Why is $S^1 \times S^1$ a Torus? 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?

Wow, its been a long time since the last post. 
Well, there's a nice way to show that it is a homeomorphism; its given in the book "Basic Topology" by M.A. Armstrong. (Pg 67 ~ 68)
I'll give a rough sketch :
Take $X = I \times I$ where $I = [0,1]$ and partition $X$ into:
1) The set {(0,0), (1,0), (0,1), (1,1)} (four corner points)
2) Sets consisting of pairs of points $(x, 0), (x, 1), x \in I$
3) Sets consisting of pairs of points $(0, y), (1, y), y \in I$
4) Sets consisting of a single point $(x, y)$ where $x, y \in I$
Then the resulting identification space is the torus. If you think of $S^1$ as inside the complex plane, define a map $f : S^1 \times S^1 \rightarrow I \times I$ by
\begin{align*}
(x,y) \mapsto (e^{2\pi ix}, e^{2\pi iy})
\end{align*}
Now look at $f^{-1}(z)$ for $z \in S^1 \times S^1$. They partition $I \times I$ - and they're the sets mentioned above.
$\Rightarrow f$ is an identification map. (If $f : X\rightarrow Y$ is a surjective (continuous) map, $X$ is compact and $Y$ is Hausdorff, then $f$ is an identification map) 
Thus, the two are homeomorphic.
