Call a "torus" that geometric shape that "looks like" a doughnut. Frequently, one encounters the assertion that $S^1 \times S^1$ is a "torus" where $S^1$ is the unit circle. Now, if I think about this, I can understand the justification for calling this a torus, but I'm trying to understand how one would go about actually proving this. Indeed, there exist analytical descriptions of the torus such as this one provided by Wikipedia. So one could theoretically, with enough inspiration, find a homemorphism $h: S^1 \times S^1 \rightarrow G(T)$ where $G(T)$ denotes the graph of the torus as realized by the analytical description. This approach, assuming it works, uses coordinates and in any event wouldn't be very enlightening.
So, my question is, is there a coordinate-free way to prove that $S^1 \times S^1$ is homeomorphic to this thing we call a doughnut?
My thoughts on this are: I believe what is key is how one chooses to define a torus. I am familiar with constructing a torus by identifying opposite sides of a rectangle and this seems like a pretty natural definition. It is intuitively clear that if the rectangle is configured as
$$
\begin{align*}
A—- & B \\
|\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&| \\
C—- & D
\end{align*}
$$
then one can map one circle homeomorphically onto the identification of AB and CD and then the other circle onto the identification of AC and BD. However, this really isn't a very precise argument and it seems to me that making it precise would eventually involve coordinates. Am I onto the right track with this approach or is there a better way of looking at the problem?
Best Answer
I think you're thinking of it in the right way, but you're also trying to have your cake and eat it here. If you want a completely rigorous argument, you can't get around having a rigorous description of the 3d shape that you're trying to prove homeomorphic to $S^1\times S^1$ in the first place. But that means, in particular, that you need a well-defined way of speaking on particular points on the geometrical torus, and how would you do that if not by coordinates?
In fact, one doesn't need that much inspiration to find a concrete coordinate-based homeomorphism. Let's define $S^1$ as the set $\{(x,y)\mid x^2+y^2=1\}\subset \mathbb R^2$ with the subspace topology. Then, $$h((x,y),(z,w)) = z(0,0,1)+(w+2)(x,y,0)$$ is your desired homeomorphism, into the set $$\{(x,y,z)\mid z^2 + \Bigl(2-\sqrt{x^2+y^2}\Bigr)^2 = 1\}\subset \mathbb R^3$$ which one ought to recognize as our geometrical torus.
To describe this as "not very enlightening" would be fair if it was about purely formal algebraic manipulation or black-box recipes for computation -- but in fact it has a clear geometric significance. One might say, instead:
However, such a verbal description risks getting complex and hard to grasp, and it is all too easy for the writer to introduce ambiguities where it can be understood only if the reader already understands which construction is meant. The coordinate formulas, on the other hand, are unambiguous, and with a bit of practice one can see the geometric construction directly in them.