Difference between revisions of "Exercises:Mond - Topology - 1/Question 7"
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====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. | ||
| + | * Lastly, we will show that {{M|\bar{f} }} is a [[homeomorphism]] using the [[compact-to-Hausdorff theorem]] | ||
| + | '''Solution:''' | ||
<div style="clear:both;"></div> | <div style="clear:both;"></div> | ||
<noinclude> | <noinclude> | ||
Revision as of 13:19, 8 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.
- Lastly, we will show that [ilmath]\bar{f} [/ilmath] is a homeomorphism using the compact-to-Hausdorff theorem
Solution:
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]!
References
