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M is a compact connected n-manifold,if for any point $p\in M,M\backslash\{p\}\cong R^n$ then M is homeomorphic to $S^n$.

I have the guess from, but I don't know how to proof it.Beside if there are any grammatical mistake,I apologize for my poor english.

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Do you know Stereographic Projection? – Berci Jan 20 '13 at 2:35
@Berci :i know it ,but i want to proof it is $S^n$,not $S^n$ have the quality that for any point p∈M,M∖{p}≅Rn – lanse Jan 20 '13 at 3:06
up vote 2 down vote accepted

The open sets around $p$ are the complement of closed set of $M$ that does not contain $p$. Since closed subset of compact set is compact, it's the complement of compact sets of $M$ that does not contain $p$. This means that $M$ is a one-point compactification of $\mathbb{R}^n$, which is unique up to homeomorphism. Since $S^n$ is one such compact manifold, $M$ must be homeomorphic to $S^n$.

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thank you,I think I need to find the define of one-point compactification.Not long ago ,someome give me a hint that What is the universal property of the one point compactification. – lanse Jan 20 '13 at 7:53
See . Your friend's hint is about the uniqueness of one point compactification. See the third property of my link here. – user27126 Jan 20 '13 at 8:03
I have saw ,i think i can solve the question ,but now ,i must to make dinner ;) ,thank you – lanse Jan 20 '13 at 8:25

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