$D^n/S^{n-1} \cong S^n$

Question

In Hatcher's book on algebraic topology, it is claimed that

$$D^n/S^{n-1}\cong S^n$$

Here $D^n = \{x\in \mathbb{R}^n: \Vert x \Vert \leq 1\}$ and $S^{n} = \{x \in \mathbb{R}^{n+1}: \Vert x \Vert =1\}$ But how is this quotient formed? I am aware of a quotient space $X/R$ where $R$ is an equivalence relation on $X$ but I have not seen something like this before.

Answer 1

Given the topological space $D^n$ we define an equivalence relation $\sim$ on $D^n$ by $$x \sim x' \Leftrightarrow x,x' \in S^{n-1}.$$

Geometrically, we collapse the boundary $\partial D^n = S^{n-1}$ of $D^n$ to a single point $p$ which gives us precisely $S^{n}$. You can visualize it for $n=1,2$.

Note that the open disk $\mathring{D^n}$ is homeomorphic to the sphere $S^n$ minus a point $N = (0,...,0,1)^\top$.

Now define a map $$f\colon (\mathring{D^n}, \partial D^n)\to (S^n, N)$$

that maps the open disk $\mathring{D^n}$ homeomorphically onto $S^n\setminus N$ and the boundary $\partial D^n$ onto $N$. Thus $f$ is clearly surjective and homeomorphic on the restriction $f\mid_{\mathring{D^n}}$.

Let $\sim$ be the equivalence relation we've defined previously and $\pi\colon D^n \to D^n/S^{n-1}$ the canonical projection. Then, by the universal property of the quotient space, $f$ induces a homeomorphism $$\overline{f}\colon D^n/S^{n-1} \to S^n.$$

$\hskip2in$

Answer 2

The quotient described refers to the quotient space with the following equivalence relation: $x \sim y$ if both $x$ and $y$ are on the boundary of the disk (the sphere $S^{n-1}$), and inequivalent otherwise. In effect, the quotient identifies the entire boundary $S^{n-1}$ to a single point and leaves all other points untouched.

Answer 3

It is a somewhat shorthand way to write an equivalence relation.

Note that $S^{n-1}$ is the boundary of $D^n$. The notation $D^n/S^{n-1}$ means that two points in $D^n$ are considered equivalent if they're both in $S^{n-1}$. In other words, the entire boundary is identified and forms a single equivalence class.

You can convince yourself that this makes sense intuitively for $n=2$: The 2D disc with the boundary $S^1$ identified is isomorphic to the 2-sphere -- the boundary $S^1$ can be considered as the north pole of the $S^2$.