How do I show that $S^1$ is the suspension of $S^0$?

Question

How do I show that $S^1$ is the suspension of $S^0$?

I have all the definitions here, I'm just bad at applying them. The suspension of a topological space $X$ is the quotient $CX / (X × ${$1$}$)$, where $CX$ is the cone on $X$, the quotient space $(X × [0, 1])/(X × ${$0$}$)$.

Answer 1

I think it would be helpful for you to draw a picture yourself, then write a proof. If you cannot do it after you draw the picture, I am sure others will be happy to help you with the proof writing process. By the way, your definition of suspension is not (but equivalent to) the usual one.

Answer 2

For convenience let $0$ and $1$ be the two points of $S^0$. Then the suspension of $S^0$ is the union of two line segments $x = 0$ and $x = 1$ ($0 \le y \le 1$) in the plane, where $\{(0,0)\}$ is identified with $\{(1,0)\}$ and $\{(0,1)\}$ is identified with $\{(1,1)\}$. This is homeomorphic to a circle.