Why is $D^{n+1}/S^{n} = S^{n+1}$ true?

Question

I went to my first lecture in Algebraic Topology and managed to get really confused. It seems like they assumed that the following statement was "obvious":

$D^{n+1}/S^{n} = S^{n+1}$

Where $D^{n}$ is the unit disk/ball in $\mathbb{R}^{n}$ and $S^{n}$ is the unit sphere in $\mathbb{R}^{n+1}$.

The notation $X/A$ is, as far as I could understand, a way of denoting that $A\subseteq X$ collapse to one point. So we divide $X$ into equivalence classes where the points in $A$ are in the same equivalence class and the points $X-A$ is in their "own" equivalence class.

My questions:

• Why is this statement true? and do $=$ indicate that there exists a homeomorphism between the two spaces?

Answer 1

Yes, $=$ is being used to denote "is homeomorphic to", which is a (slight) abuse of notation; really the notation $\cong$ should be used, but it's not a big deal here.

Intuitively, I find the fact $D^{n+1}/S^n\cong S^{n+1}$ to be quite clear. As usual in math, we may not be able to see large numbers of dimensions, but we can get a good intuition from the dimensions we can visualize. Certainly, collapsing $S^0$ (the points $1$ and $-1$) in $D^1=[-1,1]$ produces a circle. And here is a nice visualization for $n=1$:

              

It should be clear that, in general, collapsing the "edge" of a disk to a point, and leaving everything else the same, "closes up" the disk into a sphere of the next highest dimension.

---

\documentclass{standalone}
\usepackage{tikz}
\tikzstyle{mypoint}=[inner sep=0pt,outer sep=0pt,minimum size=5pt,fill,circle]
\colorlet{blue1}{blue!90!white}
\colorlet{blue2}{blue1!80}

\begin{document}
\begin{tikzpicture}[scale=1.5]
\node[mypoint] [label=285:$p$]  at (0,1.1) {};
\fill[color=blue2!80!gray] (0,-0.7) ellipse (1.3 and 0.6);
\draw[color=blue2!50!black,ultra thick] (0,-0.7) ellipse (1.3 and 0.6);
\begin{scope}[dashed,thick,->,shorten >=6pt,shorten =stealth]
\path (-1.3,-0.7) edge[out=90,in=180] (0,1.1);
\path (1.3,-0.7)  edge[out=90,in=0]   (0,1.1);
\end{scope}

\fill[ball color=blue!60] (4,0) circle (1.3);
\node[mypoint,white] [label={[white]285:$p$}] at (4,1.1) {};
\end{tikzpicture}
\end{document}

Answer 2

Here's one way to set up a concrete homeomorphism.

The sphere $S^{n+1}$ is homeomorphic to the $1$-point compactification of $\mathbb{R}^{n+1}$. This is witnessed by the stereographic projection maps. Take the disc $D^{n+1} \subset \mathbb{R}^{n+1}$, map the inner open disc to $\mathbb{R}^{n+1}$ (you have probably seen this done in a point-set topology class), and map the boundary of $D^{n+1}$ (which is $S^{n}$ by definition) to the point at infinity. This is not injective, but if you define the same map on $D^{n+1} / S^{n}$, this map is bijective and continuous (not too hard to see, recall which sets are open in the compactified $\mathbb{R}^{n+1}$), and goes from a compact space to a Hausdorff space, hence is a homeomorphism (another standard point-set result).

Another idea:

It is relatively easy to show that the sphere is two discs glued at their boundaries.

Let $X_1$ be an inner disc of half the radius of $D^{n+1}$ inside $D^{n+1}/S^{n}$. Let $X_2 = D^{n+1}/S^{n} - X_{1}$. We know that $X_1$ is a disc, and $X_2$ looks like an annular region with collapsed boundary. Define a map from $X_2$ to a disc in the following way. Map rings on the inside of $X_2$ to rings on the outside of a disc, where by rings I mean copies of $S^n$ defined as level sets of the Euclidean norm.

Do this inside-out, so that rings on the outside of $X_2$ get mapped to smaller rings in the interior of a disc. Finally, send the outer, collapsed boundary of $X_2$ to the center of the disc. This is clearly a bijection. It is not too hard to show it is continuous. It is a map from a compact space to a Hausdorff space, hence a homeomorphism, so $X_2$ is actually a disc.

Thus, $D^{n+1}/S^n$ was just two discs glued together at their boundaries, hence is homeomorphic to $S^{n+1}$.