Continuous function from $\mathbb{R}^2$ to $\mathbb{S}^2$ 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 $\mathbb{S}^2$?
 A: 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 $\mathbb{R}^2 \to S^2$.
A: Here is a proof requiring much less machinery than invariance of domain.  Knowledge of general topology and some basic knowledge of fundamental groups or homology is enough.  (No topology coursework and just a course in complex analysis may also suffice.)  For convenience I used $\mathbb C\cong \mathbb R^2$.
A continuous bijection $\phi:\mathbb C \longrightarrow S^2$ implies the existence of a continuous bijection
$\mathbb C \longrightarrow S^2\longrightarrow \mathbb C\cup\big\{\infty\big\}$, where (composing with a rotation as needed) the first map sends $0\mapsto N$ the north pole, and the second sends $N\mapsto \infty$ via stereographic projection.
Geometrically speaking, this implies a a continuous bijection from an annulus to a convex set, which creates a contradiction.
To finish, notice $\phi$ implies the existence of continuous bijection from the punctured plane, $\mathbb C^*= \mathbb C -\big\{0\big\}$, to the plane, i.e. some continuous bijective $f:\mathbb C^*\longrightarrow \mathbb C$. And $\mathbb C^*$ deformation retracts to $S^1$, the unit circle, which implies a deformation retract from $f\big(\mathbb C^*\big)\to f\big(S^1\big)$ where $f\big(S^1\big)\cong S^1$ since we are applying a continuous injective function to a compact set (in metric spaces).  Conclude $\mathbb C \simeq S^1$ which is a contradiction, since the former is simply connected and the latter is not.
