Difference between revisions of "Exercises:Mond - Topology - 1/Question 7"

Latest revision as of 23:26, 11 October 2016

Section B

Question 7

Let $D^2$ denote the closed unit disk in $\mathbb{R}^2$ and define an equivalence relation on $D^2$ by setting $x_1\sim x_2$ if $\Vert x_1\Vert=\Vert x_2\Vert=1$ ("collapsing the boundary to a single point"). Show that $\frac{D^2}{\sim}$ is homeomorphic to $\mathbb{S}^2$ - the sphere.

• Hint: first define a surjection $(:D^2\rightarrow\mathbb{S}^2)$ mapping all of $\partial D^2$ to the north pole. This may be defined using a good picture or a formula.

Solution

The idea is to double the radius of $D^2$, then pop it out into a hemisphere, then pull the rim to a point

Picture showing the "expanding $D^2$", the embedding-in-$\mathbb{R}^3$ part, and the "popping out"

Definitions:

• $H$ denotes the hemisphere in my picture.
• $E:D^2\rightarrow H$ is the composition of maps in my diagram that take $D^2$, double its radius, then embed it in $\mathbb{R}^3$ then "pop it out" into a hemisphere. We take it as obvious that it is a homeomorphism
• $f':H\rightarrow\mathbb{S}^2$, 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. $f'(\partial H)=(0,0,1)\in\mathbb{R}^3$, it should be clear that for all $x\in H-\partial H$ that $f'(x)$ is intended to be a point on the red sphere and that $f'\big\vert_{H-\partial H}$ is injective. It is also taken as clear that $f'$ is surjective
• Note: Click the pictures for a larger version
• $\frac{D^2}{\sim}$ and $D^2/\sim$ denote the quotient space, with this definition we get a canonical projection, $\pi:D^2\rightarrow D^2/\sim$ given by $\pi:x\mapsto [x]$ where $[x]$ denotes the equivalence class of $x$
• Lastly, we define $f:D^2\rightarrow\mathbb{S}^2$ to be the composition of $E$ and $f'$, that is: $f:=f'\circ E$, meaning $f:x\mapsto f'(E(x))$

The situation is shown diagramatically below:

• $\xymatrix{ D^2 \ar[d]_\pi \ar[r]^E \ar@/^1.5pc/[rr]^{f:=f'\circ E} & H \ar[r]^{f'} & \mathbb{S}^2 \\ \frac{D^2}{\sim} }$

Outline of the solution:

• We then want apply the passing to the quotient theorem to yield a commutative diagram: $\xymatrix{ D^2 \ar[d]_\pi \ar[r]^E \ar@/^1.5pc/[rr]^{f} & H \ar[r]^{f'} & \mathbb{S}^2 \\ \frac{D^2}{\sim} \ar@{.>}[urr]_{\bar{f} } }$
  • The commutative diagram part merely means that $f=\bar{f}\circ\pi$^{[Note 1]}. We get $f=\bar{f}\circ\pi$ as a result of the passing-to-the-quotient theorem.
  • We take this diagram as showing morphisms in the TOP category, meaning all arrows shown represent continuous maps. (Obviously...)
• Lastly, we will show that $\bar{f}$ is a homeomorphism using the compact-to-Hausdorff theorem

Solution body

First we must show the requirements for applying passing to the quotient are satisfied.

• We know already the maps involved are continuous and that $\pi$ is a quotient map. We only need to show:
  • $f$ is constant on the fibres of $\pi$, which is equivalent to:
    • $\forall x,y\in D^2[\pi(x)=\pi(y)\implies f(x)=f(y)]$
• Let us show this remaining condition:
  • Let $x,y\in D^2$ be given.
    • Suppose $\pi(x)\ne\pi(y)$, then by the nature of logical implication the implication is true regardless of $f(x)$ and $f(y)$'s equality. We're done in this case.
    • Suppose $\pi(x)=\pi(y)$, we must show that in this case $f(x)=y(y)$.
      • Suppose $x\in D^2-\partial D^2$ (meaning $x\in D^2$ but $x\notin \partial D^2$, ie $-$ denotes relative complement)
        • In this case we must have $x=y$, as otherwise we'd not have $\pi(x)=\pi(y)$ (for $x\in D^2-\partial D^2$ we have $\pi(x)=[x]=\{x\}$, that is that the equivalence classes are singletons. So if $\pi(x)=\pi(y)$ we must have $\pi(y)=[y]=\{x\}=[x]=\pi(x)$; so $y$ can only be $x$)
        • If $x=y$ then by the nature of $f$ being a function we must have $f(x)=f(y)$, we're done in this case
      • Suppose $x\in \partial D^2$ (the only case not covered) and $\pi(x)=\pi(y)$, we must show $f(x)=f(y)$
        • Clearly if $x\in\partial D^2$ and $\pi(x)=\pi(y)$ we must have $y\in\partial D^2$.
          • $E(x)$ is mapped to the boundary/rim of $H$, as is $E(y)$ and $f'(\text{any point on the rim of }H)=(0,0,1)\in\mathbb{R}^3$
          • Thus $f'(E(x))=f'(E(y))$, but $f'(E(x))$ is the very definition of $f(x)$, so clearly:
            • $f(x)=f(y)$ as required.

