Does there exist a continuous surjection from $\Bbb R^3-S^2$ to $\Bbb R^2-\{(0,0)\}$? 
Prove or disprove : There exists a continuous surjection from $\mathbb{R}^3- S^2$ to $\mathbb{R}^2-\{(0,0)\}$ (here $S^2\subset
\mathbb{R}^3$ denotes the unit sphere defined by the equation
  $x^2+y^2+z^2=1$),.


This question had appeared in TIFR GS-2018 exam for PhD admissions. How should I think about such a map?
 A: There is such a map :


*

*project $\mathbb{R}^3 \setminus \mathbb{S}_2$ onto $\mathbb{R}^2$ (projection on the first two coordinates, for instance);

*apply the exponential map from $\mathbb{R}^2 \simeq \mathbb{C}$ onto $\mathbb{R}^2  \setminus \{(0,0)\} \simeq \mathbb{C}^*$.
Both are continuous surjections, so their composition still is.

Given the bounty, I think this deserves at least some additional material (or: how may someone find this answer). Let $X$ and $Y$ be two topological spaces. If there is no continuous surjective map $f$ from $X$ to $Y$, then $X$ must have some property which is preserved by continuous maps, and which $Y$ doesn't have. It turns out that there are not many such properties. On top of my head, the only general ones I can see are :


*

*Cardinality : If Card(X) < Card(Y), then there is no such $f$. Example : $X = \{0\}$, $Y = \{0,1\}$.

*Compactness : If $X$ is compact and $Y$ is separable but not compact, then there is no such $f$. Example : $X = [0,1]$, $Y = \mathbb{R}$.

*Connectedness : If $X$ is connected and $Y$ isn't, then there is no such $f$. More generally, if $X$ has less (in the sense of cardinality) connected components than $Y$, then there is no such $f$. Example : $X = (0,1)$ and $Y = \mathbb{Z}$.
Although, I am sure, one may find examples with other, less obvious, obstructions. Outside of these, things may get wild. For instance, in all the following cases, there exists a continuous surjective map from $X$ to $Y$:


*

*$X = C$, any Cantor set, and $Y$ is any compact metric space.

*$X=[0,1]$ and $Y = [0,1]^2$ : Peano curve.

*More generally, $X = \mathbb{R}$ and $Y = \mathbb{R}^n$, with $n \geq 1$, using variants of the Peano curve.

*$X = \mathbb{R}^k$ and $Y = \mathbb{R}^n$, with $k \geq n$, using projections. Combining with he previous example, we get all $k \geq 1$ and $n \geq 0$.

*$X = \mathbb{R}^k$ and $Y$ is a (separable) connected, $n$-dimensional topological manifold, with $k \geq 1$. This is not easy to formalize and I have no reference at hand, but the basic idea is to take $k=n$, and make $\mathbb{R}^n$ a thin tape and wrap it around the manifold (see e.g. this related discussion). I also think that one may replace $X$ by any non-compact $k$-dimensional manifold.
With that, we have the tools to answer the question. First, since $X$ ha dimension $3$ and $Y$ has dimension $2$, we reduce the dimension by projecting. We get the plane $\mathbb{R}^2$. Then we wrap the plane around the origin in $\mathbb{R}^2$ ; and we are lucky, since this can be done very explicitly (thanks to the exponential map).
