Collapsing an arc on a sphere (Reduced suspension and unreduced suspension are the same for spheres) Let $n$ be a nonnegative integer, and fix a point $x_0 \in S^n$. Consider the suspension $SS^n$ of $S^n$, which is a quotient space of $S^n \times I$. $SS^n$ contains the line segment $x_0\times I$. How can I show that $SS^n$ is homeomorphic to the quotient space $SS^n/ x_0\times I$?
Since $SS^n$ is homeomorphic to $S^{n+1}$, we can think of this as collapsing an arc on $S^{n+1}$ yields $S^{n+1}$ itself (for $n=0,1$ this is obvious when we draw some pictures). Also, since the space $SS^n / x_0 \times I$ is by definition the reduced suspension $\Sigma S^n$ of $S^n$, this question is equivalent to asking how to show that $SS^n$ and $\Sigma S^n$ are the same. (I know that these two are homotopy equivalent, but I want the stronger result)
Thanks in advance.
 A: Instead of $S^n$ and $SS^n$ we consider $S^{n-1}$ and $SS^{n-1} \approx S^n$.
On $D^n \times \{-1,1\}$ define $(x,-1) \sim (x,1)$ for $x \in S^{n-1}$. Then $\Sigma^n = (D^n \times \{-1,1\})/\sim \phantom{.}$ is the quotient space obtained by gluing two copies of $D^n$ along their boundary. Let  $p : D^n \times \{-1,1\} \to \Sigma^n$ denote the quotient map. Clearly $\Sigma^n \approx S^n \approx SS^{n-1}$. This gives us an identification
$$h : \Sigma^n \stackrel{\approx}{\to} SS^{n-1} .$$
Let $\xi_0 = (0,\ldots,0,1) \in S^{n-1}$ and $L$ be the line segment connecting $0$ and $\xi_0$. The images of $(0,\pm 1)$ under $h$ are the two "suspension points" of $SS^{n-1}$ and the image of $L' = p(L \times \{-1,1\})$ under $h$ is the line segment $\{\xi_0\} \times I \subset SS^{n-1}$.
The quotient map $q : D^n \to D^n/L$ maps $S^{n-1}$ homeomorphically onto $q(S^{n-1})$ and by an abuse of notation we write $S^{n-1} =   q(S^{n-1}) \subset D^n/L$. Thus we can identify $SS^{n-1}/\{\xi_0\} \times I$ with the quotient space obtained from $(D^n/L) \times \{-1,1\}$ by identifying $(x,-1)$ and $(x,1)$ for $x \in S^{n-1}$.
We shall show that there exists a homeomorphism $h : D^n/L \to D^n$ such that $h([x]) = x$ for $x \in S^{n-1}$. This gives us a homeomorphism $(D^n/L) \times \{-1,1\} \to D^n \times \{-1,1\}$ which is compatible with the quotient maps and thus gives us the desired homeomorphism $SS^{n-1}/\{\xi_0\} \times I \to SS^{n-1}$.
Write the points $x \in D^n$ as $x = (\xi,t)$ with $\xi \in D^{n-1}$ and $t \in [-1,1]$ where $\lVert \xi \rVert^2 + t^2 \le 1$. Define $d(\xi) = \sqrt{1 - \lVert \xi \rVert^2}$ and
$$\phi : D^n \to D^n, \phi(\xi,t) = \begin{cases} (\xi,(1 - \lVert \xi \rVert)d(\xi) + t\lVert \xi \rVert) & t \ge 0 \\ (\xi,(1 - \lVert \xi \rVert)d(\xi) + t(2-\lVert \xi \rVert)) & t \le 0 \end{cases}$$
Note that both parts of the definition produce the same value for $t = 0$. Moreover we have in fact $\phi(\xi,t) \in D^n$ since each line segment $S(\xi) = \{\xi\} \times [-d(\xi),d(\xi)] = D^n \cap (\{\xi\} \times [-1,1])$ is mapped by $\phi$ onto itself: The line segment  $S^+(\xi) = \{\xi\} \times \left[0,d(\xi)\right]$ is mapped by $\phi$ linearly onto the line segment $\{\xi\} \times \left[(1- \lVert \xi \rVert) d(\xi),d(\xi)\right]$ and the line segment $S^-(\xi) = \{\xi\} \times \left[-d(\xi),0\right]$ is mapped by $\phi$ linearly onto the line segment $\{\xi\} \times \left[-d(\xi),(1- \lVert \xi \rVert) d(\xi)\right]$; this means $\phi(S(\xi)) = S(\xi)$. Note that $\phi$ is bijective on $S(\xi)$ if $\xi \ne 0$ because then $0 \le (1- \lVert \xi \rVert) d(\xi) < 1$. For $\xi = 0$ the map $\phi$ stretches $S^-(0)$ to $S(0)$ and collapses  $L = S^+(0)$ to $\xi_0$.
Thus $\phi$ induces a map $h : D^n/L \to D^n$ which is a homeomorphism as desired.
We have considered a special $\xi_0 \in S^{n-1}$. Given an arbitrary $x_0 \in S^{n-1}$ we can find an orthogonal linear map $\phi : \mathbb R^n \to \mathbb R^n$ such that $\phi(x_0) = \xi_0$. It induces a homeomorphism $\phi' : S^{n-1} \to S^{n-1}$  which reduces the general case of an arbitrary $x_0 \in S^{n-1}$ to the above special case.
