Difference between revisions of "Exercises:Mond - Topology - 1/Question 7"

From Maths
Jump to: navigation, search
m (→‎Solution)
(Forgot chunk of proof)
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
<noinclude>
 
<noinclude>
 +
__TOC__
 
==Section B==
 
==Section B==
 
===Question 7===
 
===Question 7===
Line 20: Line 21:
 
** We take this diagram as showing morphisms in the [[TOP]] category, meaning all arrows shown represent continuous maps. (Obviously...)
 
** We take this diagram as showing morphisms in the [[TOP]] category, meaning all arrows shown represent continuous maps. (Obviously...)
 
* Lastly, we will show that {{M|\bar{f} }} is a [[homeomorphism]] using the [[compact-to-Hausdorff theorem]]
 
* Lastly, we will show that {{M|\bar{f} }} is a [[homeomorphism]] using the [[compact-to-Hausdorff theorem]]
−
'''Solution:'''
+
=====Solution body=====
 +
First we must show the requirements for applying {{link|passing to the quotient|topology}} are satisfied.
 +
* We know already the maps involved are continuous and that {{M|\pi}} is a {{link|quotient map|topology}}. We only need to show:
 +
** {{M|f}} is constant on the {{plural|fibre|s}} of {{M|\pi}}, which is equivalent to:
 +
*** {{M|1=\forall x,y\in D^2[\pi(x)=\pi(y)\implies f(x)=f(y)]}}
 +
* Let us show this remaining condition:
 +
** Let {{M|x,y\in D^2}} be given.
 +
*** Suppose {{M|\pi(x)\ne\pi(y)}}, then by the nature of [[logical implication]] the implication is true regardless of {{M|f(x)}} and {{M|f(y)}}'s equality. We're done in this case.
 +
*** Suppose {{M|1=\pi(x)=\pi(y)}}, we must show that in this case {{M|1=f(x)=y(y)}}.
 +
**** Suppose {{M|x\in D^2-\partial D^2}} (meaning {{M|x\in D^2}} but {{M|x\notin \partial D^2}}, ie {{M|-}} denotes [[relative complement]])
 +
***** In this case we must have {{M|1=x=y}}, as otherwise we'd not have {{M|1=\pi(x)=\pi(y)}} (for {{M|x\in D^2-\partial D^2}} we have {{M|1=\pi(x)=[x]=\{x\} }}, that is that the {{plural|equivalence class|es}} are singletons. So if {{M|1=\pi(x)=\pi(y)}} we must have {{M|1=\pi(y)=[y]=\{x\}=[x]=\pi(x) }}; so {{M|y}} can only be {{M|x}})
 +
***** If {{M|1=x=y}} then by the nature of {{M|f}} being a [[function]] we must have {{M|1=f(x)=f(y)}}, we're done in this case
 +
**** Suppose {{M|x\in \partial D^2}} (the only case not covered) and {{M|1=\pi(x)=\pi(y)}}, we must show {{M|1=f(x)=f(y)}}
 +
***** Clearly if {{M|x\in\partial D^2}} and {{M|1=\pi(x)=\pi(y)}} we must have {{M|y\in\partial D^2}}.
 +
****** {{M|E(x)}} is mapped to the boundary/rim of {{M|H}}, as is {{M|E(y)}} and {{M|1=f'(\text{any point on the rim of }H)=(0,0,1)\in\mathbb{R}^3}}
 +
****** Thus {{M|1=f'(E(x))=f'(E(y))}}, but {{M|f'(E(x))}} is the very definition of {{M|f(x)}}, so clearly:
 +
******* {{M|1=f(x)=f(y)}} as required.
 +
We may now apply the {{link|passing to the quotient|topology}} theorem. This yields:
 +
* A [[continuous]] map, {{M|\bar{f}:D^2/\sim\rightarrow\mathbb{S}^2}} where {{M|1=f=\bar{f}\circ\pi}}
 +
<div style="float:right;overflow:hidden;">
 +
{| class="wikitable" border="1"
 +
|-
 +
| style="font-size:1.2em;" | <center><m>\xymatrix{ D^2 \ar[d]_\pi \ar[r]^E \ar@/^1.5pc/[rr]^{f} & H \ar[r]^{f'} & \mathbb{S}^2 \\ \frac{D^2}{\sim} \ar@{.>}[urr]_{\bar{f} } }</m></center>
 +
|-
 +
! Commutative diagram of situation<br/>(all maps are continuous)
 +
|}
 +
</div><!--
 +
-->We now have the situation of the diagram on the right, restated from above for convenience.
 +
 
