Prove a function is open in a specific topological space Let $\mathbf{R}^{n}$ be given the usual (metric) topology, and let $S^{1}=\left\{(x, y) \subset \mathbf{R}^{2} \mid x^{2}+y^{2}=1\right\}$ be given the subspace topology as a subset of $\mathbf{R}^{2} .$ Consider the function $f: \mathbf{R} \rightarrow S^{1}$ given by $f(\vartheta)=(\cos 2 \pi \vartheta, \sin 2 \pi \vartheta)$
You may assume $f$ is continuous. Let $H=H_{x>0}$ denote the subset of $S^{1}$ consisting of points where $x>0$. Prove that $f$ is open.
I know the following proposition:
Suppose $X$ is a topological space equipped with local bases $\mathscr{P}_{x}$ at each point $t$
x. The following are equivalent:
1.$f$ is open


*For all $x \in X$ and all $B \in \mathscr{P}_{x}$, the set $f(B)$ contains an open neighbourhood of $f(x)$.

Since $\{(x-1/n,x+1/n)\}$ is a local base of $x$, I am trying to prove the second statement. Is this a right strategy? Or, there is other smart way?
Thank you very much!
Update:
After I figured out this question, I still have a next question:
Deduce the map $h:[0,1]/\{0,1\}\rightarrow S^1$: $h(\vartheta)=(\cos 2 \pi \vartheta, \sin 2 \pi \vartheta)$ is homeomorphism.
By universal property of the quotient space, I know that h is bijective and continuous. I do not know what to do next. To prove the inverse image is continuous? or, prove h is open?
 A: First note that we can write $S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. Using this representation, we have $f(x) = \cos (2\pi x) + i\sin (2 \pi x) = e^{2\pi i x}$.
It is easy to verify that for each $w \in S^1$ the complex multiplication function $\mu_w : S^1 \to S^1, \mu_w(z) = w \cdot z$, is continuous. We have $\mu_{1/w} \circ \mu_w = id$ and $\mu_{w} \circ \mu_{1/w} = id$, thus each $\mu_w$ is a homeomorphism.
Similarly the translation $\tau_r : \mathbb R \to \mathbb R, \tau_r(x) = r + x$, is is a homeomorphism for each $r \in \mathbb R$.
We have
$$f \circ \tau_r = \mu_{f(r)} \circ f .$$
Just observe $(f \circ \tau_r(x) = f(r + x) = e^{2 \pi i (r + x)} = e^{2 \pi i r} \cdot e^{2\pi i x} = f(r) \cdot f(x) = (\mu_{f(r)} \circ f)(x)$.
The maps $r : H \to (-1,1), r(x,y) = x$, and $i : (-1,1) \to H, i(x)= (x,\sqrt{1-x^2})$, are continuous. We have $r \circ i = id$ and $i \circ r = id$, thus they are homeomorphisms which are inverse to each other.

*

*If $(a, b)$ is an open interval contained in $(0,1/2)$, then $f((a,b))$ is open in $S^1$: We have $f((a,b)) \subset H$. Since $H$ is open in $S^1$, it suffices to show that $f((a,b))$ is open in $H$. Therefore it suffices to show $r(f((a,b))) = (r \circ f)((a,b))$ is open in $(-1,1)$. But $(r \circ f)(x) = \cos (2 \pi x)$ which is known to establish a homeomorphism $(0,1/2) \to (-1,1)$. This proves 1.


*If $(a, b)$ is any open interval of length $\le 1/2$, then $f((a,b))$ is open in $S^1$: We have $(a,b) = \tau_a ((0,b-a))$, thus $f((a,b)) = (f \circ \tau_a)((0,b-a)) = (\mu_{f(a)} \circ f)((0,b-a)) = \mu_{f(a)}(f((0,b-a)))$. By 1. $f((0,b-a))$ is open in $S^1$. Now use the fact that $\mu_{f(a)}$ is a homeomorphism.


*For each $x \in \mathbb R$, the open intervals $(x-1/n,x+1/n)$ with $n \ge 4$ form a local base at $x$ and have length $\le 1/2$.  Now 2. applies.
