Given spaces X and Y, let X, Y denote the set of homotopy classes of maps of Xinto Y. Let I = 0, 1.(a) Show that for any X, the set X, I has a single element.Define ϕ˜ : X → I to be the zero map: ϕ˜(x) = 0, ∀x ∈ X. Let ϕ : X → I by anycontinuous map. We show ϕ ϕ˜: define Φ : X × I → I byΦ(x, t) = (1 − t)ϕ(x).This is evidently a homotopy ϕ ϕ˜.Since is an equivalence relation, this shows all maps ϕ : X → I are equivalentunder , i.e., there is only one equivalence class.
§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 Show that if h, h : X → Y are homotopic and k, k : Y → Z are homotopic, then k◦h k ◦h Since h h , we have a homotopy H :X ×I →Y and since k with H(x, 0) = h(x), H(x, 1) = h (x), 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)), ≤ t ≤ 21 K(h (x), 2t − 1), 21 ≤ t ≤ Then check that F does what it oughta do, at different times t:1 t=0 t= t=1 F (x, 0) = k(H(x, 0)) = k(h(x)) = k ◦ h(x) F x, 12 = k(H(x, 1)) = k(h (x)) = K(h (x), 0) F (x, 1) = K(h (x), 1) = k (h (x)) = k ◦ h (x) 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 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 , ≤ t ≤ 12 γ(2t − 1), ≤ t ≤ Φ(x, t) = 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 ≤ t ≤ 21 , then moving that point to p along γ(I) while 21 ≤ t ≤ Check that Φ does what it oughta do: t=0 t= Φ(x, 0) = ϕ(x) Φ x, 12 = ϕ(0) = a = γ(0) t=1 Φ(x, 1) = γ(1) = p 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) = (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 γ(t) = H(y, 2t), ≤ t ≤ 21 H(z, − 2t), 12 ≤ t ≤ And check it: t=0 t= γ(t) = H(y, 0) = idX (y) = y γ(t) = H(y, 1) = p = H(z, 1) 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 and some 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 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 : 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 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 Topology (2nd ed.) — James R Munkres Now we can set up the requisite homotopy Φ(x, t) = ϕ(H(x, 2t)), ≤ t ≤ 12 γ(2t − 1), ≤ t ≤ And check it: t=0 t= t=1 Φ(x, 0) = ϕ(idX (x)) = ϕ(x) Φ x, 12 = ϕ(p) = γ(0) Φ(x, 1) = γ(1) = q = ϕ(x) ˜ Solutions by Erin P J Pearse §52 The Fundamental Group A subset A of Rn is star convex iff for some point a0 ∈ A, all the line segments joining a0 to other points of A lie in A, i.e., (1 − λ)a + λa0 ∈ A, ∀λ ∈ (0, 1) (a) Find a star convex set that is not convex A six-pointed star like the Star of David, or a pentacle will work if you let a0 be the center (Hence the name “star convex”.) The set {(x, y) x = or y = 0} ⊆ R2 is star convex with respect to the origin Or let I = I × I ⊆ R2 and let X = {(x, y) y = 0} ⊆ R2 Then A = I ∪ 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) ≤ 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 , A] Now let [S , A]a ⊆ [S , A] be those maps which send at least one point of S to a Then [S , A]a also consists of a single element But [S , A]a = π1 (A, x0 )! So π1 (A, x0 ) is trivial Let α be a path in X from x0 to x1 ; let β be a path in X from x1 to x2 Show that if γ = α ∗ β, then γˆ = βˆ ◦ α ˆ We show γˆ = βˆ ◦ α ˆ by showing that γˆ ([f ]) = (βˆ ◦ α)([f ˆ ]), 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 def of ∗, p.326 = β¯ ∗ α([f ˆ ]) ∗ [β] ˆ α([f = β( ˆ ])) def of α ˆ def of βˆ def of the reverse α ¯ = (βˆ ◦ α)([f 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 Topology (2nd ed.) — James R Munkres 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 x1 , we have α ˆ = β (⇒) Suppose π1 (X, x0 ) is abelian, and let α, β be paths from x0 to x1 Since α ˆ and ˆ β are both homomorphisms from π1 (X, x0 ) to π1 (X, x1 ), we need to prove that they 