Proving an arc is open on a unit circle. Given a map like this:
$f: \underline{[0,1]} \rightarrow \underline S^1$, with $f(t) = (\cos(2\pi t), \sin(2\pi t))$ and $S^1 = \{ (x,y) \in \mathbb{R}^2 \ | \ x^2 + y^2 = 1 \}$.
How can one prove that with $0 \le t_1 \lt t_2 \le 1$ an arc like this: $A = f[(t_1, t_2)]=\{(\cos(2\pi t), \sin(2\pi t))\} \ | \ t \in (t_1, t_2) \}$ is open in $S^1$?
I have a difficulty finding an open disk in $\mathbb R^2$ that intersected with $S^1$ would yield $A$.
 A: Take $B=\{(x,y) : tg(t_1)x< y< tg(t_2)x\}$
This is the sector area identified by the two angles $t_1,t_2$ and is an open subset of $\mathbb{R}^2$.
$B$ is open because if you take the continuos function $f\colon \mathbb{R}^2\to \mathbb{R}^2$ sending $(x,y)\mapsto (y-tg(t_1)x,tg(t_2)x-y)$ then
$ B=f^{-1}((0,\infty)\times (0,\infty))$
Here $f$ is continuos because each factor is a linear map.
Interesect $\mathbb{S}^1$ with $B$ and you get $A$
Pay attention that if one of $t_1,t_2$ is $\frac{1}{4}$ or $\frac{3}{4}$, then the $tg(t_i)$ does not makes sense and is better to use the cotangent
$B=\{(x,y) : cot(t_2)y<x<cot(t_1)y\}$
Moreover our sets are empty if
$t_1=0$, $t_2=\frac{1}{2}$. Here take $ B=\mathbb{R}\times (0,\infty)$ if you want the upper-plane section.
$t_1=\frac{1}{4}, t_2=\frac{3}{4}$. Here take
$B=(0,\infty)\times \mathbb{R}$ if you want the right-plane section.
You can also use the fact (as said Paul Frost) that your map is a local homeomorphism and you can prove that is an homeomorphism onto a open set for each open interval $(t_1,t_2)$. Thus $f((t_1,t_2))$ is open in $\mathbb{S}^1$ without seeing $\mathbb{S}^1$ in $\mathbb{R}^2$
A: You do not need to find an open disk in the plane.
Let $J  = (t_1, t_2)$ be the open interval in your question. Its complement $C  = [0,1] \setminus J$ is compact, thus $f(C)$ is compact, hence closed in $S^1$. Therefore $S^1 \setminus f(C)$ is open in $S^1$.
Cleary $f(J) \cup f(C) =  f(J \cup C) = f([0,1]) = S^1$. Moreover $f(J) \cap f(C) = \emptyset$: Assume to the contrary that there exists $z \in f(J) \cap f(C)$. Write $z = f(t) = f(t')$ with $t \in J$ and $t' \in C$. Then $t \ne t'$ since $J \cap C = \emptyset$. The only two distint points of $[0,1]$ which have the same image under $f$ are $0$ and $1$. None of them is contained in $J$, and this gives the desired contradiction.
We conclude that $f(J) = S^1 \setminus f(C)$ which is open in $S^1$.
Update:
Here is an alternative proof (which repeats some of the arguments above).

*

*$f$ is a continuous closed surjection: Each closed subset $C \subset [0,1]$ is compact, thus $f(C)$ is compact, hence closed in $S^1$.
Therefore $f$ is a quotient map.


*$f^{-1}(f(M)) = M$ for any subset $M \subset (0,1)$: $M \subset f^{-1}(f(M))$ is true for any function. Let $t \in f^{-1}(f(M))$. Then $f(t) \in f(M)$, thus $f(t) = f(t')$ with $t' \in M$. The only two distinct points of $[0,1]$ which have the same image under $f$ are $0$ and $1$ which do not belong to $M$, thus we must have $t = t' \in M$.


*By 1. $f(J)$ is open in $S^1$ because $f^{-1}(f(J)) = J$ which is open in $[0,1]$.
