# Is it an open function?

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.

• This is correct, Gio. – Mathy Oct 23 at 10:46

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.

• Nice answer! thank you :) – user723846 Oct 24 at 9:10