Source:A first course on topology. Robert Conover
Ex 6,pg 90
Problem
Let $S^1$ denote the unit circle as subspace of the Euclidean plane. Let T denote the (hollow)torus as subspace of 3-space with product topology that gets by
virtue of being
R$\times$ R $\times$ R
(Where each copy or R has its usual topology)
Let $S^1 \times S^1$ have the product topology and show that $S^1 \times S^1$ and T are homeomorphic
What I know
Here is a possible analytic def of T and some ideas I found
on how to do it.
http://www.homepages.ucl.ac.uk/~ucahjde/tg/html/topsp07.html
I know how to do homeomorphic problems
when simple,but this one baffles me.
Procedure is:
1.show it is bijection
2.Continuous
3.inverse exists
I can say $S^1 \times S^1$ are 2 closed circles and is continuous within the subspace of the torus. I don’t know how to show the math for it. There does not seem any neat way to do
a bijection between a sphere and a torus that I can find.
Any help in showing the proof would be appreciated
Best Answer
Hint: $T$ is presumably given as a surface of revolution with a circle as the generatrix and a suitable axis of rotation. This description will immediately give the map you are looking for from $S^1 \times S^1$ to $T$. To prove that it is a homeomorphism, show that it is bijective and continuous and observe that an bijective continuous mapping between two compact hausdorff spaces is closed and hence its inverse is continuous.