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Is the following true: $S^n-S^0\cong S^{n-1}\times\mathbb{R}$ How to prove it simply? Is there a formula for $S^n-S^r$, $0\leq r\leq n$?

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    What does $S^0$ stand for?2012-11-05

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Here $S^{0}\cong \{1,-1\} \subseteq \mathbb{R}.$

Stereographic projection give an homeomorphism of $S^n-\{P\}$ to $\mathbb{R}^n$. Since $\mathbb{R}^n-\{Q\}$ and $S^{n-1}\times\mathbb{R}$ are homeomorphic ( for any $Q \in \mathbb{R}^n$)we get that $S^n-S^0\cong S^{n-1}\times\mathbb{R}.$

One homeomorphism $f:\mathbb{R}^n-\{0\} \to S^{n-1}\times\mathbb{R}$ is $f(X)=(\frac{X}{\|X\|},\ln{(\|X\|)}).$