Difference between revisions of "Exercises:Mond - Topology - 1/Question 7"
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==Section B== | ==Section B== | ||
===Question 7=== | ===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}}. Show that {{M|\frac{D^2}{\sim} }} is [[homeomorphic]] to {{M|\mathbb{S}^2}} - the [[sphere]]. | + | </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. | * '''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==== | ====Solution==== | ||
{{float-right|{{Exercises:Mond - Topology - 1/Pictures/Q7 - 1}}}} | {{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> | <div style="clear:both;"></div> | ||
<noinclude> | <noinclude> | ||
| + | |||
==Notes== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
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
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]
- [ilmath]f[/ilmath] is constant on the fibres of [ilmath]\pi[/ilmath], which is equivalent to:
- 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.
- 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].
- 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)
- Let [ilmath]x,y\in D^2[/ilmath] be given.
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]
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:
- 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.
- 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
- 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:
- 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])
- 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
- 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.
- This means [ilmath]a\sim b[/ilmath], and that means [ilmath]\pi(a)=\pi(b)[/ilmath]
- Then to have [ilmath]f(a)=f(b)[/ilmath] we must have [ilmath]b\in\partial D^2[/ilmath]
- Suppose [ilmath]a\in D^2-\partial D^2[/ilmath]
- We now have 2 cases, [ilmath]a\in D^2-\partial D^2[/ilmath] and [ilmath]a\in \partial D^2[/ilmath] respectively:
- 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.
- 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
- Let [ilmath]x,y\in\frac{D^2}{\sim} [/ilmath] be given.
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
- ↑ 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]!
- ↑ 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
