Homeomorphism between $S^1$ and $[0,1]/ 0\sim1$ I'm studying quotient spaces in my topology course.
I want to prove that circle $S^1=\{(x,y)\mid x^2+y^2=1 \}$ homeomorphic to $[0,1]/ 0\sim1$ (segment with the identified points). 
To solve this problem, I use the following theorem:

If $f:X\longrightarrow Y$ is continuous and surjective map, $X$ is compact, $Y$ is Hausdorff then the map $f/S(f): X/S(f) \longrightarrow Y$ is homeomorphism.

So if we assume the map $f: [0,1]\longrightarrow S^1$ by the rule $f(t)=(\cos(2\pi t),\sin(2\pi t))$. This map and spaces satisfy all the conditions of theorem. So $S^1\approx [0,1]/ 0\sim1$
But I think that the use of this theorem is not necessary here. Maybe you know any other ways to solve this problem.  
Edit:
A map $f : X \longrightarrow Y$ determines the partition of the set X into nonempty
preimages of the elements of Y . This partition is denoted by $S(f)$.
 A: Try to go in the other direction if you want to avoid the universal property of quotient spaces (which is the most natural approach for the direction $[0,1]/\sim\longrightarrow S^1$, and can not really be avoided).
Let $\log z$ denote the principal branch of $\log$ on$\mathbb{C}\setminus (-\infty,0]$. Recall that in particular, $\log$ is continuous and $\log(e^{i\theta})=i\theta $ for all $\theta$ in $(-\pi,\pi)$.
Now consider the map
$$
f:z\longmapsto \frac{\frac{1}{i}\log z +\pi}{2\pi}.
$$
It is a continuous bijection from $S^1\setminus \{-1\}$ onto $(0,1)$.
By composition with the canonical surjection $x\longmapsto [x]$ of $[0,1]$ onto $[0,1]/\sim$, we get
$$
\bar{f}:z\longmapsto \left[\frac{\frac{1}{i}\log z +\pi}{2\pi}\right]
$$
continuous from $S^1\setminus \{-1\}$ to $\left[0,1\right]/\sim$.
Now 
$$
\lim_{z\rightarrow -1} \bar{f}(z)=[0]=[1].
$$
So we can extend $\bar{f}$ by continuity to a continuous $\bar{f}:S^1\longrightarrow [0,1]/\sim $.
It is clear that $\bar{f}$ is a continuous bijection.
Since $S^1$ is compact, it is a homeomorphism.
