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Hello. I have two questions.

  1. Does there exist an exactly 2-fold covering map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ ?

  2. Does there exist an exactly 2-fold covering map $g:S^{n}\rightarrow S^{n}$ ?

Here $S^{n}$ is the unit $n$-sphere, $S^{n}=\{x\in\mathbb{R}^{n+1}: \|x\|=1\}$.

Great thanks.

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    $\begingroup$ No on $\mathbb{R}^n$ for all $n$ and $S^n$ for $n \geq 2$: these spaces are simply connected. Yes for $S^1$: $z \mapsto z^2$. $\endgroup$
    – Pete L. Clark
    Commented Mar 16, 2010 at 14:21
  • $\begingroup$ What about a related question (which I think was also asked in MO once...) Does there exist a continuous f : R^n -> R^n that is exactly two-to-one ? $\endgroup$
    – Gerald Edgar
    Commented Mar 16, 2010 at 15:00
  • $\begingroup$ "covering map" implies local diffeomorphism which means that this is a strictly weaker question than that one as it assume strictly stronger conditions. $\endgroup$
    – Andrew Stacey
    Commented Mar 16, 2010 at 15:05
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    $\begingroup$ Is this homework? $\endgroup$
    – Andrea Ferretti
    Commented Mar 16, 2010 at 16:05
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    $\begingroup$ There's always the disconnected cover by two copies of the base. $\endgroup$
    – Kim Morrison
    Commented Mar 16, 2010 at 16:59

1 Answer 1

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I think Pete should have made his comment an answer, so I'll do it for him.

Theorem 1.38 of Hatcher's Topology says that connected coverings of a (locally path-connected, and semilocally simply-connected) topological space $X$ are in bijection with conjugacy classes of subgroups of $\pi_1(X)$.

Since $\pi_1(X)$ is trivial for $X=\mathbb{R}$ or $X=S^n$ ($n>1$), there are no connected coverings.

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    $\begingroup$ The question seemed too close to homework for me to record an actual answer. $\endgroup$
    – Pete L. Clark
    Commented Mar 16, 2010 at 19:12
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    $\begingroup$ If it is homework, saying that there's a big theorem that does it probably doesn't finish the homework ... you'd have to run the proof or something. On the other hand, I can imagine somebody with no topology background wanting to know this for some other application, in which case a direct reference is exactly what they'd need to keep making progress. $\endgroup$
    – Anton Geraschenko
    Commented Mar 17, 2010 at 0:27
  • $\begingroup$ No non-trivial connected coverings, anyway :) $\endgroup$
    – David Roberts ♦
    Commented Mar 17, 2010 at 2:54
  • $\begingroup$ @David: the trivial connected covering too :) $\endgroup$
    – Anton Geraschenko
    Commented Mar 17, 2010 at 3:24
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    $\begingroup$ I suspect Anton is thinking big here. There's a valid argument that in the broad church of mathematics, there are some researchers, perhaps not yet well represented here, who may run into minor points of algebraic topology while working on something very far removed. They should be welcome, and in fact if we fail to welcome them then we doom mathoverflow to its present parochial specialisations. I'm not sure I agree, but it's a point worth discussing, on meta tea.mathoverflow.net/discussion/291. $\endgroup$
    – Kim Morrison
    Commented Mar 17, 2010 at 5:03

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