A new example of a fixed-point free map of a tree-like continuum

Transcription

1 A new example of a fixed-point free map of a tree-like continuum 2nd Pan-Pacific Topology Conference Busan, South Korea Rodrigo Hernández-Gutiérrez rod@xanum.uam.mx joint work with Logan C. Hoehn Universidad Autónoma Metropolitana - Iztapalapa November 14, 2017

2 The good pictures in this talk were drawn by Logan.

3 Fixed point property Definition A space X has the fixed point property if given any continuous function f : X X, there is a point x X with f (x) = x.

4 Fixed point property Definition A space X has the fixed point property if given any continuous function f : X X, there is a point x X with f (x) = x. The most important problem in continuum theory about the fpp is: Question (Unsolved) Does every continuum that does not separate the plane have the fixed-point property?

5 Fixed point property Definition A space X has the fixed point property if given any continuous function f : X X, there is a point x X with f (x) = x. The most important problem in continuum theory about the fpp is: Question (Unsolved) Does every continuum that does not separate the plane have the fixed-point property? If X R 2 is 1-dimensional, then X does not separate the plane if and only if X is tree-like.

6 Fixed point property Definition A space X has the fixed point property if given any continuous function f : X X, there is a point x X with f (x) = x. The most important problem in continuum theory about the fpp is: Question (Unsolved) Does every continuum that does not separate the plane have the fixed-point property? If X R 2 is 1-dimensional, then X does not separate the plane if and only if X is tree-like. Question (Unsolved) Does every tree-like continuum in the plane have the fixed-point property?

7 First counterexamples In 1980, Bellamy gave the first example of a tree-like continuum which does not have the fpp. A geometric description of Bellamy s example was given by Fearnley and Wright: Minc has produced several variants of Bellamy s example with a variety of additional properties.

8 Oversteegen-Rogers examples Oversteegen and Rogers gave an example of a tree-like continuum with a fixed-point-free map (1980) using an inverse limit construction. (And then they gave another example in 1982.) X 0 X 1 X 2

9 Oversteegen-Rogers examples Oversteegen and Rogers gave an example of a tree-like continuum with a fixed-point-free map (1980) using an inverse limit construction. (And then they gave another example in 1982.) X 0 X 1 X 2 Our objective: to simplify their construction (both trees and maps)

10 Inverse limits and fixed-point free maps Let X = lim {X n, f n }. X 0 f 0 X 1 g 0 f 0 g 1 f 1 f 1 X 2 g 2 f 2 f 2 X 3 X X 0 X 1 X 1 X 3 X The function is defined by g(x)(n) = g n (x(n + 1)). g

11 Inverse limits and fixed-point free maps Let X = lim {X n, f n }. X 0 f 0 X 1 g 0 f 0 g 1 f 1 f 1 X 2 g 2 f 2 f 2 X 3 X X 0 X 1 X 1 X 3 X The function is defined by g(x)(n) = g n (x(n + 1)). We will say that the coincidence set of two functions ϕ and ψ is {x : ϕ(x) = ψ(x)}. g

12 Inverse limits and fixed-point free maps Let X = lim {X n, f n }. X 0 f 0 X 1 g 0 f 0 g 1 f 1 f 1 X 2 g 2 f 2 f 2 X 3 X X 0 X 1 X 1 X 3 X The function is defined by g(x)(n) = g n (x(n + 1)). We will say that the coincidence set of two functions ϕ and ψ is {x : ϕ(x) = ψ(x)}. ϕ, ψ have no coincidence points if their coincidence set is empty. g

13 Inverse limits and fixed-point free maps Let X = lim {X n, f n }. X 0 f 0 X 1 g 0 f 0 g 1 f 1 f 1 X 2 g 2 f 2 f 2 X 3 X X 0 X 1 X 1 X 3 X The function is defined by g(x)(n) = g n (x(n + 1)). We will say that the coincidence set of two functions ϕ and ψ is {x : ϕ(x) = ψ(x)}. ϕ, ψ have no coincidence points if their coincidence set is empty. Lemma If f 0 and g 0 have no coincidence points, then g has no fixed points. g

