TITLE: How should I visualize $S^n$ as the reduced suspension of $S^{n-1}$?
QUESTION [11 upvotes]: Or is there a particularly canonical homeomorphism between $S^n\subset \mathbb{R} ^{n+1}$ and $SS^{n-1}$ (the reduced suspension)?  Right now the only thing that comes to mind is to think of the sets $\{t\}\times S^{n-1}, 0 \leq t \leq 1$  in $SS^{n-1}$ as corresponding to the sets $p^{-1}(K)$ where K is a hyperplane in $\mathbb{R}^n$ and $p$ is stereographic projection $S^n\rightarrow \mathbb{R}^n$ from the north pole of $S^n$ (as the $t$ parameter moves from $0$ to $1$ the $K$ sweep out $R^n$ through parallel hyperplanes).

REPLY [19 votes]: It is convenient to think of this in terms of the smash product $X \# Y$ of two spaces with base point. This is the space  $X\times Y$ with $X \vee Y= X \times * \cup * \times Y$ shrunk to a point. Then the reduced suspension of $X$ is really $X \# S^1$. The general result is that 
$$S^m \# S^n \cong S^ {m+n}.$$
One way of seeing this is to say that if $S^m$ has the cell structure $e^0 \cup e^m$, and similarly $S^n$, then $S^m \times S^n$ has the cell structure 
$$e^0 \cup e^m \cup e^n \cup e^{m+n}, $$
and the first three cells determine $S^m \vee S^n$. So when you form the smash product $\ldots$
Here is a figure illustrating $S^1 \# S^1$ from Topology and Groupoids.