# Can $\mathbb{S}^2$ wind around $\mathbb{S}^2$ twice, just like you would wind $\mathbb{S}^1$ around $\mathbb{S}^1$ $n$ times

NOTE: This question is a duplicate of this and this. These questions are answered.

I know this topic is about algebraic topology, but my knowledge there is rather weak.
I essentially want to know if there exists a continuous function $$\varphi:\mathbb{S}^2\rightarrow\mathbb{S}^2$$ such that $$(\forall x\in\mathbb{S}^2)|\varphi^{-1}(x)|=2$$?

Background thinking:
If we replaced $$\mathbb{S}^2$$ with $$\mathbb{S}^1$$ and understood $$\mathbb{S}^1$$ as $$R=\{z\in\mathbb{C}|1=|z|\}$$ then $$x\mapsto x^2$$ would satisfy my condition. I wanted to do it with $$\mathbb{S}^2$$ too but it seemed that some point would be covered infinitely many times (here I had layering into longitutes in mind). Then I wanted to avoid that infinity and I reduced the problem to inverting $$R$$ to itself but with reversed orientation with bounded covering, or more precisely a function $$f:\mathbb{S}^1\times[0,1]\rightarrow\mathbb{R}^2$$ such that $$f(\mathbb{S}^1\times\{0\})=R$$, $$f(\mathbb{S}^1\times\{1\})=R$$ but reversed and $$|f^{-1}(x)|\le2$$ for every $$x$$. I found it and induced a related $$\varphi$$ but then I got two points which I covered only once. Then I remembered that I could simply twist the sphere just like $$\mathbb{S}^1$$ at all latitudes and leave the poles in place and I would have the exact same situation: Every point is covered twice except the poles which are covered only once. This made me thinking that I will always have such 2 points that I covered only once, either that or two infinitely covered points as with longitudes. How would I even go about proving it?

EDIT: Comments gave some help towards formalizing my examples I gave in the discussion, so here is the clear definition of those 2 maps, after we identified $$\mathbb{S}^2$$ with $$\mathbb{C}P^1$$:

The longitudal one: $$z\mapsto\frac{z}{|z|(1-|z|)}$$; it covers the $$\infty$$ infinitely many times.

The latitudal one: $$z\mapsto z^2$$; it covers $$0$$ and $$\infty$$ only once.

EDIT 2: I have progressed a tiny little bit: If $$X$$ and $$Y$$ are metric spaces, $$X$$ being compact and $$f:X\rightarrow Y$$ is a continuous bijection then $$f$$ is a homeomorphism.
Now I thought I could find some meaningful subset of $$\mathbb{S}^2$$ for which $$\varphi$$ restricted to it has this property, but I failed...

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Dec 9, 2022 at 18:59
• I repost the duplicate here for ease of everyone's reference Commented Dec 9, 2022 at 19:13

As explained in the MO post linked in the question, there is no such map. The key step is to construct out of $$\varphi$$ a continuous involution of $$S^2$$, which must be conjugate in the homeomorphism group to either a reflection or a rotation by $$\pi$$.