Question about an example of Quotient space. 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:


*

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

*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 $\mathbb{R}/\mathbb{Z}$.)

*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 $\mathbb{R}/2\pi\mathbb{Z}$?
 A: *

*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).

*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.

*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 $\Bbb R^2$).
