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.

• What is your definition of the torus? For me, $\Bbb{S}^1 \times \Bbb{S}^1$ is the definition of the torus. – Sammy Black Nov 30 '15 at 9:23
• For me it's $C/f$, i.e. the cylinder where you glue the two circle of boundary. – Rick Nov 30 '15 at 9:25
• But anyway, haw can $\mathbb S^1\times \mathbb S^1$ (as draw in my picture) can be the torus ? – Rick Nov 30 '15 at 9:26
• @Rick You've drawn two copies of $S^1 \times \mathbb{R}$ for some reason. $S^1$ is the result of gluing together the ends of the interval $[0,1]$. Similarly, $S^1 \times S^1$ is the result of gluing together the ends of the hollow tube $S^1 \times [0,1]$. – angryavian Nov 30 '15 at 9:27
• Yes sorry, I was thinking about the intersection of the violet tube and the blue tube. It isn't $\mathbb S^1\times \mathbb S^1$ ? – Rick Nov 30 '15 at 9:30

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.
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 $\mathbb{R}^2/\mathbb{Z}^2$.