A quotient map from $[0,1]$ to $S^1$ I would like to show that the function $f(x) = (\textrm{cos}2 \pi x, \textrm{sin}2 \pi x)$ is a quotient map; I have already shown that it is surjective and continuous (the latter by invoking the universal property for functions into a product topology (I am considering $S^1$ as a subspace of $\mathbb{R} \times \mathbb{R}$)). However, I'm stuck trying to prove the final condition for $f$ to be a quotient map. Here's my work so far: 
Let $U \subset S^1$. Suppose $f^{-1}[U]$ is open in $[0,1]$. Then we need to show that $U$ is open in $S^1$. Let $\bar{y} \in U$; either $\bar{y} = (1,0)$ or $\bar{y} \neq (1,0)$. If $\bar{y} = (1,0)$, then by using the openness of $f^{-1}[U]$, we can deduce that there exists $\epsilon, \epsilon' \in \mathbb{R}_{> 0}$ such that $[0,\epsilon) \cup (\epsilon',1] \subset f^{-1}[U]$ (since $f^{-1}[\{ (1,0) \}] = \{0, 1 \}$). Similarly, if $\bar{y} \neq (1,0)$, there exist $\delta, \delta'\in \mathbb{R}_{> 0}$ such that $(\delta,\delta') \subset f^{-1}[U]$. Of course these sets, $[0,\epsilon) \cup (\epsilon',1]$ and $(\delta,\delta')$ are open in $[0,1]$. 
How do I (or can I) proceed from here to show that $U$ is open in $S^1$? On the other hand, I'd also love to hear of some more general methods for showing that the function is a quotient map (maybe certain theorems or something). Thanks in advance for the help!
 A: Maybe here is a general method:
Note that $[0,1]$ is compact and $S^1$ is Hausdorff. So, actually, your function is a closed map. Therefore, your function is a quotient map.
Edit:
Let $f:X\to Y$ be a continuous function where $X$ is compact and $Y$ is Hausdorff. Then, $f$ is a closed map.
Proof:
Pick a closed set $U$ in $X$. Since $X$ is compact, $U$ is compact. Hence, $f(U)$ is compact. Since $Y$ is Hausdorff, $f(U)$ is closed.
A: You can use the fact that a map $q:X\rightarrow Y $ is a quotient map, if it is continuous, and if it sends saturated open sets to saturated open sets; a subset is saturated if (def.) it containes every fiber {$f^{-1}(y)$} that it intersects.
A: Fix a set $U\subset S^1$ such that $f^{-1}(U)$ is open, and a point $z \in U$. Suppose $z$ has positive second coordinate for the moment. Then there is a unique point in $f^{-1}(z)$ which we'll call $x$, and by the openness of $f^{-1}(U)$ an interval $I=(x-\epsilon, x+\epsilon)\subseteq f^{-1}(U)\cap (0,1/2)$. Consider $f$ restricted to $[x-{\epsilon}/2, x+{\epsilon}/2]$. The function $\cos(2\pi x)$ has a minimum and maximum on this interval, as it is closed and bounded. In fact, by the properties of $I$ above, there are no critical points in the interval so the min and max must occur on the boundary. We will call the min and max $x_0$ and $x_1$. Write $z=(z_0, z_1)$. We claim that, $V=S^1 \cap [(x_0, x_1)\times (0, z_1+1)]$ is an open set containing $z$ and contained in $U$. Clearly it is open by the definition of the subspace topology.  By the discussion above, $z_0$ is not the min or max so $x_0<z_0<x_1$, and this gives $z\in V$. Now take $(a, \sqrt{1-a^2})\in V\subseteq S^1$. Since $x_0<a<x_1$, the intermediate value theorem gives that $\cos(2\pi c)=a$ for some $c\in I$. But then $\sin(2\pi c)=\pm \sqrt{1-a^2}$ and it must be the plus sign since $c\in I$. This gives $(a, \sqrt{1-a^2})=f(c)$, and since $c\in I\subseteq f^{-1}(U)$, we have shown that $V\subseteq U$. What we have shown is that $z$ has an open neighborhood around it which is contained in $U$. We want to extend this to all possible $z$, not just ones with positive second coordinate. $Mutatis$ $mutandis$, the same proof goes through for $z$ with second coordinate negative. (E.g., use $(z_1-1, 0)$ in the definition of $V$ instead.) I will leave it to you to modify the details for the remaining two points. You will want to use sine instead of cosine if you do the above min/max argument in these cases. This will show $U$ is open and complete the proof.
