Difference between revisions of "Exercises:Saul - Algebraic Topology - 1/Exercise 1.2"

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</noinclude>Arrange the capital letters of the Roman alphabet thought of as graphs into [[homeomorphism]] classes.  
 
</noinclude>Arrange the capital letters of the Roman alphabet thought of as graphs into [[homeomorphism]] classes.  
 
====Solutions====
 
====Solutions====
−
First note I will use the font provided by {{C|\sf}}, giving the following letters: {{M|\sf{ABCDEFGHIJKLMNOPQRSTUVWXYZ} }}
+
First note I will use the font provided by {{C|\sf}}, giving the following letters: {{M|\sf{ABCDEFGHIJKLMNOPQRSTUVWXYZ} }}. I notice that {{M|\sf{G} }} here is homeomorphic to a {{M|C}}, so I have included {{M|\underline{\text{G} } }}, this represents {{M|G}} with a {{M|\top}} shape where the {{M|\sf{G} }} only has a {{M|\rceil}}. I've done both because the {{M|\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 {{M|\mathcal{Z} }} representing a {{M|\sf{Z} }} with a {{C|-}} through the middle, again due to how common this form is
 +
* I also include {{M|\text{I} }}, as this is also really common for a capital "I", along side {{M|\sf{I} }}, the font's version of a capital "I"
 +
* I also include {{M|\text{J} }}, as this is again really common, and how I write them.
 +
* I also include {{M|\sf{k} }}, as it's very common to see them as a {{M|<}} joined at the point to a {{M|\vert}}
 +
 
 
The homeomorphism classes are:
 
The homeomorphism classes are:
 
* <span style="font-size:1.5em;">{{M|\{\sf{A, R}\} }}</span>
 
* <span style="font-size:1.5em;">{{M|\{\sf{A, R}\} }}</span>
Line 10: Line 15:
 
* <span style="font-size:1.5em;">{{M|\{\sf{C, G, I, J, L, M, N, S, U, V, W, Z} \} }}</span>
 
* <span style="font-size:1.5em;">{{M|\{\sf{C, G, I, J, L, M, N, S, U, V, W, Z} \} }}</span>
 
* <span style="font-size:1.5em;">{{M|\{\sf{D, O} \} }}</span>
 
* <span style="font-size:1.5em;">{{M|\{\sf{D, O} \} }}</span>
−
* <span style="font-size:1.5em;">{{M|\{\sf{E, F, } }}{{M|\underline{\text{G} } }}{{M|\sf{, T, Y} \} }}</span>
+
* <span style="font-size:1.5em;">{{M|\{\sf{E, F, } }}{{M|\underline{\text{G} } }}{{M|\sf{, } }}{{M|\text{J} }}{{M|\sf{, T, Y} \} }}</span>
−
* <span style="font-size:1.5em;">{{M|\{\sf{H, K} \} }}</span>
+
* <span style="font-size:1.5em;">{{M|\{\sf{H, } }}{{M|\text{I} }}{{M|\sf{, K} \} }}</span>
 
* <span style="font-size:1.5em;">{{M|\{\sf{P} \} }}</span>
 
* <span style="font-size:1.5em;">{{M|\{\sf{P} \} }}</span>
 
* <span style="font-size:1.5em;">{{M|\{\sf{Q} \} }}</span>
 
* <span style="font-size:1.5em;">{{M|\{\sf{Q} \} }}</span>
−
* <span style="font-size:1.5em;">{{M|\{\sf{X} \} }}</span>
+
* <span style="font-size:1.5em;">{{M|\{\sf{k, X, } }}{{M|\mathcal{Z}\} }}</span>
 
=====Reasoning=====
 
=====Reasoning=====
 
{| class="wikitable" border="1"
 
{| class="wikitable" border="1"
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|-
 
|-
 
! {{M|\sf{G} }}
 
! {{M|\sf{G} }}
−
| {{M|\{\sf{E, F, \underline{G} }\} }} OR {{M|\{\sf{C, G}\} }}
+
| {{M|\{\sf{C, G}\} }}
 
| If you "retract" the {{M|-}} part of the {{M|\sf{G} }} and shorten the resulting {{M|\sf{C} }} like shape until it is a {{M|\sf{C} }} - clearly the inverse of this map involves extending the bottom arc of a {{M|\sf{C} }} then bending it to a right angle is also continuous, thus homeomorphism. {{M|\top}}
 
| If you "retract" the {{M|-}} part of the {{M|\sf{G} }} and shorten the resulting {{M|\sf{C} }} like shape until it is a {{M|\sf{C} }} - clearly the inverse of this map involves extending the bottom arc of a {{M|\sf{C} }} then bending it to a right angle is also continuous, thus homeomorphism. {{M|\top}}
 
