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: $\sf{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$. I notice that $\sf{G}$ here is homeomorphic to a $C$, so I have included $\underline{\text{G} }$, this represents $G$ with a $\top$ shape where the $\sf{G}$ only has a $\rceil$. I've done both because the $\top$-like form is so common it is worth doing and apparently a sans-serif font lacks this "T" part of the G.

• I also include $\mathcal{Z}$ representing a $\sf{Z}$ with a - through the middle, again due to how common this form is
• I also include $\text{I}$, as this is also really common for a capital "I", along side $\sf{I}$, the font's version of a capital "I"
• I also include $\text{J}$, as this is again really common, and how I write them.
• I also include $\sf{k}$, as it's very common to see them as a $<$ joined at the point to a $\vert$

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{, }$$\text{J}$$\sf{, T, Y} \}$
• $\{\sf{H, }$$\text{I}$$\sf{, K} \}$
• $\{\sf{P} \}$
• $\{\sf{Q} \}$
• $\{\sf{k, X, }$$\mathcal{Z}\}$

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.
$\sf{G}$	$\{\sf{C, G}\}$	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. $\top$	$\sf{G}$
$\underline{\text{G} }$	$\{\sf{E, F, }$$\underline{\text{G} }$$\}$	\rceil}] bit in the middle, we note that it is homeomorphic to a $\sf{T}$, if you then take the tail (the $\vert$) of the $\sf{T}$ and wrap it around in a $C$ shape and we have what we represent by $\underline{\text{G} }$, this transformation of $\sf{T}$ into $\underline{\text{G} }$ is obviously a homeomorphism	$\text{G}$ - shown as it is similar to the G being described.
$\sf{H}$	$\{\sf{H}\}$	We can rule out $\{\sf{A}\}$, $\{\sf{B}\}$ and $\{\sf{D}\}$ by noting the absence of holes of $\sf{H}$ We can rule out $\{\sf{C,\ldots}\}$ by Crafty point removal, notice that removing any point from $\sf{C}$ results in one (the tips of the arc) or two (anywhere else) components. Yet removing one point from $\sf{H}$ can yield 3 components. (For example the intersection of the horizontal and vertical parts of the $\vdash$ part of $\sf{H}$) We can rule out $\{\sf{E,\ldots}\}$ by Two-step point removal, suppose ${\sf{H} }\cong{\sf E}$, pick the two points to be where the horizontal - of the $\sf{H}$ meets the vertical sides. Then we have 5 components. However removing any two points from $\sf{E}$ leaves us with at most 4 components. Contradicting that they're homeomorphic. Thus we give $\sf{H}$ its own class
$\sf{I}$	$\{\sf{C, G, I}\}$	Trivial - just take the $\sf{I}$ and bend it slightly, obviously reversible and continuous each way, therefore homeomorphism
$\text{I}$	$\{\sf{H, }$$\text{I}\}$	Trivial - just take the $\text{I}$ rotate it and stretch it a little into shape so it's a $\sf{H}$, this is obviously continuous, as is the inverse of contracting the "height" of the $\sf{H}$ then rotating it, so it's an $\sf{I}$
$\sf{J}$	$\{\sf{C, G, I, J }\}$	Obviously homeomorphic to $\sf{I}$ as if you take the $\sf{I}$ and bend the bottom round you have a $\sf{J}$, obviously continuous, as is the inverse of "straightening out" the $\sf{J}$ to yield $\sf{I}$
$\text{J}$	$\{\sf{E, F, }$$\underline{\text{G} }$$\sf{, }$$\text{J}$$\}$	We shall see later that a $\sf{T}$ is homeomorphic to an $\sf{E}$ (by rotation and extending the branches and bending them to be parallel to the "trunk"). I claim now that $\sf{T}$ is homeomorphic to $\text{J}$, we do this by extending the "serifs" at the top of the $\text{J}$ to form the branches at the top of the $\sf{T}$ and meanwhile we shorten and straighten out the curved bottom part of the $\text{J}$ (like we did with $\sf{I}$ and $\sf{J}$ above). Clearly this is continuous, and its inverse - of shrinking the branches of the $\sf{T}$, lengthening and bending round the "trunk" to form a $\text{J}$ - is also continuous. As homeomorphism is an equivalence relation we see $\text{J}\cong$$\sf{E}$, as required.
$\sf{K}$	$\{\sf{H, I, K}\}$	Upon careful inspect we see that the the $\sf{K}$ is really a $\sf{Y}$ rotated $180^\circ$ joined onto a $\vert$. This is easily seen to be homeomorphic to a $\sf{H}$ by a little rotating of the edges of the $\sf{Y}$ part and a little stretching, as usual, the inverse of this map is also obviously continuous.
$\sf{k}$	$\{\sf{k}\}$	Removing the point where the $<$ meets the $\vert$ we have 4 components. No other letter done so far has four components with one point removed (highest is 3) - therefore must have a new class made.
$\sf{L}$	$\{\sf{C, G, I, J, L}\}$	If you take the $\sf{I}$, extend it, then put a right-angled kink in the bottom we obtain an $\sf{L}$, this is clearly continuous, the inverse, which straightens out then shortens is also easily seen to be continuous, thus homeomorphic.
$\sf{M}$	$\{\sf{C, G, I, J, L, M}\}$	Take the $I$, rotate it so its horizontal, stretch it, then put some kinks in it to make the $\sf{M}$ shape, obviously continuous, the inverse of unkinking, shortening, then rotating so its vertical again is also easily seen to be continuous, thus we have a homeomorphism
$\sf{N}$	$\{\sf{C, G, I, J, L, M, N}\}$	Take the $I$, rotate it so its horizontal, stretch it, then put some kinks in it to make the $\sf{N}$ shape, obviously continuous, the inverse of unkinking, shortening, then rotating so its vertical again is also easily seen to be continuous, thus we have a homeomorphism
$\sf{O}$	$\{\sf{D, O}\}$	Take the $\mathsf{D}$ and curve the straight edge slightly until it becomes an $\sf{O}$ (they look to be the same size, but any "nudging" around to make it identical to the $\sf{O}$ is a homeomorphism itself), the inverse of the nudging homeomorphism (if required) followed by the straightening that curved edge yields a $\sf{D}$ and is also obviously continuous, thus a homeomorphism
$\sf{P}$	$\{\sf{P}\}$	$\sf{P}$ clearly has a hole, so we can rule out all the classes of hole-less letters. It only has one hole, so we can rule out $\{\sf{B}\}$. We're left with the $\{\sf{A}\}$ and the $\{\sf{D, O}\}$ classes. Take $\sf{O}$ and remove one point, we are left with one component. Take $\sf{P}$ and remove any point (except the end) of the "stem" coming off the "o" part. We now have 2 components. So these cannot be homeomorphic. Take $\sf{A}$, suppose it is homeomorphic (via $f$) to $\sf{P}$, then take $p\in{\sf A}$ to be the point where the $/$ part means the $-$ part; this splits $\sf{A}$ into two components. By hypothesis $\mathcal{O}(f(p))$ must also be 2, so pick $f(p)$ to be anywhere on the stem on the $\sf{P}$ - up to and including where the loop and the stem meet, but not the end of the stem, then that splits $\sf{P}$ into two components also. Indeed any other point would split $\sf{P}$ into only one connected component Now delete the point, say $q$ where the $\backslash$ meets the $-$. $\sf{A}$ now has 4 connected components. However for all $f(q)\in {\sf P}-\{f(p)\}$ we only have 2, or at most 3 path components. For none of these points can we have 4 This contradicts that $\sf{A}$ and $\sf{P}$ are homeomorphic. So $\sf{P}$ does not belong in $\{\sf{A} \}$ Thus we give $\sf{P}$ its own class
$\sf{Q}$

