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]. I notice that [ilmath]\sf{G} [/ilmath] here is homeomorphic to a [ilmath]C[/ilmath], so I have included [ilmath]\underline{\text{G} } [/ilmath], this represents [ilmath]G[/ilmath] with a [ilmath]\top[/ilmath] shape where the [ilmath]\sf{G} [/ilmath] only has a [ilmath]\rceil[/ilmath]. I've done both because the [ilmath]\top[/ilmath]-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 [ilmath]\mathcal{Z} [/ilmath] representing a [ilmath]\sf{Z} [/ilmath] with a - through the middle, again due to how common this form is
• I also include [ilmath]\text{I} [/ilmath], as this is also really common for a capital "I", along side [ilmath]\sf{I} [/ilmath], the font's version of a capital "I"
• I also include [ilmath]\text{J} [/ilmath], as this is again really common, and how I write them.
• I also include [ilmath]\sf{k} [/ilmath], as it's very common to see them as a [ilmath]<[/ilmath] joined at the point to a [ilmath]\vert[/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{, } [/ilmath][ilmath]\text{J} [/ilmath][ilmath]\sf{, T, Y} \} [/ilmath]
• [ilmath]\{\sf{H, } [/ilmath][ilmath]\text{I} [/ilmath][ilmath]\sf{, K} \} [/ilmath]
• [ilmath]\{\sf{P} \} [/ilmath]
• [ilmath]\{\sf{Q} \} [/ilmath]
• [ilmath]\{\sf{k, X, } [/ilmath][ilmath]\mathcal{Z}\} [/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{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]
[ilmath]\underline{\text{G} } [/ilmath]	[ilmath]\{\sf{E, F, } [/ilmath][ilmath]\underline{\text{G} } [/ilmath][ilmath]\} [/ilmath]	\rceil}] bit in the middle, we note that it is homeomorphic to a [ilmath]\sf{T} [/ilmath], if you then take the tail (the [ilmath]\vert[/ilmath]) of the [ilmath]\sf{T} [/ilmath] and wrap it around in a [ilmath]C[/ilmath] shape and we have what we represent by [ilmath]\underline{\text{G} } [/ilmath], this transformation of [ilmath]\sf{T} [/ilmath] into [ilmath]\underline{\text{G} } [/ilmath] is obviously a homeomorphism	[ilmath]\text{G} [/ilmath] - shown as it is similar to the G being described.
[ilmath]\sf{H} [/ilmath]	[ilmath]\{\sf{H}\} [/ilmath]	We can rule out [ilmath]\{\sf{A}\} [/ilmath], [ilmath]\{\sf{B}\} [/ilmath] and [ilmath]\{\sf{D}\} [/ilmath] by noting the absence of holes of [ilmath]\sf{H} [/ilmath] We can rule out [ilmath]\{\sf{C,\ldots}\} [/ilmath] by Crafty point removal, notice that removing any point from [ilmath]\sf{C} [/ilmath] results in one (the tips of the arc) or two (anywhere else) components. Yet removing one point from [ilmath]\sf{H} [/ilmath] can yield 3 components. (For example the intersection of the horizontal and vertical parts of the [ilmath]\vdash[/ilmath] part of [ilmath]\sf{H} [/ilmath]) We can rule out [ilmath]\{\sf{E,\ldots}\} [/ilmath] by Two-step point removal, suppose [ilmath]{\sf{H} }\cong{\sf E} [/ilmath], pick the two points to be where the horizontal - of the [ilmath]\sf{H} [/ilmath] meets the vertical sides. Then we have 5 components. However removing any two points from [ilmath]\sf{E} [/ilmath] leaves us with at most 4 components. Contradicting that they're homeomorphic. Thus we give [ilmath]\sf{H} [/ilmath] its own class
[ilmath]\sf{I} [/ilmath]	[ilmath]\{\sf{C, G, I}\} [/ilmath]	Trivial - just take the [ilmath]\sf{I} [/ilmath] and bend it slightly, obviously reversible and continuous each way, therefore homeomorphism
[ilmath]\text{I} [/ilmath]	[ilmath]\{\sf{H, } [/ilmath][ilmath]\text{I}\} [/ilmath]	Trivial - just take the [ilmath]\text{I} [/ilmath] rotate it and stretch it a little into shape so it's a [ilmath]\sf{H} [/ilmath], this is obviously continuous, as is the inverse of contracting the "height" of the [ilmath]\sf{H} [/ilmath] then rotating it, so it's an [ilmath]\sf{I} [/ilmath]
[ilmath]\sf{J} [/ilmath]	[ilmath]\{\sf{C, G, I, J }\} [/ilmath]	Obviously homeomorphic to [ilmath]\sf{I} [/ilmath] as if you take the [ilmath]\sf{I} [/ilmath] and bend the bottom round you have a [ilmath]\sf{J} [/ilmath], obviously continuous, as is the inverse of "straightening out" the [ilmath]\sf{J} [/ilmath] to yield [ilmath]\sf{I} [/ilmath]
[ilmath]\text{J} [/ilmath]	[ilmath]\{\sf{E, F, } [/ilmath][ilmath]\underline{\text{G} } [/ilmath][ilmath]\sf{, } [/ilmath][ilmath]\text{J} [/ilmath][ilmath]\} [/ilmath]	We shall see later that a [ilmath]\sf{T} [/ilmath] is homeomorphic to an [ilmath]\sf{E} [/ilmath] (by rotation and extending the branches and bending them to be parallel to the "trunk"). I claim now that [ilmath]\sf{T} [/ilmath] is homeomorphic to [ilmath]\text{J} [/ilmath], we do this by extending the "serifs" at the top of the [ilmath]\text{J} [/ilmath] to form the branches at the top of the [ilmath]\sf{T} [/ilmath] and meanwhile we shorten and straighten out the curved bottom part of the [ilmath]\text{J} [/ilmath] (like we did with [ilmath]\sf{I} [/ilmath] and [ilmath]\sf{J} [/ilmath] above). Clearly this is continuous, and its inverse - of shrinking the branches of the [ilmath]\sf{T} [/ilmath], lengthening and bending round the "trunk" to form a [ilmath]\text{J} [/ilmath] - is also continuous. As homeomorphism is an equivalence relation we see [ilmath]\text{J}\cong [/ilmath][ilmath]\sf{E} [/ilmath], as required.
