Help to understand the notion of quotient space I'm reading up on quotient spaces and quotient maps. For the sake of simplicity please refer to the wikipedia article.
I want to see how this works by looking at an example, specifically the map $q: [0,1] \to [0,1]/\{0,1\}$. A couple of questions:  


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*I understand the domain to be the set of real numbers from 0 to 1 (inclusive), and the co-domain to be the set of real numbers from 0 to 1 (exclusive), union with the set $\{0,1\}$, i.e. $img(q) = \{\{0,1\},0.000...,...,0.999...\}$. Is this correct?

*Are we considering $[0,1] \text{ and } [0,1]/\{0,1\}$ as subspaces of $\mathbb{R}^2$? If so, are they endowed with the subspace topology?  

*Irrespective of the answer to the question above, what are the open sets in the domain and the co-domain?  

*The article say that $[0,1]/\{0,1\}$ is homeomorphic to the circle $S^1$. How is it so? Can a specific homeomorphism be given here?  


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Edit: the questions above have become more clear to me, thanks to the help of some users. I'd like to cite another example on the same topic, with a few questions:

I understand the co-domain to be 
$\mathbb{R}/\mathbb{Z} = \{[0],...,[r]|\text{where } r \lt 1\}$,
while the domain $\mathbb{R}$ is equipped with the usual open sets $(a,b)$.
If we equip $\mathbb{R}/\mathbb{Z}$ with the quotient topology (i.e. $O/p$ as it is denoted in the definition below), it would follow that the open sets in this space $\mathbb{R}/\mathbb{Z}$ should also be sets of the form $(a,b), 0 \leq a \lt b \leq 1$.
But if we just consider it as a space without any relation to the domain $\mathbb{R}$ (i.e. its topology is $O_1$, as it is denoted in the definition below), then it seems the open sets would be of the form $[a,b), 0 \leq a \lt b \leq 1$.
Hence $O/p \neq O_1$.
Where did I go wrong with my analysis?

 A: The geometric picture you should have in mind is gluing together the two endpoints of a line segment to get a circle.
Yes the image is $(0,1) \cup \{\{0,1\}\}$.
$[0,1]$ is is being considered under the subspace topology (of either $\mathbb{R}$ or $\mathbb{R}^2$). A priori, the quotient space is not a subspace of $\mathbb{R}^2$, we have to prove that it is homeomorphic to $\mathbb{S}^1$.
The open sets in $[0,1]$ follow from the above definition.
Here is what defines the quotient topology on the image: $V$ is open in the image iff $q^{-1}[V]$ is open in the domain.
Consider a subset $V$ of $(0,1) \cup \{\{0,1\}\} $. 
If $\{0,1\} \notin V$, then $V$ is open in $(0,1) \cup \{\{0,1\}\}$ iff $V$ is open in $[0,1]$.
If  $\{0,1\} \in V$, then $V$ is open in $(0,1) \cup \{\{0,1\}\}$ iff $(V \setminus \{\{0,1\}\}) \cup \{0,1\} $ is open in $[0,1]$.
The homeomorphism  $(0,1) \cup \{\{0,1\}\} \rightarrow \mathbb{S}^1$ is given by
$x \mapsto (\cos{2\pi x},\sin{2\pi x})$ for $x \neq \{0,1\}$ and $x \mapsto (1,0)$ when $ x = \{0,1\}$ 
Edit: In fact the image of $q$ should be $\{\{x\} : x \in (0,1)\} \cup \{\{0,1\}\}$. But since we are going to give it a topology, we might as well have the image  to be $(0,1) \cup \{\{0,1\}\}$ 
A: img q = { {0,1}, {r} : r in (0,1) } which are the equivalence classes of the equivalence relation defining the quotient space:
x ~ y iff x,y in {0,1} or x = y.  
[0,1] is a closed subspace of R.  The quotient is not a subspace of R nor R$^2$.  
U is an open set of the quotient iff f$^{-1}$(U) in an open set of [0,1].
