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: $\sf{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ The homeomorphism classes are:

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

Reasoning

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