Continuous function from $\mathbb{R}^2$ to $\mathbb{S}^2$

Question

Good afternoon, I am currently going through an introductory topology class. An exercise asked us to prove there is no homeomorphism between $\mathbb{R}^2$ and $\mathbb{S}^2$ (and the reason is no continuous bijection from $\mathbb{S}^2$ to $\mathbb{R}^2$ exists, as $\mathbb{S}^2$ is compact and $\mathbb{R}^2$ is not in the topology induced by $\mathbb{R}^3$). My question now is, relaxing the requirement of having a continuous inverse, are there continuous bijections from $\mathbb{R}^2$ to $\mathbb{S}^2$?

Answer 1

There is no continuous bijection $f: \mathbb{R}^2 \to S^2$, but this is the only proof I can produce (it uses invariance of domain which is certainly not an elementary theorem):

Suppose there is. I claim that $f$ is an open map.

Write $S^2 = S^2 \smallsetminus \{ N\} \cup S^2 \smallsetminus \{S\}$ for the north and south poles $N$ and $S$, respectively. Then each of these subsets is homeomorphic to $\mathbb{R}^2$ by stereographic projection.

Let $A_1 = f^{-1}(S^2 \smallsetminus \{ N\})$ and $A_2 = f^{-1}(S^2 \smallsetminus \{S\})$. Now, regard each $f_i = f|_{A_i}$ as a map $A_i \to \mathbb{R}^2$ (by composing with stereographic projection. We can be precise and include stereographic projection where we should but for notation simplicity let's not).

Since $f$ is a continuous bijection, $f_i$ is a continuous injection, so they are open maps by invariance of domain. But then $f$ is an open map, since any open set $U$ in $\mathbb{R}^2$ is $(U \cap A_1) \cup (U \cap A_2)$, hence $f(U) = f_1(U \cap A_1) \cup f_2(U \cap A_2)$ (thinking of $f_1$ and $f_2$ here as maps to $S^2$) and the latter two sets are open since $f_i$ is open. Hence $f(U)$ is open.

This gives a contradiction, since any open continuous bijection is a homeomorphism, and there is not a homeomorphism $\mathbb{R}^2 \to S^2$.