| <span style="font-size:2.5em;">{{M|\sf{G} }}</span>
 
| <span style="font-size:2.5em;">{{M|\sf{G} }}</span>
 +
|-
 +
! {{M|\underline{\text{G} } }}
 +
| {{M|\{\sf{E, F, } }}{{M|\underline{\text{G} } }}{{M|\} }}
 +
| Remembering that a {{M|\underline{\text{G} } }} is supposed to represent a {{M|\sf{G} }} with a {{M|\tee}} rather than a {{M|\rceil}] bit in the middle, we note that it is homeomorphic to a {{M|\sf{T} }}, if you then take the tail (the {{M|\vert}}) of the {{M|\sf{T} }} and wrap it around in a {{M|C}} shape and we have what we represent by {{M|\underline{\text{G} } }}, this transformation of {{M|\sf{T} }} into {{M|\underline{\text{G} } }} is obviously a homeomorphism
 +
| <span style="font-size:2.5em;">{{M|\text{G} }}</span> - shown as it is similar to the G being described.
 +
|-
 +
! {{M|\sf{H} }}
 +
| {{M|\{\sf{H}\} }}
 +
|
 +
* We can rule out {{M|\{\sf{A}\} }}, {{M|\{\sf{B}\} }} and {{M|\{\sf{D}\} }} by noting the absence of holes of {{M|\sf{H} }}
 +
* We can rule out {{M|\{\sf{C,\ldots}\} }} by '''''Crafty point removal''''', notice that removing any point from {{M|\sf{C} }} results in one (the tips of the arc) or two (anywhere else) components. Yet removing one point from {{M|\sf{H} }} can yield 3 components. (For example the intersection of the horizontal and vertical parts of the {{M|\vdash}} part of {{M|\sf{H} }})
 +
* We can rule out {{M|\{\sf{E,\ldots}\} }} by '''''Two-step point removal''''', suppose {{M|{\sf{H} }\cong{\sf E}  }}, pick the two points to be where the horizontal {{C|-}} of the {{M|\sf{H} }} meets the vertical sides. Then we have 5 components. However removing any two points from {{M|\sf{E} }} leaves us with at most 4 components. Contradicting that they're homeomorphic.
 +
Thus we give {{M|\sf{H} }} its own class
 +
|
 +
|-
 +
! {{M|\sf{I} }}
 +
| {{M|\{\sf{C, G, I}\} }}
 +
| Trivial -  just take the {{M|\sf{I} }} and bend it slightly, obviously reversible and continuous each way, therefore homeomorphism
 +
|
 +
|-
 +
! {{M|\text{I} }}
 +
| {{M|\{\sf{H, } }}{{M|\text{I}\} }}
 +
| Trivial - just take the {{M|\text{I} }} rotate it and stretch it a little into shape so it's a {{M|\sf{H} }}, this is obviously continuous, as is the inverse of contracting the "height" of the {{M|\sf{H} }} then rotating it, so it's an {{M|\sf{I} }}
 +
|
 +
|-
 +
! {{M|\sf{J} }}
 +
| {{M|\{\sf{C, G, I, J }\} }}
 +
| Obviously homeomorphic to {{M|\sf{I} }} as if you take the {{M|\sf{I} }} and bend the bottom round you have a {{M|\sf{J} }}, obviously continuous, as is the inverse of "straightening out" the {{M|\sf{J} }} to yield {{M|\sf{I} }}
 +
|-
 +
! {{M|\text{J} }}
 +
| {{M|\{\sf{E, F, } }}{{M|\underline{\text{G} } }}{{M|\sf{, } }}{{M|\text{J} }}{{M|\} }}
 +
| We shall see later that a {{M|\sf{T} }} is homeomorphic to an {{M|\sf{E} }} (by rotation and extending the branches and bending them to be parallel to the "trunk"). I claim now that {{M|\sf{T} }} is homeomorphic to {{M|\text{J} }}, we do this by extending the "serifs" at the top of the {{M|\text{J} }} to form the branches at the top of the {{M|\sf{T} }} and meanwhile we shorten and straighten out the curved bottom part of the {{M|\text{J} }} (like we did with {{M|\sf{I} }} and {{M|\sf{J} }} above). Clearly this is continuous, and its inverse - of shrinking the branches of the {{M|\sf{T} }}, lengthening and bending round the "trunk" to form a {{M|\text{J} }} - is also continuous. As [[homeomorphism is an equivalence relation]] we see {{M|\text{J}\cong }}{{M|\sf{E} }}, as required.
 +
|
 +
|-
 +
! {{M|\sf{K} }}
 +
| {{M|\{\sf{H, I, K}\} }}
 +
| Upon careful inspect we see that the the {{M|\sf{K} }} is really a {{M|\sf{Y} }} rotated {{M|180^\circ}} joined onto a {{M|\vert}}. This is easily seen to be homeomorphic to a {{M|\sf{H} }} by a little rotating of the edges of the {{M|\sf{Y} }} part and a little stretching, as usual, the inverse of this map is also obviously continuous.
 +
|
 +
|-
 +
! {{M|\sf{k} }}
 +
| {{M|\{\sf{k}\} }}
 +
| Removing the point where the {{M|<}} meets the {{M|\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.
 +
|
 +
|-
 +
! {{M|\sf{L} }}
 +
| {{M|\{\sf{C, G, I, J, L}\} }}
 +
| If you take the {{M|\sf{I} }}, extend it, then put a right-angled kink in the bottom we obtain an {{M|\sf{L} }}, this is clearly continuous, the inverse, which straightens out then shortens is also easily seen to be continuous, thus homeomorphic.
 +
|
 +
|-
 +
! {{M|\sf{M} }}
 +
| {{M|\{\sf{C, G, I, J, L, M}\} }}
 +
| Take the {{M|I}}, rotate it so its horizontal, stretch it, then put some kinks in it to make the {{M|\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
 +
|
 +
|-
 +
! {{M|\sf{N} }}
 +
| {{M|\{\sf{C, G, I, J, L, M, N}\} }}
 +
| Take the {{M|I}}, rotate it so its horizontal, stretch it, then put some kinks in it to make the {{M|\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
 +
|
 +
|-
 +
! {{M|\sf{O} }}
 +
| {{M|\{\sf{D, O}\} }}
 +
| Take the {{M|\mathsf{D} }} and curve the straight edge slightly until it becomes an {{M|\sf{O} }} (they look to be the same size, but any "nudging" around to make it identical to the {{M|\sf{O} }} is a homeomorphism itself), the inverse of the nudging homeomorphism (if required) followed by the straightening that curved edge yields a {{M|\sf{D} }} and is also obviously continuous, thus a homeomorphism
 +
|-
 +
! {{M|\sf{P} }}
 +
| {{M|\{\sf{P}\} }}
 +
| {{M|\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 {{M|\{\sf{B}\} }}. We're left with the {{M|\{\sf{A}\} }} and the {{M|\{\sf{D, O}\} }} classes.
 +
* Take {{M|\sf{O} }} and remove one point, we are left with one component. Take {{M|\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 {{M|\sf{A} }}, suppose it is homeomorphic (via {{M|f}}) to {{M|\sf{P} }}, then take {{M|p\in{\sf A} }} to be the point where the {{M|/}} part means the {{M|-}} part; this splits {{M|\sf{A} }} into two components. By hypothesis {{M|\mathcal{O}(f(p))}} must also be 2, so pick {{M|f(p)}} to be anywhere on the stem on the {{M|\sf{P} }} - up to and including where the loop and the stem meet, but not the end of the stem, then that splits {{M|\sf{P} }} into two components also.
 +
** Indeed any other point would split {{M|\sf{P} }} into only one connected component
 +
** Now delete the point, say {{M|q}} where the {{M|\backslash}} meets the {{M|-}}. {{M|\sf{A} }} now has 4 connected components.
 +
*** However for all {{M|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 {{M|\sf{A} }} and {{M|\sf{P} }} are homeomorphic.
 +
** So {{M|\sf{P} }} does not belong in {{M|\{\sf{A} \} }}
 +
Thus we give {{M|\sf{P} }} its own class
 +
|
 +
|-
 +
! {{M|\sf{Q} }}
 
|}
 
|}
 +
====[[/Lemmas|Lemmas]]====
 +
{{Begin Notebox}}Lemmas:{{Begin Notebox Content}}
 +
{{:Exercises:Saul - Algebraic Topology - 1/Exercise 1.2/Lemmas}}{{End Notebox}}{{End Notebox Content}}
 
<noinclude>
 
<noinclude>
 
==Notes==
 
==Notes==

Revision as of 20:50, 17 January 2017

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[Note 1] point removal[Note 2] - 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:

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.

Notes

  1. ↑ T
  2. ↑ H
  3. ↑ This is a slight abuse of notation for a restriction, for a restriction we would have [ilmath]f\vert_{X-\{x\} }:X-\{x\}\rightarrow Y[/ilmath] - notice the codomain has changed to [ilmath]Y-\{f(x)\} [/ilmath]

References


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