Lemmas

[Expand]Lemmas:

Caveat:The following listed here are for reference to someone looking at the exercises only. They are done from memory and have no reference (at the time of writing) - use at your own peril.

Homeomorphisms and point removal

Suppose $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ are topological spaces and $f:X\rightarrow Y$ is a homeomorphism between them (so $X\cong Y$), then for any $x\in X$ we have:

• $f\vert_{X-\{x\} }:X-\{x\}\rightarrow Y-\{f(x)\}$ is a homeomorphism we of course consider these spaces with the subspace topology

Point removal

Let $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces and let $f:X\rightarrow Y$ be any map (possibly continuous) between them. Then

• $f:X\rightarrow Y$ being a homeomorphism implies $\forall p\in X[\mathcal{O}(p)\eq\mathcal{O}(f(p))]$ - where $\mathcal{O}(p)$ is the number of path-connected components of the space $X-\{p\}$
  • We may write $\mathcal{O}(p,X)$ to mean $p$ removed from the space $X$, this makes things clearer when dealing with subsets.

Specifically, by contrapositive, if $\exists p\in X[\mathcal{O}(p)\neq\mathcal{O}(f(p))]$ then $f:X\rightarrow Y$ is not a homeomorphism.

Two-step point removal

This is a corollary of the two claims above. Two-step point removal means that:

• Suppose $X\cong_f Y$, then $X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\}$ (by "homeomorphisms and point removal")
• Suppose $X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\}$ are indeed homeomorphic, then $\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})]$

We conclude:

• $X\cong_f Y$ implies $\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})]$

Which will be helpful for reaching contradictions.