Suspension of interval is homeomorphic to $D^2$

Question

I want to show that the suspension of $X= [-1,1]$ is homeomorphic to $D^2$ by giving an explicit bijection. I can visualize it, because the cylinder of [-1,1] is a disc when you collaps $X\times\{0\}$ and $X\times\{1\}$, but I haven't found a function yet. Once I have a continuous bijection, I know that it is a homeomorphism. Can someone maybe help me?

Answer 1

Let be $f: X=[-1,1]\times[0,1]-> Y=D^2$ $f(x,t)-> (x\sqrt{t(t-1)},t)$ It is a quotient map(identification map). The only real matter is to show that it is closed, but f is continous, X is compact(Tychonoff) and Y is T2.