Difference between revisions of "Exercises:Mond - Topology - 1/Question 6"

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(→‎Solution: Added proof of injective case)
m ("we shall define a map f to be such a map" -> "we shall define f to be such a map")
 
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Find a ''[[surjective]]'' [[continuous]] ''[[mapping]]'' from {{M|[-1,1]\subset\mathbb{R} }} to the unit [[circle]], {{M|\mathbb{S}^1}} such that it is injective except for that it sends {{M|-1}} and {{M|1}} to the same point in {{M|\mathbb{S}^1}}. Definitions may be explicit or use a picture
 
Find a ''[[surjective]]'' [[continuous]] ''[[mapping]]'' from {{M|[-1,1]\subset\mathbb{R} }} to the unit [[circle]], {{M|\mathbb{S}^1}} such that it is injective except for that it sends {{M|-1}} and {{M|1}} to the same point in {{M|\mathbb{S}^1}}. Definitions may be explicit or use a picture
 
=====Solution=====
 
=====Solution=====
−
{{float-right|{{Exercises:Mond - Topology - 1/Pictures/Q6P1 - 1}}}}We shall define a map: {{M|f:[-1,1]\rightarrow\mathbb{S}^1}} to be such a map:
+
{{float-right|{{Exercises:Mond - Topology - 1/Pictures/Q6P1 - 1}}}}We shall define {{M|f:[-1,1]\rightarrow\mathbb{S}^1}} to be such a map:
−
* {{M|f:t\mapsto\begin{pmatrix}\cos(\pi(t+1))\\\sin(\pi(t+1))\end{pmatrix} }}, this starts at the point {{M|(1,0)}} and goes anticlockwise around the circle of unit radius once.
+
* {{M|f:t\mapsto\begin{pmatrix}-\sin(\pi(t+1))\\-\cos(\pi(t+1))\end{pmatrix} }}, this starts at the point {{M|(1,0)}} and goes anticlockwise around the circle of unit radius once.
 
** '''Note: ''' I am not asked to show this is [[continuous]], merely exhibit it.  
 
** '''Note: ''' I am not asked to show this is [[continuous]], merely exhibit it.  
 +
** '''Note: ''' The reason for the odd choice of {{M|\sin}} for the {{M|x}} coordinate, and the minus signs is because my first choice was <span style="font-size:0.7em;">{{M|f:t\mapsto\begin{pmatrix}\cos(\pi(t+1))\\\sin(\pi(t+1))\end{pmatrix} }}</span>, however that didn't match up with the picture. The picture goes clockwise from the south pole, this would go anticlockwise from the east pole.
 
====Part 2====
 
====Part 2====
−
Define an [[equivalence relation]] on {{M|[-1,1]}} by declaring {{M|-1\sim 1}}, use part 1 above and applying the ''[[passing to the quotient (topology)|topological version]]'' of [[passing to the quotient]] to find a ''[[continuous]]'' [[bijection]]: {{M|(:[-1,1]/\sim\rightarrow\mathbb{S}^1)}}
+
Define an [[equivalence relation]] on {{M|[-1,1]}} by declaring {{M|-1\sim 1}}, use part 1 above and applying the ''[[passing to the quotient (topology)|topological version]]'' of [[passing to the quotient]] to find a ''[[continuous]]'' [[bijection]]: {{M|(:\frac{[-1,1]}{\sim}\rightarrow\mathbb{S}^1)}}
 
=====Solution=====
 
=====Solution=====
 
We wish to apply {{link|passing to the quotient|topology}}. Notice:
 
We wish to apply {{link|passing to the quotient|topology}}. Notice:
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#**** Now we have two cases, {{M|a\in(-1,1)}} and {{M|a\in\{-1,1\} }}, we shall deal with them separately.
 
#**** Now we have two cases, {{M|a\in(-1,1)}} and {{M|a\in\{-1,1\} }}, we shall deal with them separately.
 
