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?

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    $\begingroup$ Hint: Think about (path-)connectedness $\endgroup$
    – B. Mehta
    Dec 11, 2017 at 20:31
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    $\begingroup$ $\mathbb{R}^3-S^2$ is Euclidean space with the unit sphere removed. So there are no paths between the interior of the sphere and the exterior. $\endgroup$
    – user326210
    Dec 11, 2017 at 20:46
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    $\begingroup$ @B.Mehta I don't think it's going to be as simple as that $\endgroup$ Dec 13, 2017 at 15:47
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    $\begingroup$ @user326210 But how would that prove your point? There are examples of continuous maps which map not-path-connected spaces to path connected spaces. $\endgroup$
    – Error 404
    Dec 13, 2017 at 16:28
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    $\begingroup$ @B.Mehta Can you elaborate your approach? Thanks in advance. $\endgroup$
    – Error 404
    Dec 14, 2017 at 7:25

1 Answer 1


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).

  • $\begingroup$ Same as my comment here: math.stackexchange.com/questions/2565030/… $\endgroup$
    – zhw.
    Dec 14, 2017 at 23:40
  • $\begingroup$ @zhw: yes, it's pretty close. I didn't notice that this question had a duplicate. $\endgroup$
    – D. Thomine
    Dec 15, 2017 at 13:49
  • $\begingroup$ Right, not exactly the same, as I was addressing someone else's proof. Looks like you're headed to bounty nirvana in one of the stranger bounty situations I've seen. $\endgroup$
    – zhw.
    Dec 15, 2017 at 22:54

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