MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

closed as off-topic by Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, Daniel Moskovich Dec 21 '13 at 14:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, Daniel Moskovich
If this question can be reworded to fit the rules in the help center, please edit the question.

No on $\mathbb{R}^n$ for all $n$ and $S^n$ for $n \geq 2$: these spaces are simply connected. Yes for $S^1$: $z \mapsto z^2$. – Pete L. Clark Mar 16 '10 at 14:21
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 ? – Gerald Edgar Mar 16 '10 at 15:00
"covering map" implies local diffeomorphism which means that this is a strictly weaker question than that one as it assume strictly stronger conditions. – Loop Space Mar 16 '10 at 15:05
Is this homework? – Andrea Ferretti Mar 16 '10 at 16:05
There's always the disconnected cover by two copies of the base. – Scott Morrison Mar 16 '10 at 16:59

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.

share|cite|improve this answer
The question seemed too close to homework for me to record an actual answer. – Pete L. Clark Mar 16 '10 at 19:12
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. – Anton Geraschenko Mar 17 '10 at 0:27
No non-trivial connected coverings, anyway :) – David Roberts Mar 17 '10 at 2:54
@David: the trivial connected covering too :) – Anton Geraschenko Mar 17 '10 at 3:24
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 – Scott Morrison Mar 17 '10 at 5:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.