Covering maps on Euclidean spaces and spheres - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T10:53:22Zhttp://mathoverflow.net/feeds/question/18379http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/18379/covering-maps-on-euclidean-spaces-and-spheresCovering maps on Euclidean spaces and spheresJulgyz Harzum2010-03-16T14:13:40Z2010-03-16T17:17:20Z
<p>Hello. I have two questions.</p>
<ol>
<li><p>Does there exist an exactly 2-fold covering map
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ ?</p></li>
<li><p>Does there exist an exactly 2-fold covering map
$g:S^{n}\rightarrow S^{n}$ ?</p></li>
</ol>
<p>Here $S^{n}$ is the unit $n$-sphere,
$S^{n}={x\in\mathbb{R}^{n+1}: \|x\|=1}$.</p>
<p>Great thanks.</p>
http://mathoverflow.net/questions/18379/covering-maps-on-euclidean-spaces-and-spheres/18401#18401Answer by Anton Geraschenko for Covering maps on Euclidean spaces and spheresAnton Geraschenko2010-03-16T17:17:20Z2010-03-16T17:17:20Z<p>I think Pete should have made his comment an answer, so I'll do it for him.</p>
<p>Theorem 1.38 of Hatcher's <a href="http://www.math.cornell.edu/~hatcher/AT/ATpage.html" rel="nofollow">Topology</a> 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 $\pi_1(X)$.</p>
<p>Since $\pi_1(X)$ is trivial for $X=\mathbb{R}$ or $X=S^n$ ($n>1$), there are no connected coverings.</p>