 +
In order to apply the [[compact-to-Hausdorff theorem]] and show {{M|\bar{f} }} is a [[homeomorphism]] we must show it is [[continuous]] and [[bijective]]. We already have continuity (as a result of the passing-to-the-quotient theorem), we must show it is bijective.
 +
 
 +
We must show {{M|\bar{f} }} is both [[surjective]] and [[injective]]:
 +
# [[Surjectivity]]: We can get this from the definition of {{M|\bar{f} }}, recall on the [[passing to the quotient (function)]] page that:
 +
#* if {{M|f}} is surjective then {{M|\bar{f} }} (or {{M|\tilde{f} }} as the induced function is on that page) is surjective also
 +
#** I've already done it "the long way" once in this assignment, and I hope it is not frowned upon if I decline to do it again.
 +
# [[Injectivity]]: This follows a similar to gist as to what we've done already to show we could apply "passing to the quotient" and for the other questions where we've had to show a factored map is injective. As before:
 +
#* Let {{M|x,y\in\frac{D^2}{\sim} }} be given.
 +
#** Suppose {{M|1=\bar{f}(x)\ne\bar{f}(y)}} - then by the nature of [[logical implication]] we do not care about the RHS and are done regardless of {{M|x}} and {{M|y}}'s equality
 +
#*** Once again I note we must really have {{M|x\ne y}} as if {{M|1=x=y}} then by definition of {{M|\bar{f} }} being a function we must also have {{M|1=\bar{f}(x)=\bar{f}(y)}}, anyway!
 +
#** Suppose that {{M|1=\bar{f}(x)=\bar{f}(y)}}, we must show that in this case {{M|1=x=y}}.
 +
#*** Note that by [[surjectivity]] of {{M|\pi}} that: {{M|1=\exists a\in D^2[\pi(a)=x]}} and {{M|1=\exists b\in D^2[\pi(b)=y]}}, so {{M|1=\bar{f}(x)=\bar{f}(\pi(a))}} and {{M|1=\bar{f}(y)=\bar{f}(\pi(b))}}, also, as {{M|\bar{f} }} was the result of factoring, we have {{M|1=f=\bar{f}\circ\pi }}, so we see {{M|1=\bar{f}(\pi(a))=f(a)}} and {{M|1=\bar{f}(\pi(b))=f(b)}}, since {{M|1=\bar{f}(x)=\bar{f}(y)}} we get {{M|1=f(a)=f(b)}} and {{M|1=\bar{f}(\pi(a))=\bar{f}(\pi(b))}} also.
 +
#**** We now have 2 cases, {{M|a\in D^2-\partial D^2}} and {{M|a\in \partial D^2}} respectively:
 +
#****# Suppose {{M|a\in D^2-\partial D^2}}
 +
#****#* As we have {{M|1=f(a)=f(b)}} we must have {{M|1=b=a}}, as if {{M|b\ne a}} then {{M|f(b)\ne f(a)}}, because {{M|f\big\vert_{D^2-\partial D^2}:(D^2-\partial D^2)\rightarrow\mathbb{S}^2}} is injective<ref group="Note">Actually:
 +
* {{M|f\big\vert_{D^2-\partial D^2}:(D^2-\partial D^2)\rightarrow(\mathbb{S}^2-\{(0,0,1)\}) }} is bijective in fact!</ref> by construction
 +
#****#** If {{M|1=b=a}} then {{M|1=y=\pi(b)=\pi(a)=x}} so {{M|1=y=x}} as required (this is easily recognised as {{M|1=x=y}})
 +
#****# Suppose {{M|a\in\partial D^2}}
 +
#****#* Then to have {{M|1=f(a)=f(b)}} we must have {{M|1=b\in\partial D^2}}
 +
#****#** This means {{M|1=a\sim b}}, and that means {{M|1=\pi(a)=\pi(b)}}
 +
#****#*** But {{M|1=y=\pi(b)}} and {{M|1=x=\pi(a)}}, so we arrive at: {{M|1=x=\pi(a)=\pi(b)=y}}, or {{M|1=x=y}}, as required.
 +
We now know {{M|\bar{f} }} is a continuous bijection.
 +
 
