$S^n \times [0,2\pi]/\sim \cong S^{n+1}?$

Question

I have a question: "Is it true that $S^n \times [0,2\pi]/\sim \cong S^{n+1}$ where $S^{n}\times \{0\}$ collapses to a point and $S^n\times \{2\pi\}$ collapses to another point? "

Geometrically, this is true if $n=1$. The left hand side is a collection of circle, being the surface of a cylinder. After we collapse the top base and bottom base to points, we obtain a sphere.

Can we prove the general case with an explicit homeomorphism?

Answer 1

Yes, this is called a suspension https://en.wikipedia.org/wiki/Suspension_(topology)

in particular the suspension of $S^n$ is $S^{n+1}$.

Answer 2

An explicit homeomorphism can be constructed as follows.

Clearly $[0,2\pi]$ is homeomorphic to $\displaystyle{[-\frac{\pi}{2},\frac{\pi}{2}]}$. Consider $S^{n}\subset \mathbb R^{n+1}$ and $S^{n+1}\subset \mathbb R^{n+2}$ where $\mathbb R^{n+1}$ has coordinates $(x_0,\dots,x_n)$ and $\mathbb R^{n+2}$ has cohordinates $(x_0,\dots,x_{n+1})$, so that $S^{n}=\{x_0^2+\dots+x_n^2=1\}$ and similarly for $S^{n+1}$. Then consider $$f:S^n\times[-\frac{\pi}{2},\frac{\pi}{2}] \to S^{n+1}$$ $$f(x_0,\dots,x_n,\theta)=(\cos\theta x_0,\dots,\cos\theta x_n,\sin\theta).$$ Such map is continuous and constant on $S^n\times\{\pi/2\}$ and $S^n\times\{-\pi/2\}$, hence defines a continuous map $F$ from $(S^n\times[-\pi/2,\pi/2])/\sim$ to $S^{n+1}$, and is immediate to check that $F$ is 1-1 and onto. Since $S^n\times[-\pi/2,\pi/2]$ is compact and $S^{n+1}$ is Hausdorff, then $F$ is in fact an homeomorphism. (Alternatively, one can easily check that $F^{-1}$ is continuous.)