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Let $f:(0,1)\rightarrow C-\{(1,0)\}$ defined by $f(t)=(cos(2\pi t),sin(2\pi t))$ and where $C=\{(x,y)\in\mathbb{R}^2|x^2+y^2=1\}$.

I would like to prove that $f$ is an open function, i.e., for any $S\subset(0,1)$ open then $f(S)$ is open in $C-\{(1,0)\}$, for the topology subspace.

As $S$ is open then $S=\cup_{z\in S}(z-\delta_z,z+\delta_z)$ for some $\delta_z>0$ such that $z\in(z-\delta_z,z+\delta_z)\subset S$. As $f(S)=\cup_{z\in S} f((z-\delta_z,z+\delta_z))$, I only need to prove that $f((z-\delta_z,z+\delta_z))$ is open, i.e, $f((z-\delta_z,z+\delta_z))=O\cap(C-\{(1,0)\})$ for some $O$ open set of $\mathbb{R}^2$.

This sketch shows that if I consider $O$ as open ball of center $f(z)$ and radius $|f(z)-f(z-\delta)|$ then $f((z-\delta_z,z+\delta_z))$ is an open set, and so $f$ is an open function.

Is it correct? Thank you in advance.

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    $\begingroup$ This is correct, Gio. $\endgroup$ – Mathy Oct 23 at 10:46
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This is essentially covered by one of my answers to Open sets on the unit circle $S^1$ .

Let us give a short proof following these lines.

As you know, it suffices to show that for any $(a,b) \subset (0,1)$ the set $f((a,b))$ is open in $C$ (because $f((a,b)) \subset C \setminus \{0\}$).

Consider the map $$ F : [0,1] \to C, F(t) = (\cos(2\pi t),\sin(2\pi t)) .$$ This is a continous surjection. We have $f((a,b)) = F((a,b))$.

The set $K = [0,1] \setminus (a,b)$ is compact, hence $F(K) \subset C$ is compact, thus closed in $C$. Therefore $C \setminus F(K)$ is open in $C$. We have $C = F([0,1]) = F(K \cup (a,b)) = F(K) \cup F((a,b))$. But $K$ and $(a,b)$ are disjoint, thus $s \in K$ and $t \in (a,b)$ cannot have the same image under $F$ (note that the only two distinct points in $[0,1]$ having the same image under $F$ are $0$ and $1$). We conclude that $F(K)$ and $F((a,b))$ are disjoint, hence $F((a,b)) = C \setminus F(K)$.

Let us now come to your sketch. You are right, your open disk $O$ has the desired property. This is intuitively clear, but if you try to give an exact proof, you will see that it is technically rather intricate.

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  • $\begingroup$ Nice answer! thank you :) $\endgroup$ – user723846 Oct 24 at 9:10

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