 +
Noting that {{M|D^2}} is [[closed set|closed]] and [[bounded (metric)|bounded]] we can apply the [[Heine–Borel theorem]] to show {{M|D^2}} is compact. As {{M|\pi:D^2\rightarrow\frac{D^2}{\sim} }} is [[continuous]] (see [[quotient topology]] for information) we can use "[[the image of a compact set is compact]]" to conclude that {{M|\frac{D^2}{\sim} }} is compact.
 +
 
 +
[[A subspace of a Hausdorff space is a Hausdorff space]], as {{M|\mathbb{S}^2}} is a [[topological subspace]] of {{M|\mathbb{R}^3}}, {{M|\mathbb{S}^2}} is [[Hausdorff]].
 +
 
 +
We may now use the [[compact-to-Hausdorff theorem]] (as {{M|\bar{f} }} is a bijective continuous map between a compact space to a Hausdorff space) to show that {{M|\bar{f} }} is a [[homeomorphism]]
 +
 
 +
As we have found a homeomorphism between {{M|\frac{D^2}{\sim} }} and {{M|\mathbb{S}^2}} we have shown they are homeomorphic, written:
 +
* {{M|1=\frac{D^2}{\sim}\cong\mathbb{S}^2}}.
 
<div style="clear:both;"></div>
 
<div style="clear:both;"></div>
 
<noinclude>
 
<noinclude>

Latest revision as of 23:26, 11 October 2016

Section B

Question 7

Let [ilmath]D^2[/ilmath] denote the closed unit disk in [ilmath]\mathbb{R}^2[/ilmath] and define an equivalence relation on [ilmath]D^2[/ilmath] by setting [ilmath]x_1\sim x_2[/ilmath] if [ilmath]\Vert x_1\Vert=\Vert x_2\Vert=1[/ilmath] ("collapsing the boundary to a single point"). Show that [ilmath]\frac{D^2}{\sim} [/ilmath] is homeomorphic to [ilmath]\mathbb{S}^2[/ilmath] - the sphere.

  • Hint: first define a surjection [ilmath](:D^2\rightarrow\mathbb{S}^2)[/ilmath] mapping all of [ilmath]\partial D^2[/ilmath] to the north pole. This may be defined using a good picture or a formula.

Solution

The idea is to double the radius of [ilmath]D^2[/ilmath], then pop it out into a hemisphere, then pull the rim to a point
Picture showing the "expanding [ilmath]D^2[/ilmath]", the embedding-in-[ilmath]\mathbb{R}^3[/ilmath] part, and the "popping out"

Definitions:

