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

Revision as of 15:12, 7 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 $\pi:[-1,1]\rightarrow[-1,1]/\sim$, $\pi:x\mapsto [x]$ 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: $\forall y\in \mathbb{S}^1\exists x\in [-1,1]/\sim[\fbar(x)=y]$ Caution:I should really factor this out into its own proposition
   • 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:
   • Caution:Do later on paper, it is easy to see though as $f\big\vert_{(-1,1)}$ is clearly injective, that covers all of $\mathbb{S}^1$ except $f(1)=f(-1)$, it is clear that if $\fbar(x)=f(1)$ that $x$ can only be $\pi(-1)=\pi(1)$

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 $\fbar:\frac{[-1,1]}{\sim}\rightarrow\mathbb{S}^2$
2. the domain space, $\frac{[-1,1]}{\sim}$, to be compact, and
3. the codomain space, $\mathbb{S}^2$, 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 $\mathbb{S}^2$ is Hausdorff.

We apply the theorem:

• $\fbar$ is a homeomorphism

Notes

References