How to visualize quotient space?

Question

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?

Answer 1

1. 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.

2. 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.

Answer 2

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.