  • [ilmath]H[/ilmath] denotes the hemisphere in my picture.
  • [ilmath]E:D^2\rightarrow H[/ilmath] is the composition of maps in my diagram that take [ilmath]D^2[/ilmath], double its radius, then embed it in [ilmath]\mathbb{R}^3[/ilmath] then "pop it out" into a hemisphere. We take it as obvious that it is a homeomorphism
  • [ilmath]f':H\rightarrow\mathbb{S}^2[/ilmath], this is the map in the top picture. It takes the hemisphere and pulls the boundary/rim in (along the blue lines) to the north pole of the red sphere. [ilmath]f'(\partial H)=(0,0,1)\in\mathbb{R}^3[/ilmath], it should be clear that for all [ilmath]x\in H-\partial H[/ilmath] that [ilmath]f'(x)[/ilmath] is intended to be a point on the red sphere and that [ilmath]f'\big\vert_{H-\partial H}[/ilmath] is injective. It is also taken as clear that [ilmath]f'[/ilmath] is surjective
  • Note: Click the pictures for a larger version
  • [ilmath]\frac{D^2}{\sim} [/ilmath] and [ilmath]D^2/\sim[/ilmath] denote the quotient space, with this definition we get a canonical projection, [ilmath]\pi:D^2\rightarrow D^2/\sim[/ilmath] given by [ilmath]\pi:x\mapsto [x][/ilmath] where [ilmath][x][/ilmath] denotes the equivalence class of [ilmath]x[/ilmath]
  • Lastly, we define [ilmath]f:D^2\rightarrow\mathbb{S}^2[/ilmath] to be the composition of [ilmath]E[/ilmath] and [ilmath]f'[/ilmath], that is: [ilmath]f:=f'\circ E[/ilmath], meaning [ilmath]f:x\mapsto f'(E(x))[/ilmath]

The situation is shown diagramatically below:

  • [ilmath]\xymatrix{ D^2 \ar[d]_\pi \ar[r]^E \ar@/^1.5pc/[rr]^{f:=f'\circ E} & H \ar[r]^{f'} & \mathbb{S}^2 \\ \frac{D^2}{\sim} }[/ilmath]

Outline of the solution:

  • We then want apply the passing to the quotient theorem to yield a commutative diagram: [ilmath]\xymatrix{ D^2 \ar[d]_\pi \ar[r]^E \ar@/^1.5pc/[rr]^{f} & H \ar[r]^{f'} & \mathbb{S}^2 \\ \frac{D^2}{\sim} \ar@{.>}[urr]_{\bar{f} } }[/ilmath]
    • The commutative diagram part merely means that [ilmath]f=\bar{f}\circ\pi[/ilmath][Note 1]. We get [ilmath]f=\bar{f}\circ\pi[/ilmath] as a result of the passing-to-the-quotient theorem.
    • We take this diagram as showing morphisms in the TOP category, meaning all arrows shown represent continuous maps. (Obviously...)
  • Lastly, we will show that [ilmath]\bar{f} [/ilmath] is a homeomorphism using the compact-to-Hausdorff theorem
Solution body

First we must show the requirements for applying passing to the quotient are satisfied.

  • We know already the maps involved are continuous and that [ilmath]\pi[/ilmath] is a quotient map. We only need to show:
    • [ilmath]f[/ilmath] is constant on the fibres of [ilmath]\pi[/ilmath], which is equivalent to:
      • [ilmath]\forall x,y\in D^2[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath]
  • Let us show this remaining condition:
    • Let [ilmath]x,y\in D^2[/ilmath] be given.
      • Suppose [ilmath]\pi(x)\ne\pi(y)[/ilmath], then by the nature of logical implication the implication is true regardless of [ilmath]f(x)[/ilmath] and [ilmath]f(y)[/ilmath]'s equality. We're done in this case.
      • Suppose [ilmath]\pi(x)=\pi(y)[/ilmath], we must show that in this case [ilmath]f(x)=y(y)[/ilmath].
        • Suppose [ilmath]x\in D^2-\partial D^2[/ilmath] (meaning [ilmath]x\in D^2[/ilmath] but [ilmath]x\notin \partial D^2[/ilmath], ie [ilmath]-[/ilmath] denotes relative complement)
          • In this case we must have [ilmath]x=y[/ilmath], as otherwise we'd not have [ilmath]\pi(x)=\pi(y)[/ilmath] (for [ilmath]x\in D^2-\partial D^2[/ilmath] we have [ilmath]\pi(x)=[x]=\{x\}[/ilmath], that is that the equivalence classes are singletons. So if [ilmath]\pi(x)=\pi(y)[/ilmath] we must have [ilmath]\pi(y)=[y]=\{x\}=[x]=\pi(x)[/ilmath]; so [ilmath]y[/ilmath] can only be [ilmath]x[/ilmath])
          • If [ilmath]x=y[/ilmath] then by the nature of [ilmath]f[/ilmath] being a function we must have [ilmath]f(x)=f(y)[/ilmath], we're done in this case
        • Suppose [ilmath]x\in \partial D^2[/ilmath] (the only case not covered) and [ilmath]\pi(x)=\pi(y)[/ilmath], we must show [ilmath]f(x)=f(y)[/ilmath]
          • Clearly if [ilmath]x\in\partial D^2[/ilmath] and [ilmath]\pi(x)=\pi(y)[/ilmath] we must have [ilmath]y\in\partial D^2[/ilmath].
            • [ilmath]E(x)[/ilmath] is mapped to the boundary/rim of [ilmath]H[/ilmath], as is [ilmath]E(y)[/ilmath] and [ilmath]f'(\text{any point on the rim of }H)=(0,0,1)\in\mathbb{R}^3[/ilmath]
            • Thus [ilmath]f'(E(x))=f'(E(y))[/ilmath], but [ilmath]f'(E(x))[/ilmath] is the very definition of [ilmath]f(x)[/ilmath], so clearly:
              • [ilmath]f(x)=f(y)[/ilmath] as required.

