The quotient space $Y = X / \sim$ as a set is just the set of equivalence classes of $X$ under $\sim$, so the set $\{ [x]: x \in \mathbb{R} \} $ in your case.
The equivalence class of a number $x$ is just (in your case) the set $\{ x+n : n \in \mathbb{Z} \}$. Now we need a topology. The standard topology that we take on $Y$ is all subsets $O$ of $Y$ (where points in $Y$ are "really" subsets of $X$, the equivalence classes) such that $q^{-1}[O]$ is open in the topology of $X = \mathbb{R}$. Here $q$ is the map that sends $x$ to its class $[x]$ in $Y$, the so-called quotient map. This is called the quotient topology on $Y$, and as you see it assumes you have a topology on $X$ already, and we give $Y$ the largest topology possible to still have $q$ continuous. (The smallest one would always be the indiscrete topology, which is not very interesting, hence the other "natural" choice.)
Now, if we have a function $f$ from $Y$, the quotient space in the quotient topology, to any space $Z$, then $f$ is continuous iff $f \circ q$ is continuous as a function from $X$ to $Z$: one way is clear, as the composition of continuous maps is continuous, and for the other side, if $O$ is open in $Z$, by definition $f^{-1}[O]$ is open in $Y$ iff $q^{-1}[f^{-1}[O]]$ is open in $X$, and this set equals $(f \circ q)^{-1}[O]$ which is open, as by assumption $f \circ q$ is continuous.
Now, consider the map $f$ that sends the class $[x]$ to the point $e^{2\pi ix}$ in $\mathbb{S}^1$, the unit circle.
This is well-defined: if $x'$ were another representative of $[x]$, then $x \sim x'$ and thus $x - x'$ is an integer and so $f(x') = f(x)$.
It is continuous, as $f \circ q$ is just the regular map sending $x$ to $e^{2\pi ix}$, and this is even differentiable etc.
It is clearly surjective and injective because the only way $[x]$ and $[y]$ will have the same value is when $2 \pi ix - 2 \pi i y$ is an integer multiple of $2 \pi i$, which happens iff $x - y$ is an integer.
One can also check that $q[X] = q[[0,1]]$ and by continuity of $q$ we have that $Y$ is compact. This makes (with $\mathbb{S}^1$ Hausdorff) the map $f$ a homeomorphism, by standard theorems.
We could also have achieved this as the quotient of $[0,1]$ under the equivalence relation that has exactly one non-trivial class, namely $\{0,1\}$. This is more intuitive, as we then glue together (consider as one point) just the points $0$ and $1$, and this geometrically gives a circle. In your example (which is a nice so-called covering map, and a group homomorphism as well) we glue a lot more points together, but all classes are now similar: just shifted versions of a point by an integer. We sort of wrap the interval $[0,1)$ infinitely many times over itself.
Try to find for a point $((a,b),z)\in S^1\times [-1,1]$ a point $(x,y,z)$ in $S^2$ with the same $z$-coordinate and $x=\lambda a,\ y=\lambda b$ for a $\lambda$ which is a function in $z$.
Then show that the map $f:((a,b),z)\mapsto (x,y,z)$ is surjective and continuous, and that it induces a map $\tilde f:X^*\to S^2$. Is $\tilde f$ injective?
Best Answer
Yes. Let $Y$ be $[0, 1]$ with the discrete topology, let $Y'$ be $[0,1]$ with the Euclidean topology, and let $X = Y \sqcup Y'$. Let the equivalence relation on $X$ be the least equivalence relation such that $a \in Y$ is equivalent to $a \in Y'$. Then $\sim$ on $Y$ is just the identity relation, so $Y/\sim$ is just $Y$; but ${X/\sim} \cong Y'$. Clearly the identity $Y \to Y'$ is not a homeomorphism.