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I am new to the concept of Quotient space, and I have an example of a Quotient space from one of my lecture notes, which I can't understand.

Here is the example quoted from the lecture note:

A circle $S^1=\{(x,y)\in\mathbb{R^2}:x^2+y^2=1\}$ can be thought of as the interval $[0,2\pi]$ with end points identified $\mathbb{R}/\mathbb{Z}=\mathbb{R}/\sim$ where $x\sim x' \iff x-x'\in\mathbb{Z}$, and $\mathbb{R}/2\pi\mathbb{Z}=\mathbb{R}/\sim'$ where $x\sim' x'\iff x-x'\in2\pi\mathbb{Z}$. Since $S^1\in\mathbb{R^2}$, it has a subspace topology. We want $[0,2\pi]/\sim$ to get its topology from $[0,2\pi]$ and $\mathbb{R}/\mathbb{Z}$ to get its topology from $\mathbb{R}$. And it should be same topology.

(This lecture note was written by a professor, not by me. And the quote above is exactly quoted from the lecture note.)

My question is:

  1. what does it mean by "...with end points identified..."?

  2. Why do we want $[0,2\pi]/\sim$ to get its topology from $[0,2\pi]$? I mean, shouldn't we want $[0,2\pi]/\sim$ to get its topology from $P([0,2\pi]/\sim)$, the power set of $[0,2\pi]/\sim$ ? (The same question holds for $\mathbb{R}/\mathbb{Z}$.)

  3. What does it mean by "... it should be same topology..."? Does it mean that the topology on $[0,2\pi]/\sim$ should be the same topology on $\mathbb{R}/\mathbb{Z}$? If so, why? And what about the topology on $\mathbb{R}/2\pi\mathbb{Z}$?

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  • $\begingroup$ How did you define the quotient space? There are several ways to do this and the answer depends on that. $\endgroup$ – Vincent Boelens Aug 24 '14 at 20:28
  • $\begingroup$ Actually, the professor who wrote this lecture note didn't defined the Quotient space before the above example. He just wrote the title "Quotient topology" and then started writing this example. So, I don't understand how to understand this example. (But he defined the Product space and open function between two topological spaces.) $\endgroup$ – User Aug 24 '14 at 20:32
  • $\begingroup$ end points identified i.e $x\sim y$ iff $x=0$ and $y=2\pi$ or $x=2\pi$ and $y=0$ or $x=y$(the unique two point identified are $0$ and $2\pi$) intuitively we obtain the circle from $[0,2\pi]$ by gluing the two points $0$ and $2\pi$ . and there are some topological properties hereditary form $[0,2\pi]$ (for example the compactness) $\endgroup$ – Hamou Aug 24 '14 at 20:41
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  1. Look at first $\Bbb R/2\pi\Bbb Z$. This is obtained from $\Bbb R$ by taking the quotient w.r.t. the equivalance relation $\sim'$.
    In other words, every $x\in\Bbb R$ is 'identified with' $x+2\pi k$ for every $k\in\Bbb Z$ in the quotient space.
    Now, each real number is represented in $[0,2\pi]$, moreover, each is represented exactly once, except for the endpoints $0$ and $2\pi$, who are identified to each other in $\Bbb R/2\pi\Bbb Z$.

    To make it visible, we just glue the two endpoints of a thread (of original form straight segment).

  2. So, when we identify (glue) some points to one another in an original space, we arrive to a quotient space, and it has a natural topology inherited from the original space: namely, let a subset $U\subseteq\, X/\sim\ $ open whenever its preimage $\{x\in X\,:\,[x]_\sim\in U\}$ is open.
    Well, you are right in a sense, as the topology is defined in terms of subsets.

  3. It means that the same subsets are open w.r.t. these topologies.
    More precisely, what we have is homeomorphisms (i.e. one-to-one continuous and open preserving mappings) between $\Bbb R/\Bbb Z$, $\ \Bbb R/2\pi\Bbb Z$, $\ [0,2\pi]/\sim'$ and the circle $S^1$ as a subset of $\Bbb R^2$.
    The thing here is that in either case the open subsets are exactly the (arbitrary) unions of open intervals (e.g., this is what you get when you intersect $S^1$ with an open disk on $\Bbb R^2$).

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  • $\begingroup$ For the first question, I understand $\mathbb{R}/2\pi\mathbb{Z}$ was used to identify two end points $0$ and $2\pi$ if I understood correctly. Then, do you have any idea why the professor mentioned $\mathbb{R}/\mathbb{Z}$ in the context of identifying two end points? $\endgroup$ – User Aug 24 '14 at 21:10
  • $\begingroup$ It is easier to work with $\Bbb R/\Bbb Z$ and by scaling ($x\mapsto 2\pi x$ with inverse $x\mapsto x/2\pi$, both continuous) we can obtain a homeomorphism between $\Bbb R/\Bbb Z$ and $\Bbb R/2\pi\Bbb Z$, meaning that these spaces are the same (at least topologically). $\endgroup$ – Berci Aug 24 '14 at 21:25
  • $\begingroup$ @Yk26 He mean $[0,2\pi]$ and not $\Bbb R/\Bbb Z$! . Look carefully this phrase: can be thought of as the interval $[0,2\pi]$ with end points identified $\endgroup$ – Hamou Aug 24 '14 at 21:25
  • $\begingroup$ @Hamou Yes I know. What I am asking is that why $\mathbb{R}/\mathbb{Z}$ is mentioned after the sentence "... can be thought of as the interval $[0,2\pi]$ with end points identified..." parallel with $\mathbb{R}/2\pi\mathbb{Z}$. (i.e. there is no period after the sentence "... can be thought of as the interval $[0,2\pi]$ with end points identified") $\endgroup$ – User Aug 24 '14 at 21:38
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    $\begingroup$ via $ t\mapsto e^{2i\pi t}$, we have $\Bbb R/\Bbb Z\simeq S^{1}$. $\endgroup$ – Hamou Aug 24 '14 at 21:51

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