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Hatcher §1.1 Ex 1.1.5 (A) =⇒ (B): the Assumption Is That Any Map S1 → X Extends to a Map CS1 → X Defined on the Cone on S1

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Hatcher §1.1 Ex 1.1.5 (A) =⇒ (B): the Assumption Is That Any Map S1 → X Extends to a Map CS1 → X Defined on the Cone on S1 JMM March 30, 2004 Hatcher §1.1 Ex 1.1.5 (a) =⇒ (b): The assumption is that any map S1 → X extends to a map CS1 → X defined on the cone on S1. But the cone on S1 is homeomorphic to the 2-disc D2. 1 2 (b) =⇒ (c): Let f :(S , 1) → (X, x0) be the restriction of a map g :(D , 1) → (X, x0). The based homotopy class of f is an element [f] of π1(X, x0) (as in Ex 1.1.8). The fundamental group functor applied to the left commutative diagram yields the right commutative diagram: f f 1 1 ∗ (S , 1) / (X, x0) π1(S , 1) / π1(X, x0) I : L 8 II tt LL qq II tt LL qq I tg LL qgq i II tt i∗ L q ∗ I tt LL qqq $ 2 & 2 (D , 1) π1(D , 1) 2 2 Here π1(D , 1) = {1} is the trivial group since D deformation retracts onto the point {1} [1, 1.17]. Hence [f] = f∗[1S1 ] = g∗i∗[1S1 ] is the unit of the group π1(X, x0). (c) =⇒ (a): Let f : S1 → X be any map. The based homotopy class of f represents an element of π1(X, x0) where x0 = f(1). Since this group is assumed to be trivial, the map f is homotopic (even relative to the base point) to the constant map with value x0. Suppose that X is simply connected. This means that X is path-connected and π1(X, x0) is 1 the trivial group for all x0 ∈ X. From (c) =⇒ (a) we see that any map S → X is homotopic to a constant map. All constant maps are homotopic since X is path-connected. Thus all maps S1 → X are homotopic. Conversely, suppose that all maps S1 → X are homotopic. In particular, all constant maps 1 S → X are homotopic. But a homotopy between the constant map x0 and the constant map x1 is simply a path from x0 to x1. Thus X is path-connected. From (a) =⇒ (c) we get that π1(X, x0) for all x0 ∈ X. Thus X is simply connected. Written succinctly we have shown that X is simply connected ⇐⇒ [S1,X] = {∗} 1 1 where [S ,X] is the set of homotopy classes of maps S → X. (The relation between π1(X, x0) and [S1,X] is described in Ex 1.1.8.) Ex 1.1.6 1 We consider π1(X, x0) as the group of based homotopy classes of based maps (S , s0) → (X, x0) and define [S1,X] to be the set of (unbased) homotopy classes of maps S1 → X. There is a forgetful 1 1 map Φ: π1(X, x0) → [S ,X]. Assume that X is a path-connected. We want to show that [S ,X] 1 1 is the set of conjugacy classes of elements of π1(X, x0). (Keep in mind the case X = S ∨ S .) Lemma 0.1. Suppose that X is path-connected. 1 (1) Φ: π1(X, x0) → [S ,X] is surjective. (2) For all [f], [g] ∈ π1(X, x0): −1 Φ([f]) = Φ([g]) ⇐⇒ ∃[h] ∈ π1(X, x0):[h][f][h] = [g] 1 1 Proof. (1) The claim is that any map f0 : S → X is homotopic to a map f0 : S → X that preserves base points, f1(s0) = x0. Since X is path-connected, there is a path h: I → X from h(0) = f0(s0) 1 to x0. By the homotopy extension property applied to the pair (S , s0) f ∪h 1 0 S × {0} ∪ s0 × I 6/ X _ m m m m m m m m S1 × I 1 we get a homotopy S × I → X from f0 to a base point preserving map f1. (2) =⇒: Suppose that H : S1 × I → X is a homotopy (maybe not preserving the base points) of 1 H0 = f to H1 = g. Define a new homotopy He : S × I → X by Het = htHtht (composition af 1 2 paths) where ht is the path from x0 to H(s0, t) (ie ht(s) = H(s0, st), s ∈ I). This new homotopy He preserves the base point, Het(s0) = x0 for all t ∈ I, and He0 = H0 = f and He1 = h1H1h1 = h1gh1 where h1 is the loop t → H(s0, t) based at x0. 