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I have to show that for any $x, y $ in the interior of the (closed) unit ball $D^n$ there is a homeomorphism of pairs $(D^n, \{ x \}) \cong (D^n, \{ y \})$.

In general, a homeomorphism $f$ of (topological) pairs $(X,A)$ and $(Y,B)$ is a homeomorphism between $X$ and $Y$ which satisfies $f(A) = B$, thus inducing a homeomorphism between $A$ and $B$.

So i have to find a homeomorphism from $D^n$ onto itself which maps $x$ to $y$ (and vice versa).

My first thought was to map everything to itself and just "switch" $x$ and $y$. However, if i am not mistaken, this map is neither open nor continuous.

I would be thankful for any ideas or hints!

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  • $\begingroup$ Do you know what is a vector field ? $\endgroup$ Oct 8 '17 at 10:34
  • $\begingroup$ I heard it before, but i am not super familiar with it. $\endgroup$
    – Felix R.
    Oct 8 '17 at 10:38
  • $\begingroup$ Ok, there is a solution without vector fields, I'll write both. $\endgroup$ Oct 8 '17 at 10:41
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First solution: The cube $C^n$ is homeomorphic to the ball $D^n$ by the map $$\phi \colon x \mapsto \frac{ \max (|x_i|)}{\sqrt{\sum x_i^2}} \cdot x$$

It's enough therefore to show for the cube. Since $C^n= [-1,1]^n$, it is enough to show for $[-1,1]$. Take $a$ with $|a|<1$. The homeomorphism of $[-1,1]$ $$\psi_a(x) = \frac{x-a}{1- a x}$$ takes $a$ to $0$.

Another solution, for the record: for $|a|<1$, $a\in D^n$, consider the map: $$\Psi_a(z) = \frac{\sqrt{1-|a|^2} z + \frac{1- \sqrt{1-|a|^2}}{|a|^2} \langle z, a\rangle a - a} { 1- \langle z, a\rangle } $$

With some calculations, one shows that $\Psi_a(\cdot)$ is a homeomorphism of $D^n$ taking $a$ to $0$ ( see analytic automorphism of the unit ball in $\mathbb{C}^n$ )

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  • $\begingroup$ Just because i am really bad at calculating; Once i got $a$ to 0, via $f(x) = (1-bx)*x+b$ i obtain the homeomorphism of $[-1,1]$ that maps 0 to $b$, thus interchangig the points $a$ and $b$ after composition. Right? $\endgroup$
    – Felix R.
    Oct 8 '17 at 12:20
  • $\begingroup$ @Ayy L.: Yes, correct, not sure about your expression for $f$, but that is the idea, move $a$ to $0$ and then to $b$ is enough. $\endgroup$
    – orangeskid
    Oct 8 '17 at 12:30
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    $\begingroup$ @Nicolas Hemelsoet: Thanks! Although your approach is much better if you want to show transitivity on the boundary... $\endgroup$
    – orangeskid
    Oct 8 '17 at 12:31
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Here is an elementary solution. If $x,y$ have the same norm you can take a rotation. If $x,y$ are on the same line, for all $t,s \in (0,1)$ it's easy to find an homeomorphism $f : (0,1) \to (0,1)$ with $f(s) = t$. Now $g(z) = f(|z|) \cdot z/|z| $ will take $x$ to $y$ with $s = |x|, t = |y|$.

Finally assume that $x,y$ are not on the same sphere and not collinear, and that $|y| < |x|$. We will use a lemma :

Let $|y| = |y'|$, where $y$ is in the interior of a ball $B$. There is an homeomorphism $h : B \to B$ with $h_{| \partial B} = \text{id}$ and $h(y) = y'$.

(Proof : let $\phi : [0,1] \to \Bbb R, x \mapsto 1 - 4(x-1/2)^2$. Notice that $\phi(x) \geq 0$ and that $\phi(0) = \phi(1) = 0, \phi(1/2) = 1$. Now define $h(z) = r(\phi(|z|)\theta)z$, with $r$ being a rotation of fixed axe, with $r(\theta)y = y'$. $h(z)$ has required properties. )

Assuming this lemma, and applying this to the ball of radius $y$ we get a homeomorphism which rotates $y$ where we want and leave $x$ invariant, in particular we can make $y,x$ collinear and apply the previous point as before.

For a more elegant and quick solution, you can take a vector field taking $x$ to $y$ and which is the identify on $\partial D^n$, integrarion of such a vector field is exactly the homeomorphism you were looking for. More details are in Milnor, "Topology from the differentiable viewpoint".

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  • $\begingroup$ Ok, so if they have the same norm or lie on the same line, i can see that there is a homeomophism (via rotating or stretching, i guess). If this is both not the case, i can combine rotating and stretching to get my homeomorphism, do i understand correctly? But is it really a homeomorphism if a rotate a smaller ball in the big ball? If a take an open set that has some bits in the small ball and some not, after this rotation of the little ball this set is not open anymore, or is it? $\endgroup$
    – Felix R.
    Oct 8 '17 at 11:03
  • $\begingroup$ @AyyL. : No you can't just combine the two, because for stretching you need $x,y$ to be on the same line. You are right that rotating the small ball and leaving invariant the rest is not continuous. But what you can do is rotating with a variable angle, and make the angle of rotation smaller and smaller as long as you are getting closer to the boundary. Is it clear how to do it ? $\endgroup$ Oct 8 '17 at 11:42
  • $\begingroup$ by combining i meant composition. rotating and stretching are each a homemorphism, so their composition is one as well. i would rotate until $x$ and $y$ are on the same line, and then stretch. Is this the general idea? I can see that it would work if a vary the angle, however i do not know how to do this. $\endgroup$
    – Felix R.
    Oct 8 '17 at 12:09
  • $\begingroup$ I have edited my answer with all the necessary details. $\endgroup$ Oct 8 '17 at 12:22

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