We may now apply the passing to the quotient theorem. This yields:

  • A continuous map, [ilmath]\bar{f}:D^2/\sim\rightarrow\mathbb{S}^2[/ilmath] where [ilmath]f=\bar{f}\circ\pi[/ilmath]
[ilmath]\xymatrix{ D^2 \ar[d]_\pi \ar[r]^E \ar@/^1.5pc/[rr]^{f} & H \ar[r]^{f'} & \mathbb{S}^2 \\ \frac{D^2}{\sim} \ar@{.>}[urr]_{\bar{f} } }[/ilmath]
Commutative diagram of situation
(all maps are continuous)
We now have the situation of the diagram on the right, restated from above for convenience.

In order to apply the compact-to-Hausdorff theorem and show [ilmath]\bar{f} [/ilmath] is a homeomorphism we must show it is continuous and bijective. We already have continuity (as a result of the passing-to-the-quotient theorem), we must show it is bijective.

We must show [ilmath]\bar{f} [/ilmath] is both surjective and injective:

  1. Surjectivity: We can get this from the definition of [ilmath]\bar{f} [/ilmath], recall on the passing to the quotient (function) page that:
    • if [ilmath]f[/ilmath] is surjective then [ilmath]\bar{f} [/ilmath] (or [ilmath]\tilde{f} [/ilmath] as the induced function is on that page) is surjective also
      • I've already done it "the long way" once in this assignment, and I hope it is not frowned upon if I decline to do it again.
  2. Injectivity: This follows a similar to gist as to what we've done already to show we could apply "passing to the quotient" and for the other questions where we've had to show a factored map is injective. As before:
    • Let [ilmath]x,y\in\frac{D^2}{\sim} [/ilmath] be given.
      • Suppose [ilmath]\bar{f}(x)\ne\bar{f}(y)[/ilmath] - then by the nature of logical implication we do not care about the RHS and are done regardless of [ilmath]x[/ilmath] and [ilmath]y[/ilmath]'s equality
        • Once again I note we must really have [ilmath]x\ne y[/ilmath] as if [ilmath]x=y[/ilmath] then by definition of [ilmath]\bar{f} [/ilmath] being a function we must also have [ilmath]\bar{f}(x)=\bar{f}(y)[/ilmath], anyway!
      • Suppose that [ilmath]\bar{f}(x)=\bar{f}(y)[/ilmath], we must show that in this case [ilmath]x=y[/ilmath].
        • Note that by surjectivity of [ilmath]\pi[/ilmath] that: [ilmath]\exists a\in D^2[\pi(a)=x][/ilmath] and [ilmath]\exists b\in D^2[\pi(b)=y][/ilmath], so [ilmath]\bar{f}(x)=\bar{f}(\pi(a))[/ilmath] and [ilmath]\bar{f}(y)=\bar{f}(\pi(b))[/ilmath], also, as [ilmath]\bar{f} [/ilmath] was the result of factoring, we have [ilmath]f=\bar{f}\circ\pi[/ilmath], so we see [ilmath]\bar{f}(\pi(a))=f(a)[/ilmath] and [ilmath]\bar{f}(\pi(b))=f(b)[/ilmath], since [ilmath]\bar{f}(x)=\bar{f}(y)[/ilmath] we get [ilmath]f(a)=f(b)[/ilmath] and [ilmath]\bar{f}(\pi(a))=\bar{f}(\pi(b))[/ilmath] also.
