Quotient Map from R to S1 Hi please could someone check my understanding here please? I've been asked to find a quotient map from $\mathbb R$ to $S^1$.
Is the quotient map the surjective function from $\mathbb R$ to $\mathbb R/\sim$ where an open set in $\mathbb R/\sim$ is open iff it's pre-image is open in $\mathbb R$, or is it the surjective function from $\mathbb R$ to $S^1$ where an open set in $S^1$ is open iff it's pre-image is open in $\mathbb R$?
I've been using my lecture notes and the book by W A Sutherland but I'm struggling to follow the phrasing for questions such as this.
TIA
 A: Let's consider the map $\exp_I:I\to S^1$ such that $t\mapsto(\cos(2\pi t),\sin (2\pi t))=e^{2\pi it}$.
First we want to show that $\exp_I$ is an identification. So, in particular we have to prove that the topology on $S^1$ is the same topology induced by $\exp_I$.
Firts we can observe that $\exp_I$ is continuous because $\exp:\mathbb R\to \mathbb R^2$ is continuous.
We call $\mathcal T_d$ the topology on $S^1$ induced by the euclidean distance and we'll show that $\mathcal T_d=\mathcal T_{\exp_I}$.
The open saturated sets in $[0,1]$ are like $[0,a)\cup(b,1]$ or $(c,d)$, with $0<a<b<1$ and $0<c<d<1$. Notice that $\exp_I^{-1}(\exp_I([0,a)))=[0,a)\cup \{1\}$, which is not open.
The open sets in $(S^1,\exp_I)$ are given by the images of the sets in $\{[0,a)\cup(b,1],(c,d)\}\overset{a,b,c,d\in \mathbb R}{\underset{0<c<d<1}{_{0<a<b<1}}}$, which are arches on $S^1$. Since open sets of $\mathcal T_d$ are given by the intersection of $\epsilon$-balls in $\mathbb R^2$ with the circle $S^1$, we have shown that $\mathcal T_{\exp_I}<\mathcal T_d$.
Conversely is quite easy because open sets for $\mathcal T_d$ are the sets $B_{\epsilon}(p)\cap S^1$ that are open for $\mathcal T_{\exp_I}$ and this shows that $\mathcal T_d<\mathcal T_{\exp_I}\implies \mathcal T_d=\mathcal T_{\exp_I}\implies\exp_I$ is an identification and we can use the universal property of quotient.
We put on $I$ the topology induced by the euclidean topology on $\mathbb R$, so $A\subseteq I$ is open if $\exists U \subseteq \mathbb R$ open such that $A=U\cap I$.
On $S^1$ we consider the topology induced by the map $\exp_I$, so a set $A$ is open for $\mathcal T_{\exp_I}\iff \exp^{-1}(\exp_I(A))=A$.
We now define an equivalence relation $\sim$ such that $$t\sim t'\iff\begin{cases}t=t'\\t,t'\in \{0,1\}\end{cases}\iff\exp_I(t)=\exp_I(t').$$
Let $\pi:I\to I/\sim$ be the quotient map, whose topology on $I/\{0,1\}$ is the topology induced by the map $\pi$.
Since $\exp_I$ is an identification and respect the property $[t]_{\sim}=[t']_{\sim}\implies \exp_I(t)=\exp_I(t')$, we say that $\bar{\exp_I}:I/\{0,1\}\to S^1$ is well-defined, hence injective. Furthermore $\bar{\exp_I}$ is surjective because $\exp_I$ is surjective and it's also continuous for the universal property of quotient $\implies\bar{\exp_I}$ is an homeomorphism whose inverse is $\bar \pi:S^1\to I/\{0,1\}$. 
We can conclude that $(I/\{0,1\},\mathcal T_{\pi})\cong(S^1,\mathcal T_d)$.