We may now apply the passing to the quotient theorem. This yields:

• A continuous map, $\bar{f}:D^2/\sim\rightarrow\mathbb{S}^2$ where $f=\bar{f}\circ\pi$

In order to apply the compact-to-Hausdorff theorem and show $\bar{f}$ is a homeomorphism we must show it is continuous and bijective. We already have continuity (as a result of the passing-to-the-quotient theorem), we must show it is bijective.

We must show $\bar{f}$ is both surjective and injective:

1. Surjectivity: We can get this from the definition of $\bar{f}$, recall on the passing to the quotient (function) page that:
   • if $f$ is surjective then $\bar{f}$ (or $\tilde{f}$ as the induced function is on that page) is surjective also
     • I've already done it "the long way" once in this assignment, and I hope it is not frowned upon if I decline to do it again.
2. Injectivity: This follows a similar to gist as to what we've done already to show we could apply "passing to the quotient" and for the other questions where we've had to show a factored map is injective. As before:
   • Let $x,y\in\frac{D^2}{\sim}$ be given.
     • Suppose $\bar{f}(x)\ne\bar{f}(y)$ - then by the nature of logical implication we do not care about the RHS and are done regardless of $x$ and $y$'s equality
       • Once again I note we must really have $x\ne y$ as if $x=y$ then by definition of $\bar{f}$ being a function we must also have $\bar{f}(x)=\bar{f}(y)$, anyway!
     • Suppose that $\bar{f}(x)=\bar{f}(y)$, we must show that in this case $x=y$.
       • Note that by surjectivity of $\pi$ that: $\exists a\in D^2[\pi(a)=x]$ and $\exists b\in D^2[\pi(b)=y]$, so $\bar{f}(x)=\bar{f}(\pi(a))$ and $\bar{f}(y)=\bar{f}(\pi(b))$, also, as $\bar{f}$ was the result of factoring, we have $f=\bar{f}\circ\pi$, so we see $\bar{f}(\pi(a))=f(a)$ and $\bar{f}(\pi(b))=f(b)$, since $\bar{f}(x)=\bar{f}(y)$ we get $f(a)=f(b)$ and $\bar{f}(\pi(a))=\bar{f}(\pi(b))$ also.
         • We now have 2 cases, $a\in D^2-\partial D^2$ and $a\in \partial D^2$ respectively:
           1. Suppose $a\in D^2-\partial D^2$
              • As we have $f(a)=f(b)$ we must have $b=a$, as if $b\ne a$ then $f(b)\ne f(a)$, because $f\big\vert_{D^2-\partial D^2}:(D^2-\partial D^2)\rightarrow\mathbb{S}^2$ is injective^{[Note 2]} by construction
                • If $b=a$ then $y=\pi(b)=\pi(a)=x$ so $y=x$ as required (this is easily recognised as $x=y$)
           2. Suppose $a\in\partial D^2$
              • Then to have $f(a)=f(b)$ we must have $b\in\partial D^2$
                • This means $a\sim b$, and that means $\pi(a)=\pi(b)$
                  • But $y=\pi(b)$ and $x=\pi(a)$, so we arrive at: $x=\pi(a)=\pi(b)=y$, or $x=y$, as required.

We now know $\bar{f}$ is a continuous bijection.

Noting that $D^2$ is closed and bounded we can apply the Heine–Borel theorem to show $D^2$ is compact. As $\pi:D^2\rightarrow\frac{D^2}{\sim}$ is continuous (see quotient topology for information) we can use "the image of a compact set is compact" to conclude that $\frac{D^2}{\sim}$ is compact.

A subspace of a Hausdorff space is a Hausdorff space, as $\mathbb{S}^2$ is a topological subspace of $\mathbb{R}^3$, $\mathbb{S}^2$ is Hausdorff.

We may now use the compact-to-Hausdorff theorem (as $\bar{f}$ is a bijective continuous map between a compact space to a Hausdorff space) to show that $\bar{f}$ is a homeomorphism

As we have found a homeomorphism between $\frac{D^2}{\sim}$ and $\mathbb{S}^2$ we have shown they are homeomorphic, written:

• $\frac{D^2}{\sim}\cong\mathbb{S}^2$.

Notes

1. Jump up ↑ Technically a diagram is said to commute if all paths through it yield equal compositions, this means that we also require $f=f'\circ E$, which we already have by definition of $f$!
2. Jump up ↑ Actually:
   • $f\big\vert_{D^2-\partial D^2}:(D^2-\partial D^2)\rightarrow(\mathbb{S}^2-\{(0,0,1)\})$ is bijective in fact!

References