          • We now have 2 cases, [ilmath]a\in D^2-\partial D^2[/ilmath] and [ilmath]a\in \partial D^2[/ilmath] respectively:
            1. Suppose [ilmath]a\in D^2-\partial D^2[/ilmath]
              • As we have [ilmath]f(a)=f(b)[/ilmath] we must have [ilmath]b=a[/ilmath], as if [ilmath]b\ne a[/ilmath] then [ilmath]f(b)\ne f(a)[/ilmath], because [ilmath]f\big\vert_{D^2-\partial D^2}:(D^2-\partial D^2)\rightarrow\mathbb{S}^2[/ilmath] is injective[Note 2] by construction
                • If [ilmath]b=a[/ilmath] then [ilmath]y=\pi(b)=\pi(a)=x[/ilmath] so [ilmath]y=x[/ilmath] as required (this is easily recognised as [ilmath]x=y[/ilmath])
            2. Suppose [ilmath]a\in\partial D^2[/ilmath]
              • Then to have [ilmath]f(a)=f(b)[/ilmath] we must have [ilmath]b\in\partial D^2[/ilmath]
                • This means [ilmath]a\sim b[/ilmath], and that means [ilmath]\pi(a)=\pi(b)[/ilmath]
                  • But [ilmath]y=\pi(b)[/ilmath] and [ilmath]x=\pi(a)[/ilmath], so we arrive at: [ilmath]x=\pi(a)=\pi(b)=y[/ilmath], or [ilmath]x=y[/ilmath], as required.

We now know [ilmath]\bar{f} [/ilmath] is a continuous bijection.

Noting that [ilmath]D^2[/ilmath] is closed and bounded we can apply the Heine–Borel theorem to show [ilmath]D^2[/ilmath] is compact. As [ilmath]\pi:D^2\rightarrow\frac{D^2}{\sim} [/ilmath] is continuous (see quotient topology for information) we can use "the image of a compact set is compact" to conclude that [ilmath]\frac{D^2}{\sim} [/ilmath] is compact.

A subspace of a Hausdorff space is a Hausdorff space, as [ilmath]\mathbb{S}^2[/ilmath] is a topological subspace of [ilmath]\mathbb{R}^3[/ilmath], [ilmath]\mathbb{S}^2[/ilmath] is Hausdorff.

We may now use the compact-to-Hausdorff theorem (as [ilmath]\bar{f} [/ilmath] is a bijective continuous map between a compact space to a Hausdorff space) to show that [ilmath]\bar{f} [/ilmath] is a homeomorphism

As we have found a homeomorphism between [ilmath]\frac{D^2}{\sim} [/ilmath] and [ilmath]\mathbb{S}^2[/ilmath] we have shown they are homeomorphic, written:

  • [ilmath]\frac{D^2}{\sim}\cong\mathbb{S}^2[/ilmath].


Notes

  1. ↑ Technically a diagram is said to commute if all paths through it yield equal compositions, this means that we also require [ilmath]f=f'\circ E[/ilmath], which we already have by definition of [ilmath]f[/ilmath]!
  2. ↑ Actually:
    • [ilmath]f\big\vert_{D^2-\partial D^2}:(D^2-\partial D^2)\rightarrow(\mathbb{S}^2-\{(0,0,1)\}) [/ilmath] is bijective in fact!

References


Retrieved from "https://www.maths.kisogo.com/index.php?title=Exercises:Mond_-_Topology_-_1/Question_7&oldid=3294"