The coordinates of torus is $((R+r\cos \theta)\cos \phi,(R+r\cos \theta)\sin \phi,r\sin \phi)$
So I let $g:[0,2 \pi] \times [0,2 \pi] \to \mathbb S^1 \times \mathbb S^1$
by $g(\theta, \phi)=(e^{i\theta},e^{i\phi})$
Then I prove that this function is continuous, bijection and open mapping. Bijection is quite obvious, but how to prove continuous and open mapping?
The open set in $[0,2 \pi] \times [0,2 \pi]$ is a square without boundary (not sure if it is right) , but what is the open set in $\mathbb S^1 \times \mathbb S^1$ ?
Also if we take an open set in $[0,2 \pi] \times [0,2 \pi]$, say $U$ and act $ g $ on it to be $g(U)$, what will it look like?
Thank you!
Best Answer
I would think about the map in the opposite way:
When you said that the coordinates of the torus are $$((R+r\cos \theta)\cos \phi,(R+r\cos \theta)\sin \phi,r\sin \phi)$$ you were providing a global parametrization of the torus.
This is a map $$f(\theta, \phi) = (x(\theta, \phi), y(\theta, \phi), z(\theta, \phi))$$
With $x(\theta, \phi) = (R+r\cos \theta)\cos \phi$, etc. etc...
Each of the coordinate maps $x,y,z$ is continuous because they simply inherit continuity from the elementary trigonometric functions (finite products and sums of continuous functions are continuous), so the parametrization $f(\theta, \phi)$ is continuous. $f$ is also clearly bijective, as you noted.
Now all you need to show is that $f^{-1}$ is continuous. Equivalently, you said, you could show that $f$ is an open map. Or, better yet, since $f$ is a bijection, it's being an open map is equivalent to being a closed map.
Let's abstract for a moment. If you have any continuous $f: X \rightarrow Y$ with $X$ compact and $Y$ a $KC$-space (compact subsets of $Y$ are closed), then you have that $f$ is a closed map. This is very general and very important! Closed subsets of compact spaces are compact, and compactness is invariant under continuous maps. If compactness implies closedness in $Y$, then $f$ maps closed subsets of $X$ to closed subsets of $Y$ just because it's continuous. So, a function $f$ on a compact space is a homeomorphism iff it is bijective and continuous (so long as $Y$ has a bit of structure).
Now in your case $Y$ is $\mathbb{R}^3$, in which compact is equivalent to closed and bounded, so certainly $Y$ is a $KC$-space. Meanwhile recall that $S^1$ is compact and therefore $S^1 \times S^1$ is compact space since the product of two compact spaces is compact. So, having already seen that $f$ is a continuous bijection, $f: X \rightarrow Y$ is a homeomorphism.