Exercises:Mond - Topology - 1/Question 6

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: