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I'm trying to compute a certain DeRham cohomology. Consider $M = S^n-C$, where $C$ is the disjoint union of closed disks $C = \cup_{i=1}^m D_i$, and $m,n \geq 1$. How can we compute the cohomology $H^{*}(M)$?

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    Ar the disks disjoint?2012-12-01
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    (Also I doubt that the dimension of $S^n$ and the number of disks you are removing being equal is of any significance :-) )2012-12-01
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    Yes, the disks should be disjoint.2012-12-01
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    Edit the question so that it contains **all** details.2012-12-01

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You can proceed by induction, using the following observation.

If $M$ is a manifold and $D\subseteq M$ is a (standardly embedded) closed disk in $M$, then there is an open set $U\subseteq M$ which is a standardly embeded open disk containing $U$ such that $U\setminus D$ is a «thick sphere», and $\{U,M-D\}$ is an open covering of $M$ from which one can get a Mayer-Vietoris long exact sequence.

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    Is the manifold $M$ here the same one as in the question above? Or are you referring to any manifold $M$?2012-12-01
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    Any manifold; this is hinted by the «if M is a manifold...»2012-12-01
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    I don't know what thick sphere means2012-12-03
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    The complement of the ball of radius $1$ inside the ball of radius $2$ is what I call a think sphere.2012-12-03
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    I apologize for disturbing. I'm trying to do the case $n=m=2$, i.e. I want the De Rham cohomology of $X=S^2 \setminus (D_1 \cup D_2)$, where $D_i$ are closed disjoint disks. Using Mayer-Vietoris on the open covering of $S^2$ given by $X$ and two thick disks $\tilde{D}_i$, I conclude only that $\dim H^1(X)+\dim H^2(X)=1$. Now I don't know how to compute exactly $H^i(X)$ (I think one is $\mathbb R$ and the other is trivial, but I don't manage to prove it). Would you give me a hint, please? Thanks in advance.2013-02-11