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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?

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    $\begingroup$ Does in $D^n$ you mean the closed disc? ($\bar{D_n} = \{x\in \mathbb{R}^n : ||x|| \le 1\}) $ $\endgroup$ – dan Sep 26 '18 at 9:21
  • $\begingroup$ @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. $\endgroup$ – Hawk Sep 26 '18 at 9:22
  • $\begingroup$ 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. $\endgroup$ – dan Sep 26 '18 at 10:19
  • $\begingroup$ 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? $\endgroup$ – Hawk Sep 26 '18 at 11:31
  • $\begingroup$ Yes because $\bar{D}$ includes the boundary $\partial \bar{D}^n$ , $\mathbb{R}^n$ is homeomorphic to $D^n$ but not to $\bar{D}^n$. $\endgroup$ – dan Sep 26 '18 at 11:49
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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 $t$.

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