$f:[0,2\pi]\rightarrow\mathbb{S}$, $f(t)=(\cos t, \sin t)$. Find a base for the finest Topology that makes $f$ continuous I am solving the following question for a topology assignment. 
Let $f:[0,2\pi]\rightarrow\mathbb{S}$, $f(t)=(\cos t, \sin t)$ where $[0,2\pi]$ is endowed with the induced topology by the euclidean topology on $\mathbb{R}$. Find a base for the finest topology on $\mathbb{S}$, the unit circle, that makes $f$ continuous. 
It is clear that such a topology on $\mathbb{S}$ is given by $\tau=\{f^{-1}(V)\;\vert\; V\text{ open in } [0,2\pi]\}$. Since a base for the topology on $[0,2\pi]$ is given by sets of the type $(a,b)$ with $0\leq a<b\leq 2\pi$, $[0,a)$ and $(b,2\pi]$, I would say that a base for $\tau$ is given by sets of the type $f^{-1}((a,b)), f^{-1}([0,a))$ and $f^{-1}((b,2\pi])$. Is that correct? 
Also, I am very tempted to say that $\tau$ is the quotient topology because $\mathbb{S}=[0,2\pi]/\sim$ where $x\sim y$ iff $x=y$ or $x=0, y=2\pi$ or $x=2\pi, y=0$ and the quotient map is exactly $f$. Also, the product topology is the finest that makes the quotient map continuous. However, $\tau$ and the quotient topology seem not to coincide because $\tau$ contains arches that are for example open on one end and closed at the other, but the product topology only contains open arches together with their intersections and unions. So, where I am going wrong? 
Many thanks for the help!
 A: Hint: the map is equivalent to the map $f:\mathbb{R}/2\pi \mathbb{Z} \rightarrow \mathbb{S}^1$ given by $t \mapsto e^{it}$.
A: Using the corrected formula in the comments
$$\tau = \{V \,\bigm|\, \text{$f^{-1}(V)$ is open in $[0,2\pi]$}\}
$$
your guess is correct, $\tau$ is indeed the quotient topology on $\mathbb{S}$ induced by the function $f$.
Certainly $\tau$ is a topology with the desired property that $f$ be continuous, certainly $\tau$ contains all topologies with this property, and certainly there cannot be any finer topology with these properties, because if $V \not\in\tau$ then $f^{-1}(V)$ is not open in $[0,2\pi]$ and so $f$ is not continuous.
You express some doubts about $\tau$ being the quotient topology induced by $f$. In some sense this is a matter of definitions; for instance, this exact formula for $\tau$ is used as the definition of "the quotient topology induced by $f$" as that definition is stated in Munkres book "Topology". But you also express a particular doubt:


*

*"$\tau$ and the quotient topology seem not to coincide because $\tau$ contains arches that are for example open on one end and closed at the other, but the product topology only contains open arches together with their intersections and unions."


I am not entirely sure that I understand your concern here, but I think perhaps this will allay your worries. Note that if $V \in \tau$ contains a half-open arc of the form $0 \le \theta < a$ for some small positive angle $a$ (here I am using polar coordinates), then it follows that $f^{-1}(V) \subset [0,2\pi]$ contains the point $0$ as well as the point $2\pi$. Since $f^{-1}(V)$ is open in $[0,2\pi]$, it must contain an interval of the form $[0,-\epsilon)$ as well as an interval of the form $(2\pi-\epsilon,2\pi]$. It follows that $V$ also contains a half-open arc of the form $b < \theta \le 0$ for some small negative angle $b$. 
