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  1. 65.2.6
  2. 65.2.12
  3. Problem 65C

65.2.6 #

Exercise 65.2.6. Show that for any path 𝛼, 𝛼∗𝛼 is homotopic to the “do-nothing” loop at 𝛼(0). (Draw apicture.)Solution by kiwiyouWe can trace back the path 𝛼.Define a path homotopy 𝑓𝑠(𝑡):[0,1]×[0,1]→𝑋 as:𝑓𝑠(𝑡)={{{{{{{𝛼(2𝑡) (0≤𝑡≤12𝑠)𝛼(𝑠)=𝛼(1−𝑠)(12𝑠≤𝑡≤1−12𝑠)𝛼(2𝑡−1)(1−12𝑠≤𝑡≤1)𝑡=0121𝑠=0𝑠=1Since 𝑓0=𝛼∗𝛼 and 𝑓1=1, 𝛼∗𝛼 is nulhomotopic.∎Definition 65.2.2. Denote the trivial do-nothing loop by 1. A loop 𝛾 is nulhomotopic if it is homotopicto 1; i.e. 𝛾≃1.

65.2.12 #

Exercise 65.2.12. Let 𝑋 be a path-connected space. Prove that 𝑋 is simply connected if and only if𝜋1(𝑋) is the trivial group. (One direction is easy; the other is a little trickier.)Solution by kiwiyou𝑋 is simply connected ⟹ 𝜋1(𝑋) is the trivial groupBy definition of simply connected space, every loop 𝛾 is (path) homotopic to the “do-nothing” loop at 𝛾(0),which is the identity element of 𝜋1(𝑋).Definition 7.7.4. A space 𝑋 is simply connected if it’s path-connected and for any points 𝑝 and 𝑞,all paths from 𝑝 to 𝑞 are homotopic.Since all elements of 𝜋1(𝑋) are homotopic to the identity element, 𝜋1(𝑋) is the trivial group.∎Example 1.1.17 (Trivial group)The trivial group, often denoted 0 or 1, is the group with only an identity element. I will use thenotation {1}.𝑋 is simply connected ⟸ 𝜋1(𝑋) is the trivial groupLet 𝑝,𝑞∈𝑋.For all path 𝑎,𝑏 both from 𝑝 to 𝑞, note that 𝑏∗𝑎∈𝜋1(𝑋), which is the trivial group.𝑏≃𝑏∗1≃𝑏∗(𝑎∗𝑎)≃(𝑏∗𝑎)∗𝑎≃1∗𝑎≃𝑎Therefore, 𝑋 is simply connected.∎

Problem 65C #

Problem 65C (RMM 2013). Let 𝑛≥2 be a positive integer. A stone is placed at each vertex of a regular2𝑛-gon. A move consists of selecting an edge of the 2𝑛-gon and swapping the two stones at the endpoints ofthe edge. Prove that if a sequence of moves swaps every pair of stones exactly once, then there is some edgenever used in any move.Solution by finalchildAssume a sequence of moves that swaps every pair of stones exactly once.Fix a stone 𝑂. Every other stone should swap with 𝑂 either counterwise or counterclockwise.Claim: the order of the stones is in the following form (clockwise):−𝐴𝑥−…−𝐴1−𝑂−𝐵1−…−𝐵𝑦−where 𝐴𝑖 swaps with 𝑂 clockwise and 𝐵𝑖 swaps with 𝑂 counterclockwise. 𝑥, 𝑦 can be zero.Proof: Assuming a counterexample leads to a violation of the rule on the sequence.Let 𝐴0=𝐵0=𝑂 for convenience.Claim: The edge between 𝐵𝑦 and 𝐴𝑥 is never used.Proof:The movement of 𝑂 is limited by the number of 𝐴𝑖s and 𝐵𝑖s, so crossing the edge is impossible.Assume the edge is first used by some two stones other than 𝑂. Every combination of 𝐴𝑖 and 𝐵𝑖 leads to aviolation of the rule on the sequence. ∎