14 Why easier? Both the Oversteegen-Rogers example and our example are given as inverse limits of trees: T 0 f 0 T 1 g 0 f 0 g 1 f 1 f 1 T 2 g 2 f 2 f 2 T 3 X T 0 T 1 T 2 T 3 X Oversteegen & Rogers New example Degree branch point increasing with n constant 5 Valence f n increasing with n constant 6 Valence g n increasing with n constant 12 # branch points 3, 9, 33, 129,... 2, 4, 7, 13,... # end points 24, 54, 138, 438,... 7, 12, 21, 39, 75,... g

15 Tent maps and their coincidence points τ 3, τ 6 : [0, 1] [0, 1] have coincidence points {0, 2/3} {2/9, 4/9, 8/9}

16 Coincidence points Looking at how coincidence points move after applying each (both) tent maps: 2/9 4/9 2/3 0 8/9

17 Coincidence points Looking at how coincidence points move after applying each (both) tent maps: 2/9 4/9 2/3 0 8/9 In oder to avoid coincidence points:

18 Coincidence points Looking at how coincidence points move after applying each (both) tent maps: 2/9 4/9 2/3 0 8/9 In oder to avoid coincidence points: we add three stickers to each image of a coincidence point.

19 X 0 X 1 1 2/3 2/3 f 0 1/3 0 X 0 0 X 1 Picture of X 0 : main stick and two tridents.

20 X 0 X 1 near 0 We look at the image of the main stick near 0. Let ɛ > 0 be small Recall that τ 3 is red and τ 6 is blue. Blue is faster. g 0 (ɛ) = 6ɛ = τ 6 (ɛ) f 0 (ɛ) = 3ɛ = τ 3 (ɛ) f 0 (0) = 0 g 0 (0) = ɛ/2 Image of f 0 Image of g 0

21 X 0 X 1 near 0 We look at the image of the main stick near 0. Let ɛ > 0 be small Recall that τ 3 is red and τ 6 is blue. Blue is faster. g 0 (ɛ) = 6ɛ = τ 6 (ɛ) f 0 (ɛ) = 3ɛ = τ 3 (ɛ) f 0 (0) = 0 g 0 (0) = ɛ/2 Image of f 0 Image of g 0 For f 0 (red), we use the identity in the trident.

22 X 0 X 1 near 0 We look at the image of the main stick near 0. Let ɛ > 0 be small Recall that τ 3 is red and τ 6 is blue. Blue is faster. g 0 (ɛ) = 6ɛ = τ 6 (ɛ) f 0 (ɛ) = 3ɛ = τ 3 (ɛ) f 0 (0) = 0 g 0 (0) = ɛ/2 Image of f 0 Image of g 0 For f 0 (red), we use the identity in the trident. For g 0 (blue), we add 1 (mod 3) to permute the trident.

23 X 0 X 1 near 2/3 Now we look near 2/3, using the same small ɛ. Near 2/3, τ 3 (red) is smaller than τ 6 (blue).

24 X 0 X 1 near 2/3 Now we look near 2/3, using the same small ɛ. Near 2/3, τ 3 (red) is smaller than τ 6 (blue). We lift the graph of g 0 (blue) to avoid f 0.

25 X 0 X 1 near 2/3 Now we look near 2/3, using the same small ɛ. Near 2/3, τ 3 (red) is smaller than τ 6 (blue). We lift the graph of g 0 (blue) to avoid f 0. g 0 ( 2 3 ± ɛ) = τ 6( 2 3 ± ɛ) f 0 ( 2 3 ± ɛ) = τ 3( 2 3 ± ɛ) f 0 ( 2 3 ) = 0 g 0 ( 2 3 ) = ɛ/2 Image of f 0 Image of g 0

26 X 0 X 1 near 2/3 Now we look near 2/3, using the same small ɛ. Near 2/3, τ 3 (red) is smaller than τ 6 (blue). We lift the graph of g 0 (blue) to avoid f 0. g 0 ( 2 3 ± ɛ) = τ 6( 2 3 ± ɛ) f 0 ( 2 3 ± ɛ) = τ 3( 2 3 ± ɛ) f 0 ( 2 3 ) = 0 g 0 ( 2 3 ) = ɛ/2 Image of f 0 Image of g 0 The trident at 2/3 is handled in a way similar to the previous case.

