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Note, that a proof of this is already done on Parametrization of $n$-spheres. However, I wanted to approach the proof differently and was wondering if my line of logic would work.

Let $S^{n}$ be the $n$ sphere. Suppose it can be covered via one parameterization. Then, $\exists$ a mapping $\phi:U\subseteq\mathbb{R}^k \to S^n$ (for some open $U$ ).

However, note that $S^n$ is compact and spherically symmetric so that it attains two (distinct) points on $S^n$ which are the farthest from the origin such that only the last $n-k$ coordinates of each values are different. (here I tried to think of the sphere in three dimensions and the points at the north and south pole, they are essentially identical except for the last coordinate). However, I cannot seem to be able to prove the existence of such two points and am just hand-waving its existence. Can someone try to help me at this step? Of course, if such two points do not necessarily have to exist, then I would like to know that as well.

If I can prove this, then it would follow that $\phi$ is not injective so it cannot be a parameterization (I am taking the definition that parameterizations need to be diffeomorphic).

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    $\begingroup$ Suppose you find two such points. Puncture the sphere by removing a point that is not either of the two points. IIUC, the punctured sphere can be parameterized, even though it still contains two points that (you say) prove it cannot be parameterized. It seems to me the thing you are trying to prove is not something you can show just by looking at a finite number of points. $\endgroup$ – David K Apr 24 '17 at 4:35
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You are complicating what is, in essence, a simple problem. Remember that, in particular, diffeomorphisms are homeomorphisms. If $\emptyset \ne U = \mathring U \subseteq \Bbb R^k$ is diffeomorphic to $S^n$, then it is homeomorphic to it. Since $S^n$ is compact, $U$ too will have to be so. In particular, it must be closed. Since $U$ is also open and non-empty, and $\Bbb R^k$ is connected, $U$ must be the whole space $\Bbb R ^k$. We deduce that $\Bbb R^k$ is compact - which is not possible, since it is unbounded, therefore $S^n$ cannot be diffeomorphic to any open subset of $\Bbb R^k$.

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  • $\begingroup$ I suppose I just wanted to answer this question another way but ended up complicating the proof unnecessarily. Thank you for your answer! $\endgroup$ – Anmol Bhullar Apr 25 '17 at 18:51
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    $\begingroup$ @AnmolBhullar: Don't worry, the previous version of my answer was, too, unnecessarily complicated! $\endgroup$ – Alex M. Apr 25 '17 at 19:26

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