[ilmath]\sf{K} [/ilmath]	[ilmath]\{\sf{H, I, K}\} [/ilmath]	Upon careful inspect we see that the the [ilmath]\sf{K} [/ilmath] is really a [ilmath]\sf{Y} [/ilmath] rotated [ilmath]180^\circ[/ilmath] joined onto a [ilmath]\vert[/ilmath]. This is easily seen to be homeomorphic to a [ilmath]\sf{H} [/ilmath] by a little rotating of the edges of the [ilmath]\sf{Y} [/ilmath] part and a little stretching, as usual, the inverse of this map is also obviously continuous.
[ilmath]\sf{k} [/ilmath]	[ilmath]\{\sf{k}\} [/ilmath]	Removing the point where the [ilmath]<[/ilmath] meets the [ilmath]\vert[/ilmath] 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.
[ilmath]\sf{L} [/ilmath]	[ilmath]\{\sf{C, G, I, J, L}\} [/ilmath]	If you take the [ilmath]\sf{I} [/ilmath], extend it, then put a right-angled kink in the bottom we obtain an [ilmath]\sf{L} [/ilmath], this is clearly continuous, the inverse, which straightens out then shortens is also easily seen to be continuous, thus homeomorphic.
[ilmath]\sf{M} [/ilmath]	[ilmath]\{\sf{C, G, I, J, L, M}\} [/ilmath]	Take the [ilmath]I[/ilmath], rotate it so its horizontal, stretch it, then put some kinks in it to make the [ilmath]\sf{M} [/ilmath] 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
[ilmath]\sf{N} [/ilmath]	[ilmath]\{\sf{C, G, I, J, L, M, N}\} [/ilmath]	Take the [ilmath]I[/ilmath], rotate it so its horizontal, stretch it, then put some kinks in it to make the [ilmath]\sf{N} [/ilmath] 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
[ilmath]\sf{O} [/ilmath]	[ilmath]\{\sf{D, O}\} [/ilmath]	Take the [ilmath]\mathsf{D} [/ilmath] and curve the straight edge slightly until it becomes an [ilmath]\sf{O} [/ilmath] (they look to be the same size, but any "nudging" around to make it identical to the [ilmath]\sf{O} [/ilmath] is a homeomorphism itself), the inverse of the nudging homeomorphism (if required) followed by the straightening that curved edge yields a [ilmath]\sf{D} [/ilmath] and is also obviously continuous, thus a homeomorphism
[ilmath]\sf{P} [/ilmath]	[ilmath]\{\sf{P}\} [/ilmath]	[ilmath]\sf{P} [/ilmath] 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 [ilmath]\{\sf{B}\} [/ilmath]. We're left with the [ilmath]\{\sf{A}\} [/ilmath] and the [ilmath]\{\sf{D, O}\} [/ilmath] classes. Take [ilmath]\sf{O} [/ilmath] and remove one point, we are left with one component. Take [ilmath]\sf{P} [/ilmath] 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 [ilmath]\sf{A} [/ilmath], suppose it is homeomorphic (via [ilmath]f[/ilmath]) to [ilmath]\sf{P} [/ilmath], then take [ilmath]p\in{\sf A} [/ilmath] to be the point where the [ilmath]/[/ilmath] part means the [ilmath]-[/ilmath] part; this splits [ilmath]\sf{A} [/ilmath] into two components. By hypothesis [ilmath]\mathcal{O}(f(p))[/ilmath] must also be 2, so pick [ilmath]f(p)[/ilmath] to be anywhere on the stem on the [ilmath]\sf{P} [/ilmath] - up to and including where the loop and the stem meet, but not the end of the stem, then that splits [ilmath]\sf{P} [/ilmath] into two components also. Indeed any other point would split [ilmath]\sf{P} [/ilmath] into only one connected component Now delete the point, say [ilmath]q[/ilmath] where the [ilmath]\backslash[/ilmath] meets the [ilmath]-[/ilmath]. [ilmath]\sf{A} [/ilmath] now has 4 connected components. However for all [ilmath]f(q)\in {\sf P}-\{f(p)\} [/ilmath] we only have 2, or at most 3 path components. For none of these points can we have 4 This contradicts that [ilmath]\sf{A} [/ilmath] and [ilmath]\sf{P} [/ilmath] are homeomorphic. So [ilmath]\sf{P} [/ilmath] does not belong in [ilmath]\{\sf{A} \} [/ilmath] Thus we give [ilmath]\sf{P} [/ilmath] its own class
[ilmath]\sf{Q} [/ilmath]

Lemmas

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

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

Point removal

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

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

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

Two-step point removal

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

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

We conclude:

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

Which will be helpful for reaching contradictions.