"Show that every closed ball in $\mathbb{R}^n$ is an $n$-dimensional manifold with boundary, as is the complement of every open ball. Assuming the theorem on the invariance of the boundary, show that the manifold boundary of each is equal to its topological boundary as a subset of $\mathbb{R}^n$, namely a sphere. Hint: for the unit ball in Rn, consider the map $\pi \circ \sigma^{-1}: \mathbb{R}^n \rightarrow \mathbb{R}^n$, where $\sigma$ is the stereographic projection and $\pi$ is a projection from $\mathbb{R}^{n + 1}$ to $\mathbb{R}^n$ that omits some coordinate other than the last."

So, I've got (the first part, anyways) of the question done, but my technique was a little different than what Prof. Lee suggested: I considered the ball as infinitely many foliated spheres, mapped each one to a plane using the stereographic projection, and then put the last coordinate as a function of the distance from the north/south pole. However, my solution seemed to neatly avoid any use of $\pi$ as mentioned, and I'm curious if anyone knows how that solution runs.

  • 1
    $\begingroup$ The main idea (as I understand it) is that $\pi \circ \sigma^{-1}$ takes the unit ball to $\mathbb{R}^{n-1}\times (-\infty, 0]$. It is the composition of continuous, open maps, and is thus continuous and open. You can show that, in a nbhd. of a boundary point (of the unit ball), it is injective, and thus defines a chart for each boundary point into (a space homeomorphic to) $\mathbb{H}^n$. $\endgroup$ – Steve D Jan 24 '18 at 22:53
  • $\begingroup$ @SteveD Thanks! I already solved this question (in the suggested manner) some time after posting it, but never bothered to post a solution. Half out of apathy, and half out of Prof. Lee's request in the intro to not post solutions. Thank you very much! $\endgroup$ – Duncan Ramage Jan 24 '18 at 22:55
  • 1
    $\begingroup$ Ah, OK, this question came up among several people today, and we could not find a nice exposition online, but only found your question! $\endgroup$ – Steve D Jan 24 '18 at 22:56
  • $\begingroup$ There's a explicit construction for $\bar{\mathbb{B}}^3$ here. $\endgroup$ – Sou May 2 '18 at 11:46

I've been work out this problem for some time, and i think i probably solved it based on the hint given on the book. Maybe this seems a little long. But it is really not. I tried my best to make this simple. It is really help to follow this argument with a picture for the case $n=2$ in mind.

It is enough to solve this for the closed unit ball $\bar{\mathbb{B}}^n$ in $\mathbb{R}^n$, since any other closed ball $\bar{B}_r(p)$ in $\mathbb{R}^n$ is homeomorphic to $\bar{\mathbb{B}}^n$ by composition of translation $T : \bar{B}_r(p) \to \bar{B}_r(0)$, defined as $x \mapsto x - p$ together with dilation $D : \bar{B}_r(0) \to \bar{\mathbb{B}}^n$, defined as $x \mapsto \frac{x}{r}$.

$\diamond \quad $ As a subspace of $\mathbb{R}^n$, $\bar{\mathbb{B}}^n$ is a second countable Hausdorff space. For any point $p\in \mathbb{B}^n$, the identity map on $\mathbb{B}^n$ serve as the homeomorphism. So we only need to construct homeomorphisms between neighbourhood of points on $\partial \bar{\mathbb{B}}^n=\mathbb{S}^{n-1}$ with open subsets in $\mathbb{H}^n$. To do this we need to consider $\bar{\mathbb{B}}^n$ as a subspace of $\mathbb{R}^{n+1}$.

Consider the stereographic projection from the south pole $\sigma : \mathbb{S}^n\smallsetminus \{S\} \to \mathbb{R}^n$, which is a homeomorphism, defined as $$ \sigma(x_1,\dots,x_{n+1}) = \frac{(x_1,\dots,x_n)}{1+x_{n+1}}. $$ For $i=1,\dots,n$, define $$U_i^{+} =\{ (x_1,\dots,x_n) \in \mathbb{R}^n : x_i > 0 \}, \quad U_i^{-} =\{ (x_1,\dots,x_n) \in \mathbb{R}^n : x_i < 0 \}$$ be $2n$-many open subsets of $\mathbb{R}^n$, and also for $i=1,\dots,n$ $$ \widetilde{U}_i^+ = \{(x_1,\dots,x_{n+1}) \in \mathbb{R}^{n+1} : x_i>0 \}, \quad \widetilde{U}_i^- = \{(x_1,\dots,x_{n+1}) \in \mathbb{R}^{n+1} : x_i<0 \} $$ are $2n$-many open subsets in $\mathbb{R}^{n+1}$.

Observe that $$\sigma^{-1}(U_i^{\pm}) = \mathbb{S}^n \cap \widetilde{U}_i^{\pm}$$ for each $i=1,\dots,n$. That is $\sigma^{-1}$ map $U_i^+$ to the open hemisphere of $\mathbb{S}^n$ where $x_i>0$, and same for $U_i^-$. In particular,

$\bullet \quad \sigma^{-1}$ is an identity map on $$\partial \bar{\mathbb{B}}^n = \mathbb{S}^{n-1} = \{(x_1,\dots,x_{n+1}) \in \mathbb{R}^{n+1} : x_{n+1} = 0 \text{ and }\sum_{i=1}^{n} x_i^2 = 1 \},$$

$\bullet \quad $ The inside region, $\mathbb{B}^n \cap U_i^+$, is maped to the upper part ($x_{n+1}>0$) of the hemisphere $\sigma^{-1}(U_i^+)$, and the outside region, $U_i^+ \smallsetminus \mathbb{B}^n$, is maped to the lower part ($x_{n+1}<0$) of hemisphere $\sigma^{-1}(U_i^+)$.

Since these hemispheres $\sigma^{-1}(U^{\pm}_i)$ homeomorphic to open unit ball $\mathbb{B}^{n-1}$ via projection map $\pi_i : (x_1,\dots,x_i,\dots,x_{n+1}) \mapsto (x_1,\dots,x_{i-1},x_{i+1},\dots,x_{n+1})$, then by restricting the composition map $\pi_i \circ \sigma^{-1} : \mathbb{R}^n \to \mathbb{R}^n$ to $U_i^{\pm} \cap \bar{\mathbb{B}}^n$, we obtain the desired homeomorphisms $$ \varphi:=\pi_i \circ (\sigma^{-1})|_{U_i^{\pm}\cap \bar{\mathbb{B}}^n} : U_i^{\pm}\cap \bar{\mathbb{B}}^n \to \mathbb{H}^n, $$ with domains cover $\partial\bar{\mathbb{B}}^n = \mathbb{S}^{n-1}$. By construction, any $p\in \mathbb{S}^{n-1}$ must contained in one such neighbourhoods, with $\varphi(p)=0 \in \partial \mathbb{H}^n$ and $\varphi(U_i^{\pm}\cap \bar{\mathbb{B}}^n)$ is an open half unit ball in $\mathbb{H}^n$. Therefore, $\bar{\mathbb{B}}^n$ is an $n$-manifold with boundary with manifold boundary is equal to its topological boundary $\mathbb{S}^{n-1}$.

Note that by similar way we can show that the complement of any open ball is an $n$-manifold with boundary, with its topological boundary as the manifold boundary. Only this time we use stereographic projection form the north pole.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.