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 ˆ is an isomorphism, by Cor 52.2 (we are given that X is ˆ path connected) Now we show α ˆ = β α([f ˆ ]) = [α] ¯ ∗ [f ] ∗ [α] def of α ˆ = [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) ¯ ]) = [f ] So we have actually proven that α Similarly, β([f ˆ and βˆ 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 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 isn’t even defined, since α ends at x1 , and f ends at x0 Try again: [f ] ∗ [α] = [f ] ∗ [α] ∗ [ex1 ] = [f ] ∗ [α] ∗ β¯ ∗ [β] Thm 51.2(2) Thm 51.2(3) = [f ] ∗ α ∗ β¯ ∗ [β] = α ∗ β¯ ∗ [f ] ∗ [β] def of ∗, p.326 π1 (X, x0 ) is abelian = [α] ∗ β¯ ∗ [f ] ∗ [β] [α] ¯ ∗ [f ] ∗ [α] = [α] ¯ ∗ [α] ∗ β¯ ∗ [f ] ∗ [β] [α] ¯ ∗ [f ] ∗ [α] = [ex ] ∗ β¯ ∗ [f ] ∗ [β] Thm 51.2(3) [α] ¯ ∗ [f ] ∗ [α] = β¯ ∗ [f ] ∗ [β] ˆ ]) α ˆ ([f ]) = β([f Thm 51.2(2) and Thm 51.2(2) def of α, ˆ βˆ def of ∗, p.326 multiply both sides on the left Whew! Note that the fourth equality above is justified because α ∗ β¯ is a loop, so we avoid the problem of the other (incorrect) solution above Solutions by Erin P J Pearse (⇐) Now suppose that α ˆ = βˆ for every pair α, β of paths from x0 to x1 We must show π1 (X, x0 ) is abelian, so pick f, g ∈ π1 (X, x0 ) and show [f ] ∗ [g] = [g] ∗ [f ] Now we have the freedom to choose our α and β, so define α = f, β = f ∗ g Then compute: α ˆ ([f ]) = f ([f ]) = [f¯] ∗ [f ] ∗ [f ] α=f def of fˆ = [ex0 ] ∗ [f ] Thm 51.2(3), ex0 = ex1 = [f ] Thm 51.2(2) and ˆ ]) = f ∗ g([f ]) β([f = f ∗ g ∗ [f ] ∗ [f ∗ g] β =f ∗g def of fˆ = [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) ˆ we know α ˆ ]), and so Since α ˆ = β, ˆ ([f ]) = β([f [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 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 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∗ : π1 (X, a0 ) → π1 (A, a0 ) is surjective Let f : I → A be a loop in A, based at x0 We must find a loop g in X, based at x0 , such that r∗ ([g]) = [r ◦ g] = [f ] Because f is also a loop in X, based at x0 , 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 ∀t ∈ I, Topology (2nd ed.) — James R Munkres 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 ] = [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 check t=0 Φ(x, 0) = g(f (x)) = g ◦ f (x) = h ◦ f (x) t=1 Φ(x, 1) = g(a0 ) = y0 = ϕ(x) ˜ Note: we don’t know that h((1 − t)f (x) + ta0 ) is defined, but since h is extendible to a continuous map on all of Rn , we konw that g((1 − t)f (x) + ta0 ) is defined, since (1 − t)f (x) + ta0 is just some point between f (x) ∈ A and a0 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 ) = y0 and h(x1 ) = y1 Let α be a path in X from x0 to x1 , and let β = h ◦ α Show that βˆ ◦ (hx )∗ = (hx )∗ ◦ α, ˆ so that the diagram commutes: (hx )∗ π1 (X, x0 ) −−−−− −−→ π1 (Y, y0 ) ˆ α ˆ β (hx )∗ π1 (X, x1 ) −−−−− −−→ π1 (Y, y1 ) We need to see that these two operations 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 βˆ ◦ (hx )∗ ([f ]) = h ◦ α ◦ (hx )∗ ([f ]) def of β 0 = h ◦ α([h ◦ f ]) = h ◦ α ∗ [h ◦ f ] ∗ [h ◦ α] def of h∗ def of βˆ = h ◦ α ∗ [h ◦ f ] ∗ [h ◦ α] ˆ def of β Solutions by Erin P J Pearse On the other hand, 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 ∗ 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) ¯ Topology (2nd ed.) — James R Munkres §53 Covering Spaces 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 Vα ⊆ X × Y such that for each α, p α : Vα → U is a homeomorphism Since Y is discrete, {y} is open in Y and so U × {y} is open in X × Y by the def of product topology Then we show p−1 (U ) = U × {y} y∈Y Note: this union is disjoint because (x, y) ∈ U × {y1 } ∩ U × {y2 } =⇒ y = y = y2 =⇒ U × {y1 } = U × {y2 } Let x ∈ p−1 (U ) 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 = {Vα }α∈A and B = {Vβ }β∈B Fix b ∈ B Then for any α, we can find the unique point (by homeomorphism) bα ∈ Vα such that p(bα ) = b, and for any β, we can find the unique point bβ ∈ Vβ such that p(bβ ) = b Note that every Vα , Vβ is connected, by homeomorphism with U We will show that there is a bijection between these partitions; i.e., A is actually just a reindexing of B Fix α0 Define f : A → B as follows: find the unique β0 such that bα0 ∈ Vβ0 and define f (α0 ) = β0 There will be such a Vβ0 , because E = ∪Vβ , and that Vβ0 will be unique by the disjointness of the Vβ Now to see that this is a bijection For bβ ∈ Vβ ⊆ E, ∃α such that bβ ∈ Vα because Vα contains E This shows surjectivity For injectivity, note that b β ∈ V α1 ∩ V α1 =⇒ by the disjointness of the partition 10 α = α2 Topology (2nd ed.) — James R Munkres Consider the covering map p : R × R+ → R20 of Example of §53:2 (x, t) → ((cos 2πx, sin 2πx), t) → t(cos 2πx, sin 2πx) Find liftings of the paths f (t) = (2 − t, 0), g(t) = ((1 + t) cos 2πt, (1 + t) sin 2πt), h(t) = f ∗ g Sketch the paths and their liftings The point here is that f ∗ g is a closed loop, but no lifting of it is t ~ f f -1 p 1 x Figure Three liftings of f f˜ is the one in the centre t g ~ g -1 p -2 -1 x Figure Three liftings of g g˜ is the one in the centre t g ~ f -1 f ~ g p -2 -1 x Figure f˜ ∗ g˜ is a lifting of f ∗ g 2I use the shorthand R20 = R2 \{(0, 0)} 14 Solutions by Erin P J Pearse Consider the covering map p × p : R × R → S × S of Example of §53 Consider the path in S × S given by f (t) = (cos 2πt, sin 2πt) × (cos 4πt, sin 4πt) = (eit , e2it ), if we consider T2 = S × S as a subspace of C × C Sketch what f looks like when S × S is identified with the doughnut surface D Find a lifting f˜ of f to R × R, and sketch it 2π ~ f p = S1 f f S1 2π S1 × S1 Figure T2 = S × S is represented two different ways on the right The path (loop) f begins at the white dot and traverses T2 For visualization, the first S axis is sketched on top of the doughnut, the second S 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 once, it traverses the vertical S twice Consider the maps g, h : S → S given g(z) = z n and h(z) = z −n Compute the induced homomorphisms g∗ , h∗ of the infinite cyclic group π1 (S , 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 = e2πint 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 or g : γ(t) → γ(nt) 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 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 by the covering map given by p(x) = (cos 2πx, sin 2πx) Then by Thm 53.3, we may define P = p × p : R → T2 , the standard cover of the torus, as in Example 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 Given [f ], [g] ∈ π1 (T2 , b0 ), let f˜ and g˜ be their respective liftings to paths in R2 beginning at e0 = (0, 0) Let (m, n) = f˜(1) and (j, k) = g˜(1) Since these points are in the preimage of b0 , we know m, n, j, k ∈ Z Then φ([f ]) + φ([g]) = (m, n) + (j, k) = (m + j, n + k) ∈ Z2 Now take g˜˜ = (m, n) + g˜ to be the translate of g˜ which begins at (m, n) ∈ R2 Then 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)), ≤ t ≤ 12 , P (g˜˜(2t − 1)), 12 ≤ t ≤ 1, def of ∗ f (2t), ≤ 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 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 Solutions by Erin P J Pearse 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 17 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} Show that if A is a retract of B , then every continuous map f : A → A has a fixed point If A is a retract, then A ⊆ B and there is a retraction