27 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph.

28 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1.

29 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 ɛ) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 ɛ)

30 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 ɛ/2) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 ɛ/2)

31 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 ɛ/4) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 ɛ/4)

32 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 ɛ/8) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 ɛ/8)

33 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9)

34 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 + ɛ/8) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 + ɛ/8)

35 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 + ɛ/4) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 + ɛ/4)

36 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 + ɛ/2) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 + ɛ/2)

37 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 + ɛ) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 + ɛ)

38 X 0 X 1 near 2/9 In this case, τ 3 (red) and τ 6 (blue) approach 2/3 from different sides of the graph. Let F 2/3 i the i-th leg of the trident at 2/3 (i { 1, 0, 1}). f 0 will climb leg 0 and g 0 will climb leg 1. g 0 (2/9 + ɛ) F 2/3 1 F 2/3 0 F 2/3 1 2/3 f 0 (2/9 + ɛ) For 4/9 and 8/9 we will use other combinations of legs as we show next.

39 X 0 X 1 near 4/9 and 8/9 We are forced to follow a certain combination of legs of the trident.

40 X 0 X 1 near 4/9 and 8/9 We are forced to follow a certain combination of legs of the trident. Assume that f 1 = f 0 and g 1 = g 0 (for now) and let us see what we require:

41 X 0 X 1 near 4/9 and 8/9 We are forced to follow a certain combination of legs of the trident. Assume that f 1 = f 0 and g 1 = g 0 (for now) and let us see what we require: Notice that 2/9 and 4/9 are sibilings :

42 X 0 X 1 near 4/9 and 8/9 We are forced to follow a certain combination of legs of the trident. Assume that f 1 = f 0 and g 1 = g 0 (for now) and let us see what we require: Notice that 2/9 and 4/9 are sibilings : τ 3 (2/27) = 2/9 and τ 6 (2/27) = 4/9.

43 X 0 X 1 near 4/9 and 8/9 We are forced to follow a certain combination of legs of the trident. Assume that f 1 = f 0 and g 1 = g 0 (for now) and let us see what we require: Notice that 2/9 and 4/9 are sibilings : τ 3 (2/27) = 2/9 and τ 6 (2/27) = 4/9. Thus, ( ) ( ) ( ) ( ) f 0 = (f 0 g 1 ) = (g 0 f 1 ) = g 0 F 2/

44 X 0 X 1 near 4/9 and 8/9 We are forced to follow a certain combination of legs of the trident. Assume that f 1 = f 0 and g 1 = g 0 (for now) and let us see what we require: Notice that 2/9 and 4/9 are sibilings : τ 3 (2/27) = 2/9 and τ 6 (2/27) = 4/9. Thus, ( ) ( ) ( ) ( ) f 0 = (f 0 g 1 ) = (g 0 f 1 ) = g 0 F 2/ In a similar way, we can find the following: x f 0 (x) g 0 (x) 2/9 F 2/3 0 F 2/3 1 4/9 F 2/3 1 F 2/3 1 8/9 F 2/3 1 F 2/3 0

45 X 0 X 1, the rest Everywhere else in the main stick, f 0 is τ 3 and g 0 is τ 6.

46 X 0 X 1, the rest Everywhere else in the main stick, f 0 is τ 3 and g 0 is τ 6. We make both f 0 and g 0 constant at the tridents at 1/3 and 1 from X 1.

47 X 0 X 1, the rest Everywhere else in the main stick, f 0 is τ 3 and g 0 is τ 6. We make both f 0 and g 0 constant at the tridents at 1/3 and 1 from X 1. Thus, f 0 and g 0 have no coincidence points

48 X 0 X 1, the rest Everywhere else in the main stick, f 0 is τ 3 and g 0 is τ 6. We make both f 0 and g 0 constant at the tridents at 1/3 and 1 from X 1. Thus, f 0 and g 0 have no coincidence points (first requirement).

