# Two topology questions regarding quotient $D^n/S^{n-1}$ and homotopy $S^{n-1} \to S^{n} - \{ a,b\}$

I got two simple questions.

1. Why is $$D^n/S^{n-1} = S^n$$? If we are quotienting out the boundary, the interior isn't empty.

2. The two spaces $$S^{n-1}$$ and $$S^{n} - \{ a,b\}$$ where $$a,b$$ are north/south pole of n-sphere are supposed to be homotopic. What is the homotopic map? I can never come up with these maps on my own. Are there some techniques to these?

All I ever can start is $$u = (x/|x|)$$ and think of some clever projection. $$(u_1,\dots,u_{n-1}) \to (u_1,\dots, u_{n-1},?)$$ Are most of these maps that involve $$S^n$$ just some clever application of stereographic projection?

• Does in $D^n$ you mean the closed disc? ($\bar{D_n} = \{x\in \mathbb{R}^n : ||x|| \le 1\})$ – dan Sep 26 '18 at 9:21
• @dan yeah. unit disk. It was just an excerpt of something I read in Hatcher. It just looked right, but I couldn't justify it in my head. – Hawk Sep 26 '18 at 9:22
• I am sorry for not having a formal answer, but for part 1 - I think of $\bar{D_n} / \partial \bar{D}^n$ as $D^n \cup {p}$ when $p\in \partial \bar{D}^n$ now the stereographic projection gives you $S^n \setminus\{point\} \mapsto \mathbb{R}^n$ and $\mathbb{R}^n$ is homeomorphic to the open disc $D^n$ so we got homeomorphism from $S^n\setminus \{point\}$ to $D^n$ . Now take your one boundary point (which represnt $\partial \bar{D}^n$) and define $point \mapsto boundary point$ this gives the wished homemorphism. – dan Sep 26 '18 at 10:19
• So basically you are saying $S^n/\{x\} = S^n/\{p\} = D^n$ and adjoining the point $\{p \}$ means $D^n/S^{n-1}= D^n \cup p = S^n$. Is there actually a difference between $\bar{D}$ and $D$ in your notation? – Hawk Sep 26 '18 at 11:31
• Yes because $\bar{D}$ includes the boundary $\partial \bar{D}^n$ , $\mathbb{R}^n$ is homeomorphic to $D^n$ but not to $\bar{D}^n$. – dan Sep 26 '18 at 11:49

We write $$S^n = \{ (x,r) \in \mathbb{R}^n \times \mathbb{R} \mid \lVert x \rVert^2 + r^2 = 1 \}$$, where $$\lVert - \rVert$$ denotes the Euclidean norm.

Question 1:

Define $$p : D^n \to S^n, p(x) = \begin{cases} (\dfrac{\sqrt{1 - (1 - 2\lVert x \rVert)^2}}{\lVert x \rVert}x, 1-2\lVert x \rVert) & x \ne 0 \\ (0,1) & x = 0 \end{cases}$$ This is well-defined because $$1-2\lVert x \rVert \in [-1,1]$$ and $$\left\lVert \dfrac{\sqrt{1 - (1 - 2\lVert x \rVert)^2}}{\lVert x \rVert}x \right\rVert^2 + (1-2\lVert x \rVert)^2 = 1 - (1 - 2\lVert x \rVert)^2 + (1-2\lVert x \rVert)^2 = 1 .$$ $$p$$ is continuous because for $$x \to 0$$ we get $$\left\lVert \dfrac{\sqrt{1 - (1 - 2\lVert x \rVert)^2}}{\lVert x \rVert}x \right\rVert = \sqrt{1 - (1 - 2\lVert x \rVert)^2} \to 0$$.

For $$x \in S^{n-1}$$ we have $$p(x) = (0,-1) = b$$.

Next define $$j : S^n \setminus \{b\} \to int(D^n) = D^n \setminus S^{n-1}, j(y,r) = \begin{cases} \dfrac{1 - r}{2\sqrt{1- r^2}}y & y \ne a = (0,1) \\ 0 & y = a = (0,1) \end{cases}$$ This is well-defined because $$\left\lVert \dfrac{1 - r}{2\sqrt{1- r^2}}y \right\rVert = \dfrac{1 - r}{2\sqrt{1- r^2}}\left\lVert y \right\rVert = \dfrac{1 - r}{2} < 1 .$$ It is easily verified that $$p(j(y,r)) = (y,r)$$ for all $$(y,r)$$ and $$j(p(x)) = x$$ for all $$x \in int(D^n)$$. Hence $$p$$ maps $$int(D^n)$$ bijectively onto $$S^n \setminus \{b\}$$.

$$D^n , S^n$$ are compact Hausdorff, hence $$p$$ is a closed map and thus a quotient map (aka identification map). By the above considerations there exists a unique function $$h : D^n/S^{n-1} \to S^n$$ such that $$h \circ \pi = p$$, and it is a bijection. By the universal property of quotient maps we see that both $$h$$ and $$h^{-1}$$ are continuous, i.e. $$h$$ is a homeomorphism.

Question 2:

Define $$i : S^{n-1} \to S^n \setminus \{a,b\}, i(x) = (x,0) ,$$ $$g : S^n \setminus \{a,b\} \to S^{n-1}, g(x,r) = \dfrac{x}{\lVert x \rVert} .$$ For $$g$$ note that $$\lVert x \rVert \ne 0$$ because $$x \ne a, b$$, i.e. $$r \in (-1,1)$$. We have $$g \circ i = id$$. Next define $$H : (S^n \setminus \{a,b\}) \times [0,1] \to S^n \setminus \{a,b\}, H(x,r,t) = (\sqrt{\dfrac{1 - t^2r^2}{1- r^2}} x,tr) .$$ This is well-defined because $$\left\lVert \sqrt{\dfrac{1 - t^2r^2}{1- r^2}} x \right\rVert^2 + (tr)^2 = \dfrac{1 - t^2r^2}{1- r^2} \lVert x \rVert^2 + t^2r^2 = \\ \dfrac{1 - t^2r^2}{1- r^2} (1 - r^2) + t^2r^2 = 1 .$$ We have $$H(x,r,1) = (x,r), H(x,r,0) = (\dfrac{x}{\sqrt{1- r^2}},0) =(\dfrac{x}{\lVert x \rVert},0) = (i \circ g)(x,r)$$, that is $$i \circ g \simeq id$$.

This means that $$S^n \setminus \{a,b\}$$ and $$S^{n-1}$$ are homotopy equivalent.

More precisely, $$i$$ embeds $$S^{n-1}$$ as a strong deformation retract into $$S^n \setminus \{a,b\}$$ because we have $$H(x,0,t) = (x,0)$$ for all $$t$$.