Show: Quotient space is homeomorphic to unit sphere 
An equivalence relation on $\mathbb{R}$ is given by
  $$
x\sim y\Leftrightarrow x-y\in\mathbb{Z}.
$$
  Show that the quotient space $(\mathbb{R}/{\sim},\tau_1)$ is homeomorphic to $(S^1,\tau_2)$, where $\tau_1$ is the quotient topology and $\tau_2$ the induced topology.

I have to find a bijective continuous function
$$
f\colon \mathbb{R}/{\sim}\to S^1
$$
with $f^{-1}$ continuous.
Do you have an idea how to find such a function?
 A: Let $S^1=\{z\in\mathbb{C}\mid|z|=1\}$. This is a common definition for $S^1$ but may not be the one you've been given.
Let $f\colon\mathbb{R}/{\sim}\rightarrow S^1$ be given by $f([t])=e^{2\pi it}$.
Can you show that this is well defined? (That is, show that if $t_1\in[t]_{\sim}$ and $t_2\in[t]_{\sim}$ then $e^{2\pi it_1}=e^{2\pi it_2}$).
Can you show that this is a homeomorphism?
A: I'm going to give a less technical answer than what others have written here, because I think you must have missed something other than that part if you didn't think of $x\mapsto e^{2\pi ix}$.
You have $x\sim y\Leftrightarrow x-y\in\mathbb{Z}$.  That means $0$ becomes the same point as $1$, while everything between $0$ and $1$ is a different point from the one point that is $0\sim1$.  And $0.1$ becomes the same point as $1.1$, and $0.2$ becomes the same as $1.2$, and so on.  In other words, moving along the interval from $0$ to $1$, you return to your starting point and start over, just as with the circle.  Therefore, the interval $[0,1]$ should get wrapped around the circle, with $0$ and $1$ getting mapped to the same point.  In then the interval $[1,2]$, being the same as $[0,1]$ in the quotient space, gets wrapped around the circle in the same way, with $2$ being mapped to the same point on the circle to which $0$ and $1$ were mapped, and so on.
When you learned trigonometry, you saw $(\cos(2\pi x),\sin(2\pi x))$ wrapping around the circle in that same way.  Since those are periodic functions with period $1$, it follows that if $x\sim y$ then $(\cos(2\pi x),\sin(2\pi x))=(\cos(2\pi y),\sin(2\pi y))$, so that is the mapping you need.
A: Try $f(t)=e^{2\pi i t}$ and find its kernel.
A: What about $f(x) = e^{2\pi i x}$? It factors through the quotient.
