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I'm studding some few topology, and there is some surprising result to me :

1) Why is $\mathbb R/\mathbb Z$ a circle ?

First of all, a logical thing would have been that $\mathbb R/\mathbb Z\cong [0,1)$ it would be so natural ! So why it can't work ? Then I tried to show that $g:[t]\longmapsto e^{2\pi it}$ is an homeomorphism, but I can't show continuity, bijectivity and the fact that $g(U)$ is open.

2) Why $\mathbb R^2/\mathbb Z$ is a cylinder and $\mathbb R^2/\mathbb Z^2$ a torus ? In other words, what would be the homeomorphism ?

3) Why $\mathbb S^2$ is homeomorph to the torus ? What is the homeomorphism

4) I know that the torus is also metrizable, so let denote $M$ the metric space of the torus. Does it mean that $\mathbb R^2/\mathbb Z^2\cong \mathbb S^2\cong M$ ?

I'm sorry, I'm sure it's obvious, but I really have problem to understand.

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If we take $\mathbb{R}/\mathbb{Z}$ as the group quotient so we mod out via the equivalence relation $x \sim y$ iff $x - y \in \mathbb{Z}$, not that we identify $\mathbb{Z}$ to a point and keep all other points separate (as is also often denoted by this notation):

Then $0 \sim 1$, so we can approach $0 = 1$ from two directions, both $\frac{1}{n} $ and $1 - \frac{1}{n}$ have the same limit, so all points have near neighbours on both sides. You have sort of bent $[0,1)$ so that the 1 gets close to 0, and eventually identical with it. You glue the $0$ and $1$ together, as it were.

To see an explicit homeomorphism, let $f(t) = e^{2\pi it}, f: \mathbb{R} \rightarrow S^1$, then $f$ is continuous. You have to check that $x \sim y$ iff $f(x) = f(y)$. Then define $\hat{f}: \mathbb{R}/\mathbb{Z} \rightarrow S^1$, by $\hat{f}([x]) = f(x)$. The previous condition on $f$ and $\sim$ is a reformulation that $f$ is well-defined (as $x \sim y \rightarrow f(x) = f(y)$ so the representative of the class does not matter for the value) and 1-1 (as $f(x)= f(y) \rightarrow x \sim y \rightarrow [x] = [y]$). It's continuous: if $O \subset S^1$ is open, then $\hat{f}^{-1}[O]$ is open by the definition of quoient topology iff $q^{-1}[\hat{f}^{-1}[O]] = f^{-1}[O]$ is open in $\mathbb{R}$, and this is just continuity of $f$.

We don't need to show openness of $\hat{f}$ as w

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The thing is that the notation $\mathbb R /\mathbb Z$ does not suffice to describe what really is going on. A quotient topology is always generated by an equivalence relation $\sim$ on the topological space $X$. The case where you will find the circle is $x\sim y:\Leftrightarrow x-y\in\mathbb Z$. The process of building the quotione space is now gluing together all points which are said to be equivalent by $\sim$. Therefore, usually one writes $\mathbb R/\!\sim$ for the quotient and the notation $\mathbb R/\mathbb Z$ is only used because the relation might be clear from the context.

Further, $\mathbb R^2/\mathbb Z$ is defined using $(x,y)\sim (x',y') :\Leftrightarrow x-x'\in\mathbb Z$ and $\mathbb R^2/\mathbb Z^2$ is defined using $(x,y)\sim (x',y') :\Leftrightarrow (x-x',y-y')\in\mathbb Z^2$. When you have done topology for some time, these definition apeare pretty natural.

So but why does $\mathbb R/\mathbb Z$ gives a circle using $\sim$. Here is an intuitive explanation: Imaging you are at $1-\varepsilon\in\mathbb R$, so close to the upper end of the interval $[0,1]$. Moving up will bring you to the $1$ and then to $1+\varepsilon$. But $1+\varepsilon$ is glued to $\varepsilon$ (because $(1+\varepsilon)-\varepsilon=1\in\mathbb Z$). So for all practical purpose, you are now in $\varepsilon$, close to the lower end of the interval $[0,1]$. By just moving a distance of $2\varepsilon$ you made it from one end to the other of the interval. So it seems like the ends are glued together. And you would agree, that gluing the ends of $[0,1]$ will give you a circle.

It is similar for the cylinder: gluing two of the opposite edges of a sheet of paper will give you a tube, or in mathematical langueage, a cylinder. Then gluing the opposite edges of the cyclinder (which are circles) closes the cylinder to a torus.