49 X 0 X 1, the rest Everywhere else in the main stick, f 0 is τ 3 and g 0 is τ 6. We make both f 0 and g 0 constant at the tridents at 1/3 and 1 from X 1. Thus, f 0 and g 0 have no coincidence points (first requirement). If the speaker has time, he will next explain why we need tridents at 1/3 and 1.

50 Extra tridents at 1/3 and 1: why Assume for the moment that X 0 = X 1 and defined as above

51 Extra tridents at 1/3 and 1: why Assume for the moment that X 0 = X 1 and defined as above, we will reach a contradiction.

52 Extra tridents at 1/3 and 1: why Assume for the moment that X 0 = X 1 and defined as above, we will reach a contradiction. 1/9 and 2/9 are sibilings: τ 3 (1/27) = 1/9 and τ 6 (1/27) = 2/9

53 Extra tridents at 1/3 and 1: why Assume for the moment that X 0 = X 1 and defined as above, we will reach a contradiction. 1/9 and 2/9 are sibilings: τ 3 (1/27) = 1/9 and τ 6 (1/27) = 2/9 g 0 ( 1 9 ) ( ) ( ) ( ) = (g 0 f 1 ) = (f 1 g 0 ) = f 1 F 2/

54 Extra tridents at 1/3 and 1: why Assume for the moment that X 0 = X 1 and defined as above, we will reach a contradiction. 1/9 and 2/9 are sibilings: τ 3 (1/27) = 1/9 and τ 6 (1/27) = 2/9 g 0 ( 1 9 ) ( ) ( ) ( ) = (g 0 f 1 ) = (f 1 g 0 ) = f 1 F 2/ /9 witnesses that 1/3 and 2/3 are sibilings

55 Extra tridents at 1/3 and 1: why Assume for the moment that X 0 = X 1 and defined as above, we will reach a contradiction. 1/9 and 2/9 are sibilings: τ 3 (1/27) = 1/9 and τ 6 (1/27) = 2/9 g 0 ( 1 9 ) ( ) ( ) ( ) = (g 0 f 1 ) = (f 1 g 0 ) = f 1 F 2/ /9 witnesses that 1/3 and 2/3 are sibilings ( ) 1 g 0 = (g 0 f 1 ) 3 ( ) ( ( )) 1 1 [ = f 0 g 0 f F 2/3 0 ] F0 0

56 Extra tridents at 1/3 and 1: why g 0 (1/3) F 0 0

57 Extra tridents at 1/3 and 1: why g 0 (1/3) F 0 0 But we can do this with other legs!

58 Extra tridents at 1/3 and 1: why g 0 (1/3) F 0 0 But we can do this with other legs! Since 5/9 and 8/9 are sibilings (witnessed by 5/27), we get: g 0 (1/3) F 0 1

59 Extra tridents at 1/3 and 1: why g 0 (1/3) F 0 0 But we can do this with other legs! Since 5/9 and 8/9 are sibilings (witnessed by 5/27), we get: g 0 (1/3) F 0 1 And 7/9 and 4/9 are sibilings (witnessed by 7/27), we get: g 0 (1/3) F 0 1

60 Extra tridents at 1/3 and 1: why g 0 (1/3) F 0 0 But we can do this with other legs! Since 5/9 and 8/9 are sibilings (witnessed by 5/27), we get: g 0 (1/3) F 0 1 And 7/9 and 4/9 are sibilings (witnessed by 7/27), we get: g 0 (1/3) F 0 1 A way to avoid this: add a trident at 1/3 (similar argument at 1).

61 The trees T n 1 1 2/3 0 X 0 f 0 2/3 1/3 0 X 1 f 1 5/6 2/3 1/2 1/3 1/6 0 X 2 For the next one, add tridents at the midpoint of every two consecutive tridents.