i.e., a continuous map r : B → A with r(a) = a, ∀a ∈ A Let j : A → B by inclusion Then j ◦ f ◦ r : B → B 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 = f (c) A is the identity j is inclusion Suppose h : S → S is nulhomotopic (a) Show that h has a fixed point h extends to k : B → S by Lemma 55.3 with X = S But S ⊆ B , so we can actually consider k as a mapping k : B → B Then k must have a fixed point by Brouwer’s Theorem (55.6), call it b Since Im(k) ⊆ S , k(b) = b implies that b ∈ S Thus h(b) is defined In fact, as an extension, k = h on S 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 → S by f (x) = −x Then f ◦ h : S → S 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 Show that if A is a nonsingular × matrix having nonnegative entries, then A has a positive real eigenvalue Following Cor 55.7, let B = S ∩ O1 ⊆ R3 All the components of x ∈ B are nonnegative, and x = 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 > 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 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 a constant map Define π : S n × I → B n+1 by 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 19 Topology (2nd ed.) — James R Munkres §58 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 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), ≤ t ≤ 21 , H(x, 0) = idX , H(x, 1) = r(x), K(r(x), 2t − 1), 21 ≤ t ≤ 1, K(x, 0) = idA , K(x, 1) = s(x), as in §51 Exercise Then s ◦ r F idX 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 × S For (x, y) ∈ B × S , the homotopy H((x, y), t) = ((1 − t)x, y) shows that S is a deformation retract of the solid torus, so π1 (B × S ) = π1 (S ) = Z (b) The torus with a point removed I \{( 12 , 12 )} has its boundary as a deformation retract, by the straight-line homotopy (from ( 21 , 12 )) Under the quotient map which makes I into the torus, ∂ I \{( 12 , 12 )} becomes a figure-eight (c) The cylinder C = S × I π1 (C) = Z, because the cylinder has S as a deformation retract by the homotopy H((x, y), t) = (x, (1 − t)y) (d) The infinite cylinder IC = S × 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 Note that α ∗ β = γ, so that the β γ α x0 Figure Loops around the axes in R3 fundamental group can be generated without γ Also, note that β ∗ α so fundamental group is not abelian γ = γ, (f) {x x > 1} This space has 2S = {x ∈ R2 x = 2} as a deformation retract by the straightline homotopy H(x, t) = (1 − t)x + 2tx/ x 2S is obviously homeomorphic to S by x → x/2, so π1 ((f)) = Z (g) {x x ≥ 1} This space has S 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 S1 ∪ (R+×0) S1 ∪ (R+×R) S1 ∪ (R×0) Figure The spaces in exercises (i)-(l) (i) S ∪ (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 ∪ (R+ × R) π1 ((j)) is trivial by H(x, t) = (1 − t)x + tx/ x (k) S ∪ (R × 0) By the homotopy H(x, t) = x (1 − t)x + tx/ x x ≤ 1, , x > we obtain the theta space of Example 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 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 = idX : 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 But this is just the same as Y X idX and f ◦ g idY (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 g ◦ (k ◦ h) g ◦ (k ◦ h) ◦ f idY and g ◦ idY §51 Exercise g ◦ idY ◦ f (x) §51 Exercise 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 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 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 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 to Similarly, there is a homotopy from idY 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 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 This is just an application of Exercise 