Continuous surjection between $\mathbb R^3 \setminus \mathbb S^2$ and $\mathbb R^2 \setminus \{0\}$ In my class, teacher says that there is a continuous surjection between $\mathbb R^3 \setminus \mathbb S^2$ and $\mathbb R^2 \setminus \{0\}$. I tried to find an example at home. I tried taking center of sphere at origin and then joining the lines any point on $\mathbb R^3\setminus \mathbb S^2$ and south pole $(0,0,-1)$ and taking the point at which the line cuts $\mathbb R^2$ but I stuck at $z$-axis as all points on $z$-axis goes to origin. Please help me. Thanks in advance.
 A: Notice that $\mathbb{R}^3\backslash S^2$ is not connected: there is the "inside" $B$ of the sphere and the "outside" $X$. We handle them separately.
First, $X\approx Z:=\mathbb{R}^3\backslash0$ by essentially a deformation retraction $f$. Let $Y=\{v\in X:\|v\|\le2\}$. For $v\in Y$, let $f(v)=(\|v\|-1)v$. This amounts to "stretching" $Y$ so that its distance-2 boundary stays fixed and its distance-1 boundary is drawn to the origin. For $v\in X\backslash Y$, let $f(v)=v$. These agree on the mutual boundary of $X$ and $X\backslash Y$, so they join continuously and surject onto $Z$.
The circle group $S^1\cong\mathbb{R}/2\pi\mathbb{Z}$ acts on $Z$ by rotating about the $z$-axis. Quotient by this action; this is a continuous map onto a space homeomorphic to the upper-half-(complex )plane $\mathbb{H}$ union the nonzero reals (notably, excluding 0). Symbolically, this is a continuous map from $X$ to $\mathbb{H}\sqcup(\mathbb{R}\backslash0)$.
Notice that all we have left to account for is the lower-half-plane (that is, excluding all of $\mathbb{R}$)! For $(a,b,c)\in B$, project onto the open unit disk $\mathbb{D}\subset\mathbb{C}$ via $(a,b,c)\mapsto a+bi$. Now, consider the (negative of the) standard conformal map (a biholomorphism) $-g:\mathbb{D}\to\mathbb{H}$ sending $z\mapsto-\frac{z-i}{z+i}$. This makes for a surjection from $B$ to $-\mathbb{H}$.
Combining these maps gives image $\mathbb{H}\sqcup(\mathbb{R}\backslash0)\sqcup-\mathbb{H}=\mathbb{C}\backslash0$.
To double-check continuity, open sets in the image are either entirely in one half-plane, or contain part of the real line.

*

*In the former case, if the half-plane is upper then the preimage is just the original set revolved about the $z$-axis, and if the half-plane is lower then the preimage is the preimage under a biholomorphism (hence a bicontinuous map), then the preimage under a projection.

*In the latter case, split the open set into that strictly below the real axis, and that weakly above it. The part above revolves about the $z$-axis, containing part of the axis, ensuring openness. The part below is still just the preimage of an open set under a bicontinuous map, then the preimage under a projection.

EDIT: Here's a "cleaner" solution (doesn't involve any conformal maps, or use $B$ at all). We can actually show that the result is true for $\mathbb{R}^3\backslash(S^2\sqcup B)$ with not much more work. Repeat the above solution up until we arrive at the surjection from $X$ to $\mathbb{H}\sqcup(\mathbb{R}\backslash0)$. Take this and identify the positive and negative reals, that is, $x$ and $-x$ for $x\in\mathbb{R}^+$. Quotienting gives a space homeomorphic to $\mathbb{C}\backslash0$. Now map $B$ via your favorite continuous map (easiest being any constant map).
