Covering maps on Euclidean spaces and spheres [closed]

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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closed as off-topic by Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, Daniel MoskovichDec 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

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.