Are $S^1$ and $\mathbb{R}/{\sim}$ the "same thing"? I am reading quotient space of topology and I am a little bit confused. I am looking at the relationship between $\mathbb{R},S^1$ and the quotient space $\mathbb{R}/{\sim}$, where the relation $\sim$ corresponds to the partition $\mathbb{R}=\mathbb{Z}\cup(\mathbb{R}-\mathbb{Z})$.
Function $f:\mathbb{R}\to S^1$ give by $f(t)=(\cos(2\pi t),\sin(2\pi t))$ is continuous, onto but not one-to-one. We can also define function $g:\mathbb{R}\to \mathbb{R}/{\sim}$ where $g(t)=[t]$.
The circle $S^1$ and the quotient space $\mathbb{R}/{\sim}$ is not the same mathematical object, I suppose. But intuitively speaking they should be the "same thing". So there should be some relationship (bijective function, I guess) between these two spaces. But I was considering the function $\pi\circ f^{-1}$ and it's not injective. Maybe I should define the equivalence relation in a different way? Like "$x\sim y$ if $x\equiv y\pmod {2\pi}$" Also, suppose we can find such a function, can it be a homeomorphism? 
 A: The relation is not given by that partition. As said in the comments, the relation is:
$x \sim y \iff x=y \mod 2\pi$
Now, consider the following map:
$f: \mathbb{R} \rightarrow S^1$; $x \mapsto e^{i x}$
Since it takes equivalents to the same image, the induced map:
$\tilde{f}: \mathbb{R} /{\sim} \rightarrow S^1$
is continuous.
Now, take the map $g: S^1 \rightarrow \mathbb{R}/{\sim}$; $e^{ix} \mapsto [x]$. It is obviously well defined, and easily seen to be continuous. Note that $g$ is the inverse of $f$. Therefore, the spaces are homeomorphic. 
A: We could also show that $f\colon\mathbb R\to S^1$ given by $$f(x)=e^{ix}$$ is an open map. To check this, it is enough to check that image of any open interval $(a,b)\subseteq\mathbb R$ is an open set in $S^1$. (If image of every set from a basis is open, then the map is open.)
The we could check continuity of $f$, which can be done in various ways. For example, we can use $f(x)=\cos x+i\sin x$ and the fact that cosine and sine are continuous functions.
It is also clear that $f$ is surjective.
Once we get this, we can use the fact that every continuous surjective open map is a quotient map.