#****# We have {{M|1=f(a)=f(b)}}, suppose {{M|a\in(-1,1)}}
 
#****# We have {{M|1=f(a)=f(b)}}, suppose {{M|a\in(-1,1)}}
−
#****#* Recall that our very definition of {{M|f}} required it to be "almost injective", specifically that {{M|f\big\vert_{(-1,1)}:(-1.1)\rightarrow\mathbb{S}^1}} was [[injective]] (and that it was only "not injective" on the endpoints)
+
#****#* Recall that our very definition of {{M|f}} required it to be "almost injective", specifically that {{M|f\big\vert_{(-1,1)}:(-1,1)\rightarrow\mathbb{S}^1}} was [[injective]] (and that it was only "not injective" on the endpoints)
 
#****#* As {{M|f}} is "injective in this range" we see that to have {{M|1=f(a)=f(b)}} means {{M|1=a=b}} (by injectiveness of {{M|f\big\vert_{(-1,1)} }})  
 
#****#* As {{M|f}} is "injective in this range" we see that to have {{M|1=f(a)=f(b)}} means {{M|1=a=b}} (by injectiveness of {{M|f\big\vert_{(-1,1)} }})  
 
#****#** As {{M|1=a=b}} we see {{M|1=y=\pi(b)=\pi(a)=x}} and conclude {{M|1=y=x}} - as required.  
 
#****#** As {{M|1=a=b}} we see {{M|1=y=\pi(b)=\pi(a)=x}} and conclude {{M|1=y=x}} - as required.  

Latest revision as of 12:26, 12 October 2016

Section A

Question 6

Part I

Find a surjective continuous mapping from [ilmath][-1,1]\subset\mathbb{R} [/ilmath] to the unit circle, [ilmath]\mathbb{S}^1[/ilmath] such that it is injective except for that it sends [ilmath]-1[/ilmath] and [ilmath]1[/ilmath] to the same point in [ilmath]\mathbb{S}^1[/ilmath]. Definitions may be explicit or use a picture

Solution
Take -1 to any point on the circle (we pick the south pole in this diagram), then go around the circle clockwise at a constant speed such that by [ilmath]f(1)[/ilmath] one has done a full revolution and is back at the starting point.

The right-hand-side is intended to demonstrate that the interval [ilmath](-1,1)[/ilmath] to the circle without the south-pole is bijective, ie it is injective and surjective, so the diagram on the left is "almost injective" in that it is injective everywhere except that it maps [ilmath]-1[/ilmath] and [ilmath]1[/ilmath] both to the south pole.
As required
We shall define [ilmath]f:[-1,1]\rightarrow\mathbb{S}^1[/ilmath] to be such a map:
  • [ilmath]f:t\mapsto\begin{pmatrix}-\sin(\pi(t+1))\\-\cos(\pi(t+1))\end{pmatrix} [/ilmath], this starts at the point [ilmath](1,0)[/ilmath] and goes anticlockwise around the circle of unit radius once.
    • Note: I am not asked to show this is continuous, merely exhibit it.
    • Note: The reason for the odd choice of [ilmath]\sin[/ilmath] for the [ilmath]x[/ilmath] coordinate, and the minus signs is because my first choice was [ilmath]f:t\mapsto\begin{pmatrix}\cos(\pi(t+1))\\\sin(\pi(t+1))\end{pmatrix} [/ilmath], however that didn't match up with the picture. The picture goes clockwise from the south pole, this would go anticlockwise from the east pole.

Part 2

Define an equivalence relation on [ilmath][-1,1][/ilmath] by declaring [ilmath]-1\sim 1[/ilmath], use part 1 above and applying the topological version of passing to the quotient to find a continuous bijection: [ilmath](:\frac{[-1,1]}{\sim}\rightarrow\mathbb{S}^1)[/ilmath]

Solution

We wish to apply passing to the quotient. Notice:

  1. we get [ilmath]\pi:[-1,1]\rightarrow\frac{[-1,1]}{\sim} [/ilmath], [ilmath]\pi:x\mapsto [x][/ilmath] automatically and it is continuous.
  2. we've already got a map, [ilmath]f[/ilmath], of the form [ilmath](:[-1,1]\rightarrow\mathbb{S}^1)[/ilmath]

In order to use the theorem we must show:

