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

Revision as of 22:01, 11 October 2016

Section A

Question 6

Part I

Find a surjective continuous mapping from $[-1,1]\subset\mathbb{R}$ to the unit circle, $\mathbb{S}^1$ such that it is injective except for that it sends $-1$ and $1$ to the same point in $\mathbb{S}^1$. 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 $f(1)$ one has done a full revolution and is back at the starting point.

The right-hand-side is intended to demonstrate that the interval $(-1,1)$ 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 $-1$ and $1$ both to the south pole.
As required

• $f:t\mapsto\begin{pmatrix}\cos(\pi(t+1))\\\sin(\pi(t+1))\end{pmatrix}$, this starts at the point $(1,0)$ and goes anticlockwise around the circle of unit radius once.
  • Note: I am not asked to show this is continuous, merely exhibit it.

Part 2

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

Solution

We wish to apply passing to the quotient. Notice:

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

In order to use the theorem we must show:

• "$f$ is constant on the fibres of $\pi$", that is:
  • $\forall x,y\in [-1,1][\pi(x)=\pi(y)\implies f(x)=f(y)]$
• Proof:
  • Let $x,y\in[-1,1]$ be given
    • Suppose $\pi(x)\ne\pi(y)$, 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 $\pi(x)=\pi(y)$, we must show that this means $f(x)=f(y)$
      • It is easy to see that if $x\in(-1,1)\subset\mathbb{R}$ then $\pi(x)=\pi(y)\implies y=x$
        • By the nature of $f$ being a function (only associating an element of the domain with one thing in the codomain) and having $y=x$ we must have: $f(x)=f(y)$
      • Suppose $x\in\{-1,1\}$, it is easy to see that then $\pi(x)=\pi(y)\implies y\in\{-1,1\}$
        • But $f(-1)=f(1)$ so, whichever the case, $f(x)=f(y)$

We may now apply the theorem to yield:

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

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

1. Surjective: $\forall y\in \mathbb{S}^1\exists x\in \frac{[-1,1]}{\sim}[\fbar(x)=y]$
   • There are two ways to do this:
     1. Note that (from passing to the quotient) that if $f$ is surjective, then the resulting $\fbar$ is surjective.
     2. Or the long way of showing the definition of $\fbar$ being a surjection, $\forall y\in\mathbb{S}^1\exists x\in\frac{[-1,1]}{\sim}[\fbar(x)=y]$
        • Let $y\in \mathbb{S}^1$ be given.
          • Note that $f$ is surjective, and $f=\fbar\circ\pi$, thus $\exists p\in[-1,1]$ such that $p=f^{-1}(y)=(\fbar\circ\pi)^{-1}(y)=\pi^{-1}(\fbar^{-1}(y))$, thus $\pi(p)=\fbar^{-1}(y)$
          • Choose $x\in[-1,1]/\sim$ to be $\pi(p)$ where $p\in[-1,1]$ exists by surjectivity of $f$ and is such that $f(p)=y$
            • Now $\fbar(\pi(p))=f(p)$ (by definition of $\fbar$) and $f(p)=y$, as required.
2. Injective:
   • Let $x,y\in\frac{[-1,1]}{\sim}$ be given. We wish to show that $\fbar(x)=\fbar(y)\implies x=y$
     • Suppose $\fbar(x)\ne\fbar(y)$, 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 $\fbar$ being a function we cannot have $x=y$ in this case! As a function must map each element of the domain to exactly one thing of the codomain. Anyway!
     • Suppose $\fbar(x)=\fbar(y)$, we must show that in this case we have $x=y$.
       • By surjectivity of $\pi:[-1,1]\rightarrow\frac{[-1,1]}{\sim}$ we see $\exists a\in [-1,1]\big[\pi(a)=x\big]$ and $\exists b\in [-1,1]\big[\pi(b)=y\big]$
         • Notice now we have $\fbar(x)=\fbar(\pi(a))$ and that (from the passing to the quotient part of obtaining $\fbar$) we have $f=\fbar\circ\pi$, this means:
           • $\fbar(x)=\fbar(\pi(a))=f(a)$, we also have $\fbar(y)=\fbar(\pi(b))=f(b)$ from the same thoughts, but using $y$ and $b$ instead of $x$ and $a$.
         • In particular: $f(a)=f(b)$
         • Now we have two cases, $a\in(-1,1)$ and $a\in\{-1,1\}$, we shall deal with them separately.
           1. We have $f(a)=f(b)$, suppose $a\in(-1,1)$
              • Recall that our very definition of $f$ required it to be "almost injective", specifically that $f\big\vert_{(-1,1)}:(-1.1)\rightarrow\mathbb{S}^1$ was injective (and that it was only "not injective" on the endpoints)
              • As $f$ is "injective in this range" we see that to have $f(a)=f(b)$ means $a=b$ (by injectiveness of $f\big\vert_{(-1,1)}$)
                • As $a=b$ we see $y=\pi(b)=\pi(a)=x$ and conclude $y=x$ - as required.
           2. We have $f(a)=f(b)$, and this time $a\in\{-1,1\}$ instead
              • Again by definition of $f$, we recall $f(-1)=f(1)$ - it maps the endpoints of $[-1,1]$ to the same point in $\mathbb{S}^1$.
              • To have $f(a)=f(b)$ clearly means that $b\in\{-1,1\}$ (regardless of what value $a\in\{-1,1\}$ takes)
                • But $\pi(a)=[a]=\{-1,1\}$ and also $\pi(b)=[b]=\{-1,1\}$
                • So we see $y=\pi(b)=\{-1,1\}=\pi(a)=x$, explicitly: $x=y$, as required
         • We have shown that in either case $x=y$
   • Since $x,y\in\frac{[-1,1]}{\sim}$ was arbitrary, we have shown this for all $x,y$. The very definition of $\fbar$ being injective.

Thus $\fbar$ is a bijection

Part 3

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

Solution

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

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

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

We apply the theorem:

• $\fbar$ is a homeomorphism

Notes

References