51.3(c) to [X, X] 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 r A = idA Since it doesn’t matter what point x0 we chose above (X is path connected by Exercise 51.3(b)), let x0 ∈ A Then H A×I defines a homotopy from idA to x0 23 Topology (2nd ed.) — James R Munkres §59 The Fundamental Group of S n Let X be the union of two copies of S 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 Let U = X ∩ (−1, ∞) × R2 and V = X ∩ (−∞, 1) × R2 , so that U and V are clearly open U has a copy of S 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 π1 (X) is trivial Criticize the following “proof” that S is simply connected: Let f be a loop in S based at x0 Choose a point p ∈ S not lying in the image of f Since S \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 (of course, the inverse will not be continuous) Consider a composition with the the Peano map (see Thm 44.1 on p 272) (I use the shorthand Rn0 = Rn \{0}.) (a) Show that R and Rn are not homeomorphic if n > 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 = 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 > 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 = this gives π1 (R20 ) = π1 (S ) = Z 24 Solutions by Erin P J Pearse For n ≥ 3, this gives π1 (Rn0 ) = π1 (S n−1 ) = 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 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, π1 (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 − < x < 0}, so that U is a punctured hemisphere, and let V be S 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 with two punctures, each of which has negative x-coordinate Let V be S with two punctures, each of which has positive x-coordinate Then each of U , V has a copy of S as a deformation retract, and thus has fundamental group Z However, X = U ∪ V is S , 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 ) to send it to) and is hence a trivial homomorphism 25 Topology (2nd ed.) — James R Munkres §60 The Fundamental Group of Some Surfaces Compute the fundamental group of the solid torus S × B and the product space S × S Applying Thm 60.1, π1 (S × B ) ∼ = π1 (S ) × π1 (B ) ∼ = Z × {e} ∼ = Z, and ∼ ∼ π1 (S × S ) = π1 (S ) × π1 (S ) = Z × {e} ∼ = Z Let X be the quotient space obtained from B by identifying each point x of S with its antipode −x Show that X is homeomorphic to the projective plane P An element of P looks like {x, −x} where x ∈ S Consider the copy of B which is the projection of S into the xy-plane, and let ∼ be the equivalence relation which identifies antipodal points of B Define a map f : P → B / ∼ by f ({x, −x}) = p(x) z(x) ≥ , p(−x) z(x) ≤ 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 lies above the xy-plane, then map {x, −x} onto the projection of x, and if x ∈ S 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) = But for such an x, p(x) = p(−x) by the equivalence relation, since p(x) and p(−x) are antipodal points of S Now note that f is bijective, by constructing the inverse For b ∈ B / ∼, let x = S+2 ∩ {b + (0, 0, t) t ∈ R} be the point in the upper hemisphere of S 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) ≥ = g(p(−x)) z(x) ≤ 26 and {x, −x} z(x) ≥ = {−x, x} {−x, x} z(x) ≤ Solutions by Erin P J Pearse Note that B / ∼ is compact, by Compactness Mantra (1), and that P 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 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 a closed map, by Exercise 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 C = f (U ) , so f (U ) is open 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 The space P and the covering map p : S → P are familiar ones What are they? For z ∈ S , define p(z) = z Check that this maps antipodal points to the same point: p(−z) = (−z)2 = z = p(z) We already know this is a covering map, by previous problems (and homework) So P = S 27 Topology (2nd ed.) — James R Munkres Consider the covering map indicated in Figure 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~ B1 A0 A1 e1 e0 e2 B0 ~ f p g f B x0 A Figure An 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 e2 , 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 e1 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 28 [...]... t, 0), g(t) = ( (1 + t) cos 2πt, (1 + t) sin 2πt), h(t) = f ∗ g Sketch the paths and their liftings The point here is that f ∗ g is a closed loop, but no lifting of it is t 2 ~ f f 1 -1 p 0 1 0 1 2 x Figure 1 Three liftings of f f˜ is the one in the centre t g 2 1 ~ g -1 0 p 1 -2 -1 0 1 2 x Figure 2 Three liftings of g g˜ is the one in the centre t g 2 ~ f -1 f ~ g 1 0 p 1 -2 -1 0 1 x Figure 3 f˜ ∗... that S 1 is a deformation retract of the solid torus, so 1 (B 2 × S 1 ) = 1 (S 1 ) = Z (b) The torus with a point removed I 1 \{( 12 , 12 )} has its boundary as a deformation retract, by the straight-line homotopy (from ( 21 , 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 1 (C) = Z, because the cylinder has S 1 as... 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 1 ((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 1 ((g)) = Z (h) {x x < 1} By H(x, t) = (1 − t)x, 1 ((h)) is trivial S1 ∪ (R+×0) S1 ∪... 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, 1 (S 1 × B 2 ) ∼ = 1 (S 1 ) × 1 (B 2 ) ∼ = Z × {e} ∼ = Z, and 1 2 ∼ 1 2 ∼ 1 (S × S ) = 1 (S ) × 1 (S ) = 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... ≤ t ≤ 21 , H(x, 0) = idX , H(x, 1) = r(x), K(r(x), 2t − 1) , 21 ≤ 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) ∈ B 2 × S 1 , the homotopy H((x, y), t) = ( (1 − t)x,... ((h)) is trivial S1 ∪ (R+×0) S1 ∪ (R+×R) S1 ∪ (R×0) Figure 6 The spaces in exercises (i)-(l) (i) S 1 ∪ (R+ × 0) 1 ((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) 1 ((j)) is trivial by H(x, t) = (1 − t)x + tx/ x (k) S 1 ∪ (R × 0) By the homotopy H(x, t) = x (1 − t)x + tx/ x x ≤ 1, , x > 1 we obtain the theta space of Example 3... (t) = = P (f˜(2t)), 0 ≤ t ≤ 12 , P (g˜˜(2t − 1) ), 12 ≤ 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 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 Solutions by Erin P J Pearse... 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 ˜ = α ∗ β By definition, p ◦ (α ˜ : I → E by We need to show p ◦ (α ˜ ∗ β) ˜ ∗ β) ˜ p ◦ (α ˜ ∗ β)(t) = = p ◦ α(2t), ˜ 0 ≤ t ≤ 12 , ˜ − 1) , 1 ≤ t ≤ 1, p ◦ β(2t 2 def of ∗ α(2t), 0 ≤ t ≤ 21 , β(2t − 1) , 21 ≤ t ≤ 1, def of α ˜ , β˜ = α ∗ β(t) def of ∗ 13 Topology (2nd ed.)... 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 ) = Vi and p Vi = pi : Vi → U is a homeomorphism, ∀i i =1 We assume that ∃b1 such that |p 1 (b1 )| = j = k, and we will contradict the fact that B is connected Define C = {b |p 1 (b)| = k} and D = {b |p 1 (b)| = k}... doughnut surface D Find a lifting f˜ of f to R × R, and sketch it 2 2π ~ f p 1 = S1 f f 0 1 2 0 S1 2π S1 × S1 Figure 4 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 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 ...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... path-connectedness of Y we have one: γ : I → Y, with γ(0) = ϕ(p), γ(1) = q Topology (2nd ed.) — James R Munkres Now we can set up the requisite homotopy Φ(x, t) = ϕ(H(x, 2t)), ≤ t ≤ 12 γ(2t − 1), ≤ t... reverse of β followed by the reverse of α, in the third line above Topology (2nd ed.) — James R Munkres Let x0 and x1 be points of the path-connected space X Show that π1 (X, x0 ) is ˆ abelian