View source for Exercises:Mond - Topology - 1/Question 7

<noinclude>
__TOC__
==Section B==
===Question 7===
</noinclude>Let {{M|D^2}} denote the [[closed unit disk]] in {{M|\mathbb{R}^2}} and define an [[equivalence relation]] on {{M|D^2}} by setting {{M|1=x_1\sim x_2}} if {{M|1=\Vert x_1\Vert=\Vert x_2\Vert=1}} ("collapsing the boundary to a single point"). Show that {{M|\frac{D^2}{\sim} }} is [[homeomorphic]] to {{M|\mathbb{S}^2}} - the [[sphere]]. 
* '''Hint: ''' first define a [[surjection]] {{M|(:D^2\rightarrow\mathbb{S}^2)}} mapping all of {{M|\partial D^2}} to the north pole. This may be defined using a good picture or a formula.
====Solution====
{{float-right|{{Exercises:Mond - Topology - 1/Pictures/Q7 - 1}}}}
Definitions:
* {{M|H}} denotes the hemisphere in my picture.
* {{M|E:D^2\rightarrow H}} is the composition of maps in my diagram that take {{M|D^2}}, double its radius, then embed it in {{M|\mathbb{R}^3}} then "pop it out" into a hemisphere. We take it as obvious that it is a [[homeomorphism]]
* {{M|f':H\rightarrow\mathbb{S}^2}}, 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. {{M|1=f'(\partial H)=(0,0,1)\in\mathbb{R}^3}}, it should be clear that for all {{M|x\in H-\partial H}} that {{M|f'(x)}} is intended to be a point on the red sphere and that {{M|1=f'\big\vert_{H-\partial H} }} is [[injective]]. It is also taken as clear that {{m|f'}} is [[surjective]]
* '''Note: '''{{Note|Click the pictures for a larger version}}
* {{M|\frac{D^2}{\sim} }} and {{M|D^2/\sim}} denote the [[quotient topology|quotient space]], with this definition we get a [[canonical projection of the quotient topology|canonical projection]], {{M|\pi:D^2\rightarrow D^2/\sim}} given by {{M|\pi:x\mapsto [x]}} where {{M|[x]}} denotes the [[equivalence class]] of {{M|x}}
* Lastly, we define {{M|f:D^2\rightarrow\mathbb{S}^2}} to be the [[function composition|composition]] of {{M|E}} and {{M|f'}}, that is: {{M|1=f:=f'\circ E}}, meaning {{M|f:x\mapsto f'(E(x))}}
The situation is shown diagramatically below:
* <span><m>\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} }</m></span> 
'''Outline of the solution:'''
* We then want apply the {{link|passing to the quotient|topology}} theorem to yield a [[commutative diagram]]: <span><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></span>
** The commutative diagram part merely means that {{M|1=f=\bar{f}\circ\pi}}<ref group="Note">Technically a diagram is said to commute if all paths through it yield equal compositions, this means that we also require {{M|1=f=f'\circ E}}, which we already have by definition of {{M|f}}!</ref>. We get {{M|1=f=\bar{f}\circ\pi}} 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 {{M|\bar{f} }} is a [[homeomorphism]] using the [[compact-to-Hausdorff theorem]]
=====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>
<noinclude>

==Notes==
<references group="Note"/>
==References==
<references/>
</noinclude>