-axis. Show that p. E 0 is not a covering map. where n ∈ Z. as well as a small open horizontal interval around every point (n. n). Consider a tiny open disc centered at x0 . 0) and a small vertical interval around every point (0. Its preimage looks like a similar open “X” centered at the origin. . Its intersection with X is a small open “X” shape. . they are not homeomorphic to the “X” in the base space. 28 . So P 1 = S 1 . We already know this is a covering map. define p(z) = z 2 . To see this. 4. note that any map from an “X” onto an interval is necessarily not injective. p E 0 is not a covering map because while these open intervals are open and disjoint. Check that this maps antipodal points to the same point: p(−z) = (−z)2 = z 2 = p(z). What are they? For z ∈ S 1 . by previous problems (and homework). The space P 1 and the covering map p : S 1 → P 1 are familiar ones. and the fundamental group is not abelian. J. 29 . so the lifting of f ∗ g looks like f˜ followed by a counterclockwise loop around B0 ending at e2 . 54. By Thm. An alternative cover for the figure-eight. respectively. so the lifting of g ∗ f looks like g˜ followed by a counterclockwise loop around A0 ending at e1 . Use this covering space to show that the fundamental group of the figure eight is not abelian. Consider the covering map indicated in Figure 8. p wraps A1 around A twice and wraps B1 around B twice. Since these liftings have different endpoints.3 indicates that f ∗ g and g ∗ f are not path homotopic. ⇐⇒ (f ∗ g)∼ . Thm. Let f loop once around A counterclockwise and let g loop once around B counterclockwise.3.Solutions by Erin P. Pearse 5. Here. Hence f ∗ g 'p g ∗ f [f ] ∗ [g] = [f ∗ g] 6= [g ∗ f ] = [g] ∗ [f ]. (g ∗ f )∼ have the same endpoint. The lifting of g is g˜ running counterclockwise from e0 to e1 . 54. p maps A0 homeomorphically onto A and B. g~ B1 A0 A1 e1 e0 e2 B0 ~ f p g f B x0 A Figure 8. The lifting of f is f˜ running counterclockwise from e0 to e2 . 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