62 The trees T n 1 1 2/3 0 X 0 f 0 2/3 1/3 0 X 1 f 1 5/6 2/3 1/2 1/3 1/6 0 X 2 For the next one, add tridents at the midpoint of every two consecutive tridents. How do we continue defining the maps?

63 Γ-sets Given a pair of functions ϕ, ψ : Y Z, let Γ ϕ,ψ = {(a, b) : ϕ(a) = ψ(b)} Y Y

64 Γ-sets Given a pair of functions ϕ, ψ : Y Z, let Γ ϕ,ψ = {(a, b) : ϕ(a) = ψ(b)} Y Y Claim If ϕ, ψ : X Y are such that (ϕ (x), ψ (x)) Γ ϕ,ψ, then we have the commutative diagram ϕ Y ψ X Z ψ Y ϕ

65 Γ-sets Given a pair of functions ϕ, ψ : Y Z, let Γ ϕ,ψ = {(a, b) : ϕ(a) = ψ(b)} Y Y Claim If ϕ, ψ : X Y are such that (ϕ (x), ψ (x)) Γ ϕ,ψ, then we have the commutative diagram ϕ Y ψ X Z ψ Y ϕ Recursion: Draw Γ-set of f n and g n and see that we can follow it in the definition of f n+1 and g n+1.

66 Γ g0,f 0 T 1 T 1

67 Γ g1,f 1 T 2 T 2

68 Rest of the recursion It turns out that we can describe the Γ-sets we need, in a precise way.

69 Rest of the recursion It turns out that we can describe the Γ-sets we need, in a precise way. And it is also easy to continue the recursive definition of the functions f n, g n.

70 Rest of the recursion It turns out that we can describe the Γ-sets we need, in a precise way. And it is also easy to continue the recursive definition of the functions f n, g n. Let s see the picture of how the inverse limit looks like:

71 How the space looks like 1 1 2/3 2/3 5/6 2/3 f 0 f 1 1/2 1/3 1/3 0 X 0 0 X 1 1/6 0 X 2

72 How the space looks like

73 How the space looks like

74 How the space looks like

75 How the space looks like

76 How the space looks like

77 How the space looks like

78 How the space looks like

79 Summary of the space From our paper: The black dots indicate where the largest several triods are attached. At each grey dot, the arcs are pulled out to follow close to one leg of a nearby triod.

80 Final questions Question Is there any commuting system of coincidence-point-free functions where T 0 f 0 T 1 g 0 f 0 g 1 f 1 f 1 T 2 g 2 f 2 T 3 g 3 f 3 T 4 T 0 T 1 T 2 T 3 f 2

81 Final questions Question Is there any commuting system of coincidence-point-free functions where T 0 f 0 T 1 g 0 f 0 g 1 f 1 f 1 T 2 g 2 f 2 T 3 g 3 f 3 T 4 T 0 T 1 T 2 T 3 the number of branch points of the trees is bounded? f 2

82 Final questions Question Is there any commuting system of coincidence-point-free functions where T 0 f 0 T 1 g 0 f 0 g 1 f 1 f 1 T 2 g 2 f 2 T 3 g 3 f 3 T 4 T 0 T 1 T 2 T 3 the number of branch points of the trees is bounded? each T n is equal to the same tree? f 2

83 Final questions Question Is there any commuting system of coincidence-point-free functions where T 0 f 0 T 1 g 0 f 0 g 1 f 1 f 1 T 2 g 2 f 2 T 3 g 3 f 3 T 4 T 0 T 1 T 2 T 3 the number of branch points of the trees is bounded? each T n is equal to the same tree? each T n is equal to the simple triod? f 2

84 Final questions Question Is there any commuting system of coincidence-point-free functions where T 0 f 0 T 1 g 0 f 0 g 1 f 1 f 1 T 2 g 2 f 2 T 3 g 3 f 3 T 4 T 0 T 1 T 2 T 3 the number of branch points of the trees is bounded? each T n is equal to the same tree? each T n is equal to the simple triod? Question Let T be the simple triod. Are there continuous functions f, g : T T with no coincidence points and f g = g f? f 2

85 Thank you 대단히감사합니다