Why $\mathbb S^1\times \mathbb S^1$ is isomorphic to the torus? Could someone explain why $\mathbb S^1\times \mathbb S^1$ is isomorphic to the torus ? I recall that $\mathbb S^1=\{ x^2+y^2=1\mid x,y\in\mathbb R \}$. To me it would be more something like the picture.

 A: Think of a unit square $[0,1] \times [0,1]$ with opposite sides identified.  We identify $[0,1] \times \{0\}$ with $[0,1] \times \{1\}$ so that $\{t_1\} \times [0,1]$ is a circle for all $t_1 \in [0,1]$.  In particular, this first gluing gives $[0,1] \times S^1$ (which is a cylinder).  Similarly, doing the second gluing, we identify $\{0\} \times S^1$ and $\{1 \} \times S^1$ so that $[0,1] \times \{t_2\}$ is a circle for all $t_2 \in S^1$.  Visually, you're gluing the two ends of a cylinder together to get the torus.  This is $S^1 \times S^1$.
Also, it might be natural to think of $S^1$ as $\mathbb{R}/\mathbb{Z}$ and the torus as $\mathbb{R}^2/\mathbb{Z}^2$. 
A: The idea is that a torus is like a circle of circles :) A nice picture that helps the imagination can be found on wikipedia:

So the idea behind $\mathbb S^1 \times \mathbb S^1$ is that one needs to provide two coordinates $(r,s) \in \mathbb S^1 \times \mathbb S^1$ to specify a point on the torus: $r$ tells you which red circle you want to be on (by specifying a point on the violet circle) and $s$ tells you where on this red circle you want to be.
Obviously this can be made more rigorous but I think this will already help with the intuition.