  • "[ilmath]f[/ilmath] is constant on the fibres of [ilmath]\pi[/ilmath]", that is:
    • [ilmath]\forall x,y\in [-1,1][\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath]
  • Proof:
    • Let [ilmath]x,y\in[-1,1][/ilmath] be given
      • Suppose [ilmath]\pi(x)\ne\pi(y)[/ilmath], by the nature of implies we do not care about the RHS of the implication, true or false, the implication holds, so we're done
      • Suppose [ilmath]\pi(x)=\pi(y)[/ilmath], we must show that this means [ilmath]f(x)=f(y)[/ilmath]
        • It is easy to see that if [ilmath]x\in(-1,1)\subset\mathbb{R} [/ilmath] then [ilmath]\pi(x)=\pi(y)\implies y=x[/ilmath]
          • By the nature of [ilmath]f[/ilmath] being a function (only associating an element of the domain with one thing in the codomain) and having [ilmath]y=x[/ilmath] we must have: [ilmath]f(x)=f(y)[/ilmath]
        • Suppose [ilmath]x\in\{-1,1\} [/ilmath], it is easy to see that then [ilmath]\pi(x)=\pi(y)\implies y\in\{-1,1\}[/ilmath]
          • But [ilmath]f(-1)=f(1)[/ilmath] so, whichever the case, [ilmath]f(x)=f(y)[/ilmath]

We may now apply the theorem to yield:

  • a unique continuous map, [ilmath]\overline{f}:[-1,1]/\sim\rightarrow\mathbb{S}^1[/ilmath] such that [ilmath]f=\overline{f}\circ\pi[/ilmath]

The question requires us to show this is a bijection, we must show that [ilmath]\newcommand{\fbar}{\bar{f} }\fbar[/ilmath] is both injective and surjective:

  1. Surjective: [ilmath]\forall y\in \mathbb{S}^1\exists x\in \frac{[-1,1]}{\sim}[\fbar(x)=y][/ilmath]
    • There are two ways to do this:
      1. Note that (from passing to the quotient) that if [ilmath]f[/ilmath] is surjective, then the resulting [ilmath]\fbar[/ilmath] is surjective.
      2. Or the long way of showing the definition of [ilmath]\fbar[/ilmath] being a surjection, [ilmath]\forall y\in\mathbb{S}^1\exists x\in\frac{[-1,1]}{\sim}[\fbar(x)=y][/ilmath]
        • Let [ilmath]y\in \mathbb{S}^1[/ilmath] be given.
          • Note that [ilmath]f[/ilmath] is surjective, and [ilmath]f=\fbar\circ\pi[/ilmath], thus [ilmath]\exists p\in[-1,1][/ilmath] such that [ilmath]p=f^{-1}(y)=(\fbar\circ\pi)^{-1}(y)=\pi^{-1}(\fbar^{-1}(y))[/ilmath], thus [ilmath]\pi(p)=\fbar^{-1}(y)[/ilmath]
          • Choose [ilmath]x\in[-1,1]/\sim[/ilmath] to be [ilmath]\pi(p)[/ilmath] where [ilmath]p\in[-1,1][/ilmath] exists by surjectivity of [ilmath]f[/ilmath] and is such that [ilmath]f(p)=y[/ilmath]
            • Now [ilmath]\fbar(\pi(p))=f(p)[/ilmath] (by definition of [ilmath]\fbar[/ilmath]) and [ilmath]f(p)=y[/ilmath], as required.
  2. Injective:
    • Let [ilmath]x,y\in\frac{[-1,1]}{\sim} [/ilmath] be given. We wish to show that [ilmath]\fbar(x)=\fbar(y)\implies x=y[/ilmath]
      • Suppose [ilmath]\fbar(x)\ne\fbar(y)[/ilmath], then we're done, as by the nature of logical implication we do not care about the right hand side.
        • Note though, by the definition of [ilmath]\fbar[/ilmath] being a function we cannot have [ilmath]x=y[/ilmath] in this case! As a function must map each element of the domain to exactly one thing of the codomain. Anyway!
      • Suppose [ilmath]\fbar(x)=\fbar(y)[/ilmath], we must show that in this case we have [ilmath]x=y[/ilmath].
        • By surjectivity of [ilmath]\pi:[-1,1]\rightarrow\frac{[-1,1]}{\sim} [/ilmath] we see [ilmath]\exists a\in [-1,1]\big[\pi(a)=x\big][/ilmath] and [ilmath]\exists b\in [-1,1]\big[\pi(b)=y\big][/ilmath]
          • Notice now we have [ilmath]\fbar(x)=\fbar(\pi(a))[/ilmath] and that (from the passing to the quotient part of obtaining [ilmath]\fbar[/ilmath]) we have [ilmath]f=\fbar\circ\pi[/ilmath], this means:
            • [ilmath]\fbar(x)=\fbar(\pi(a))=f(a)[/ilmath], we also have [ilmath]\fbar(y)=\fbar(\pi(b))=f(b)[/ilmath] from the same thoughts, but using [ilmath]y[/ilmath] and [ilmath]b[/ilmath] instead of [ilmath]x[/ilmath] and [ilmath]a[/ilmath].
          • In particular: [ilmath]f(a)=f(b)[/ilmath]
          • Now we have two cases, [ilmath]a\in(-1,1)[/ilmath] and [ilmath]a\in\{-1,1\} [/ilmath], we shall deal with them separately.
            1. We have [ilmath]f(a)=f(b)[/ilmath], suppose [ilmath]a\in(-1,1)[/ilmath]
              • Recall that our very definition of [ilmath]f[/ilmath] required it to be "almost injective", specifically that [ilmath]f\big\vert_{(-1,1)}:(-1,1)\rightarrow\mathbb{S}^1[/ilmath] was injective (and that it was only "not injective" on the endpoints)
              • As [ilmath]f[/ilmath] is "injective in this range" we see that to have [ilmath]f(a)=f(b)[/ilmath] means [ilmath]a=b[/ilmath] (by injectiveness of [ilmath]f\big\vert_{(-1,1)} [/ilmath])
                • As [ilmath]a=b[/ilmath] we see [ilmath]y=\pi(b)=\pi(a)=x[/ilmath] and conclude [ilmath]y=x[/ilmath] - as required.
            2. We have [ilmath]f(a)=f(b)[/ilmath], and this time [ilmath]a\in\{-1,1\} [/ilmath] instead
              • Again by definition of [ilmath]f[/ilmath], we recall [ilmath]f(-1)=f(1)[/ilmath] - it maps the endpoints of [ilmath][-1,1][/ilmath] to the same point in [ilmath]\mathbb{S}^1[/ilmath].
              • To have [ilmath]f(a)=f(b)[/ilmath] clearly means that [ilmath]b\in\{-1,1\}[/ilmath] (regardless of what value [ilmath]a\in\{-1,1\} [/ilmath] takes)
                • But [ilmath]\pi(a)=[a]=\{-1,1\}[/ilmath] and also [ilmath]\pi(b)=[b]=\{-1,1\}[/ilmath]
                • So we see [ilmath]y=\pi(b)=\{-1,1\}=\pi(a)=x[/ilmath], explicitly: [ilmath]x=y[/ilmath], as required
          • We have shown that in either case [ilmath]x=y[/ilmath]
    • Since [ilmath]x,y\in\frac{[-1,1]}{\sim}[/ilmath] was arbitrary, we have shown this for all [ilmath]x,y[/ilmath]. The very definition of [ilmath]\fbar[/ilmath] being injective.

Thus [ilmath]\fbar[/ilmath] is a bijection

Part 3

Show that [ilmath][-1,1]/\sim[/ilmath] is homeomorphic to [ilmath]\mathbb{S}^1[/ilmath]

Solution

To apply the "compact-to-Hausdorff theorem" we require:

  1. A continuous bijection, which we have, namely [ilmath]\fbar:\frac{[-1,1]}{\sim}\rightarrow\mathbb{S}^1[/ilmath]
  2. the domain space, [ilmath]\frac{[-1,1]}{\sim} [/ilmath], to be compact, and
  3. the codomain space, [ilmath]\mathbb{S}^1[/ilmath], to be Hausdorff

We know the image of a compact set is compact, and that closed intervals are compact in [ilmath]\mathbb{R} [/ilmath], thus [ilmath][-1,1]/\sim=\pi([1-,1])[/ilmath] must be compact. We also know [ilmath]\mathbb{R}^2[/ilmath] is Hausdorff and every subspace of a Hausdorff space is Hausdorff, thus [ilmath]\mathbb{S}^1[/ilmath] is Hausdorff.

We apply the theorem:

Notes

References


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