Let $e^r$ be a homeomorphic copy of $I^r$ in $\mathbb{R}^n$($I=[0,1]$).How to compute the homology group $H_q(\mathbb{R}^n-e^r)$?($r,n,q$ are non-negative integers)
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2$\Bbb R^n\setminus e^r$, at least when $e^r$ is the standard embedding of $I^r$, will be homotopy equivalent to $S^{n-1}$. – Olivier Bégassat May 09 '13 at 03:26
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Do you know Alexander Duality? – Cheerful Parsnip May 09 '13 at 03:57
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But, in the Alexander Duality, that's $S^n$. – Nirvanacs May 09 '13 at 06:37
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OK!I know how to solve it. Consider $R^n$ as the complementof a point in $S^n$. – Nirvanacs May 10 '13 at 10:20
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I applied the Mayer-Vietoris sequence in $\mathbb{S}^n -Y$ and the following result:
$``$Let $Y$ a subset of the $\mathbb{S}^n$ homeomorphic of $I^r$. Then the reduced homology groups is $$ \tilde{H}_j(\mathbb{S}^n -Y)=0 \quad \forall j." $$ The subsets were used in sequence: $$ U = \mathbb{S}^n -(Y \cup \{pto\}), \quad V \cong \mathbb{D}^n \quad \text{and} \quad U \cap V \cong \mathbb{D}^n - \{pto\}. $$ Comments: The $\mathbb{R}^n$ is homeomorphic of $\mathbb{S}^n-\{pto\}$, $\{pto\} \subset int (V)$ and $V \cap Y = \varnothing$.
Vasco
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