Exercises:Saul - Algebraic Topology - 1/Exercise 1.2

Exercises

Exercise 1.2

Arrange the capital letters of the Roman alphabet thought of as graphs into homeomorphism classes.

Solutions

First note I will use the font provided by \sf, giving the following letters: [ilmath]\sf{ABCDEFGHIJKLMNOPQRSTUVWXYZ} [/ilmath] The homeomorphism classes are:

• [ilmath]\{\sf{A, R}\} [/ilmath]
• [ilmath]\{\sf{B} \} [/ilmath]
• [ilmath]\{\sf{C, G, I, J, L, M, N, S, U, V, W, Z} \} [/ilmath]
• [ilmath]\{\sf{D, O} \} [/ilmath]
• [ilmath]\{\sf{E, F, } [/ilmath][ilmath]\underline{\text{G} } [/ilmath][ilmath]\sf{, T, Y} \} [/ilmath]
• [ilmath]\{\sf{H, K} \} [/ilmath]
• [ilmath]\{\sf{P} \} [/ilmath]
• [ilmath]\{\sf{Q} \} [/ilmath]
• [ilmath]\{\sf{X} \} [/ilmath]

Reasoning

Letter	Class so far	Reasoning	Comment
[ilmath]\sf{A} [/ilmath]	[ilmath]\{\sf{A}\} [/ilmath]	There are no classes yet. So [ilmath]\sf{A} [/ilmath] founds one
[ilmath]\sf{B} [/ilmath]	[ilmath]\{\sf{B}\} [/ilmath]	Crafty point removal - by removing a point from the bottom right (\) edge of the [ilmath]\sf{A} [/ilmath] or the bottom left (/) edge we end up with 2 pathwise connected components (herein: components). However removing any point from [ilmath]\sf{B} [/ilmath] results in one component.
[ilmath]\sf{C} [/ilmath]	[ilmath]\{\sf{C}\} [/ilmath]	There are no loops in [ilmath]\sf{C} [/ilmath] (it is obviously homeomorphic to just a line ([ilmath]\vert[/ilmath]) say, due to the absence of holes (of which [ilmath]\sf{A} [/ilmath] has one and [ilmath]\sf{B} [/ilmath] has two - see fundamental group) we must conclude [ilmath]\sf{C} [/ilmath] is none of the existing groups and founds its own.
[ilmath]\sf{D} [/ilmath]	[ilmath]\{\sf{D}\} [/ilmath]	By Crafty point removal we see removing any point from [ilmath]\sf{D} [/ilmath] leaves one component, where as a crafty choice of point can leave [ilmath]\sf{A} [/ilmath] with 2 components. By noticing the fundamental group of [ilmath]\sf{B} [/ilmath] would be [ilmath]\mathbb{Z}*\mathbb{Z} [/ilmath] and the fundamental group of [ilmath]\sf{D} [/ilmath] will be that of the circle, [ilmath]\mathbb{Z} [/ilmath] we see that [ilmath]\sf{D} [/ilmath] is not homeomorphic to [ilmath]\sf{B} [/ilmath]
[ilmath]\sf{E} [/ilmath]	[ilmath]\{\sf{E}\} [/ilmath]	By Crafty point removal we can have 3 components. No member of any class so far has this property. Therefore [ilmath]\sf{E} [/ilmath] must have its own class
[ilmath]\sf{F} [/ilmath]	[ilmath]\{\sf{E, F}\} [/ilmath]	A continuous map that doubles the length of the bottom [ilmath]\vert[/ilmath] of the [ilmath]F[/ilmath] and bends the latter half of it at a right angle to the right is easily seen to be an [ilmath]\sf{E} [/ilmath] and the inverse map simply shortens the [ilmath]\lfloor[/ilmath]-like part of the [ilmath]\sf{E} [/ilmath] and "unkinks" the right angle.
[ilmath]\sf{G} [/ilmath]	[ilmath]\{\sf{E, F, \underline{G} }\} [/ilmath] OR [ilmath]\{\sf{C, G}\} [/ilmath]	If you "retract" the [ilmath]-[/ilmath] part of the [ilmath]\sf{G} [/ilmath] and shorten the resulting [ilmath]\sf{C} [/ilmath] like shape until it is a [ilmath]\sf{C} [/ilmath] - clearly the inverse of this map involves extending the bottom arc of a [ilmath]\sf{C} [/ilmath] then bending it to a right angle is also continuous, thus homeomorphism. [ilmath]\top[/ilmath]	[ilmath]\sf{G} [/ilmath]