Is $f:[0,1] \to S^1$ a quotient map? Let $f:[0,1] \to S^1$ be a continuous onto function that only identifies $0$ with $1$. My question is: must $f$ be a quotient map? I was reading the proof in a book with $f(t)=e^{2 \pi i t}$, and they use the fact that this $f$ is closed, but is $f$ closed in the general case?
$\bf{EDIT:}$ The following theorem and example are taken from Willard's General Topology:

Where $\tau_f$ is the quotient topology induced by $f$. So, my question is, if I replace the function $f$ given in the example by any continuous onto function $f:[0,2 \pi] \to S^1$ that only identifies $0$ with $2 \pi$, do we still have $\tau = \tau_f$? Why? Is the same argument that $f$ is closed valid, so that we can use the theorem?
 A: Yes, f is a quotient map provided that we have the quotient topology on the quotient space. However, it's not generally true that a quotient map is closed. You can easily create some examples of the projection map (where you can define an equivalence relation to make the projection map a quotient map) such that the image of a closed set is not closed.
A: A quotient map between two topological spaces $X,Y$, or as I call them identifications, is a continuous map $f: X \to Y$ that factors into a map $g: X / \sim  \to Y$ in such a way that $g$ is an homeomorphism, i.e. $f = g \circ \pi$ for the natural projection $\pi: X\to X/\sim$. Here $x \sim y$ iff $f(x) = f(y)$.
As mentioned in the comments you only need to prove that $f$ is surjective and open or closed and that $f(x) = f(y)$ iff $x=y$ or if $x=1, y=0$ ($x=0, y=1$). Since $[0,1]$ is compact, every closed subset $F$ is compact and by continuity its image $f(F)$  is compact and therefore closed in $S^1$ since this the later is Hausdorf.
Note that the only requirement on the topology of $S^1$ is that the topology is Hausdorff.
