How to visualize quotient space? I'm trying to understand quotient spaces. 
The first example is [0,1]/~: 
a~b <=> |a-b|=1
=> [0,1]/~ = {(x,y) $\in$ $\mathbb{R}^2$|$x^2+y^2$=1}
What I get from the definition of a~b is that we "glue" together the point a = (0,0) and b = (1,0). But this definition doesn't say anything about the points "in the middle". What happens to them? How do we get a circle if we only do things to the endpoints?
 A: *

*The line segment $[0,1]$ is one dimensional, its elements have one coordinate. 
But, yes, all what is happening here is that we glue together the two endpoints $0$ and $1$ of this segment.

*The 'equation' you write should be read as homeomorphic, and it means that there are continuous functions between the two topological spaces which are inverses of each other. 
There is a standard map you can define $\varphi:[0,1]\to\Bbb R^2$ such that its image $\varphi([0,1])$ covers the circle once except for one point, $\varphi(0)=\varphi(1)$. 
Specifically, this map induces a homeomorphism from $[0,1]/\sim$ to the circle.
A: As all relations this relation induces an equivalence relation.
Note that the set of equivalence classes is $X:=\{\{x\}\mid x\in(0,1)\}\cup\{\{0,1\}\}$
Prescribe the function $\nu:[0,1]\to X$ by $x\mapsto\{x\}$ if $x\in(0,1)$ and $x\mapsto\{0,1\}$ otherwise.
Then let $X$ be equipped with quotient topology: $$\tau_X=\{U\in\wp(X)\mid\nu^{-1}(B)\text{ is open in }[0,1]\}=\{U\in\wp(X)\mid\bigcup U\text{ is open in }[0,1]\}$$
Then $\nu:[0,1]\to X$ is a quotient map.
Further prescribe $f:[0,1]\to Y:=\{\langle\cos2\pi t,\sin2\pi t\rangle\mid t\in[0,1)\}$ by $t\mapsto\langle\cos2\pi t,\sin2\pi t\rangle$.
Function $f$ is continuous, surjective and closed, hence is a quotient map.
Further $f$ and $\nu$ respect each other in the sense that: $$\nu(r)=\nu(s)\iff f(r)=f(s)$$
That implies the existence of unique  maps $h:X\to Y$ and $g:Y\to X$ such that $\nu=g\circ f$ and $f=h\circ\nu$.
Then $g\circ f$ and $h\circ\nu$ are both continuous, and the fact that $\nu$ and $f$ are both quotient maps allows the conclusion that $h$ and $g$ are both continuous. 
Further based on uniqueness it can be shown $h\circ g=\text{id}_Y$ and $g\circ h=\text{id}_X$ so we conclude that $h$ and $g$ are homeomorphisms.
So the spaces $X=[0,1]/\sim$ and $Y$ (a circle) are homeomorphic.
