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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.

enter image description here

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    $\begingroup$ What is your definition of the torus? For me, $\Bbb{S}^1 \times \Bbb{S}^1$ is the definition of the torus. $\endgroup$ – Sammy Black Nov 30 '15 at 9:23
  • $\begingroup$ For me it's $C/f$, i.e. the cylinder where you glue the two circle of boundary. $\endgroup$ – Rick Nov 30 '15 at 9:25
  • $\begingroup$ But anyway, haw can $\mathbb S^1\times \mathbb S^1$ (as draw in my picture) can be the torus ? $\endgroup$ – Rick Nov 30 '15 at 9:26
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    $\begingroup$ @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 $S^1 \times [0,1]$. $\endgroup$ – angryavian Nov 30 '15 at 9:27
  • $\begingroup$ Yes sorry, I was thinking about the intersection of the violet tube and the blue tube. It isn't $\mathbb S^1\times \mathbb S^1$ ? $\endgroup$ – Rick Nov 30 '15 at 9:30
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The idea is that a torus is like a circle of circles :) A nice picture that helps the imagination can be found on wikipedia:

enter image description here

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.

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  • $\begingroup$ Very nice. Thank you :-) $\endgroup$ – Rick Nov 30 '15 at 9:53
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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 $\mathbb{R}^2/\mathbb{Z}^2$.

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