1 1 (2) ⇐=: Let f, h: S → X be loops based at x0. Let H : S × I → X be the homotopy Ht = φtfφt (composition of paths) where φt is the path from h(t) to h(1) = x0 (given by φt(s) = h((1−s)t+s)). 1 1 Then H0 = hfh and H1 = f. (Draw a picture of the homotopy a ' bab in case X = Sa ∨ Sb ). Ex 1.1.8 3 No, the Borsuk–Ulam theorem does not hold for the torus: Imbed the torus M1 ⊂ R symmet- 2 rically around origo and let f : M1 → R be projection to the xy-plane. Then f(−p) = −f(p) for all p ∈ M1 and f(p) is never (0, 0) so f(p) 6= f(−p) for all p ∈ M1. In fact, by the same argument, the Borsuk–Ulam theorem does not hold for the compact ori- entable surface Mg when g is odd. Does the Borsuk–Ulam hold for Mg when g is even? Ex 1.1.9 (Morten Poulsen, 2004) 2 Since A1, A2 and A3 is compact they have finite measure. Every point s ∈ S determines a unit vector i R3 and hence a direction, so we can regard s as a unit vector in R3. For each Ai choose a plane Pi with s as a normal, such that Pi divides Ai in two pieces of equal measure. It is intuitively clear that Pi exists, by continuously sliding Pi along the line determined by s, and is unique, but the proof is omitted. Let di(s), i = 1, 2, denote the Euclidian distance between P3 and Pi in the direction determined by s. The situation is illustrated in the figure below. In the situation pictured below are d1(s) < 0 2 and d2(s) > 0. We want to proof that there exists s ∈ S , such that di(s) = 0, i = 1, 2, hence the three planes coincide and the result follows. Figure 1. The Borsuk-Ulam theorem in dimension 2 2 2 Now define f : S → R by f(s) = (d1(s), d2(s)). Remark that f is an odd map, i.e. f(−s) = 2 −f(s). It is clear that f is continuous. So by the Borsuk-Ulam Theorem there exists s0 ∈ S , such that f(s0) = f(−s0). But this means that di(s0) = −di(s0), hence di(s0) = 0, i = 1, 2, as desired. As a final remark, note that since the Borsuk-Ulam Theorem holds for continuous maps Sn → Rn, then this argument easily generalises to Rn for a hyperplane of dimension n − 1, hence the result also holds i Rn. That is, given that the used geometric results hold. Ex 1.1.16 (e) The space X deformation retracts onto one of the circles in its boundary A = S1 ∨ S1. The homomorphism Z∗Z = π1(A) → π1(X) = Z induced by the inclusion is not injective (all subgroups 3 b X '$ a Figure 2. A disc with&% two boundary points identified of the abelian group Z are abelian and Z ∗ Z is not abelian). Hence there is no retraction of X onto A. (f) The fundamental group of the M¨obius band MB is infinite cylic generated by the homotopy class [C] ∈ π1(MB, •) = h[C]i of the core circle. The fundamental group of the boundary circle, A, is also infinite cyclic generated by the homotopy class [A] ∈ π1(A, •) = h[A]i of the obvious map 2 I → A. Let i: A → MB be the inclusion map. Then i∗[A] = [C] . Suppose that r : MB → A is F A t C ? 6 A F t Figure 3. The M¨obius band with core circle C and boundary circle A a retraction onto the boundary circle. Applying the functor π1 to the topological diagram to the left we get the group diagram to the right: i i∗ A / MB π1(A) / π1(MB) > >>> HHHH >>>> HHHH >> r π1 HHH r∗ >>> HHHH >>> +3 HHHH >> H A π1(A) 2 a contradiction because r∗i∗[A] = (r∗[C]) is a square and [A] is not a square. References [1] Allen Hatcher, Algebraic topology, Cambridge University Press, Cambridge, 2002. MR 2002k:55001.
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